Quantum entanglement for systems of identical bosons: II. Spin squeezing and other entanglement tests

For two mode systems with mode annihilation operators $\widehat{a}$, $\widehat{b}$ associated with the two single particle states $| {\phi }_{a}\rangle$, $| {\phi }_{b}\rangle$, and where the non-zero bosonic commutation rules are $[\widehat{e},{\widehat{e}}^{\dagger }]=\widehat{1}$ ($\widehat{e}=\widehat{a}$ or $\widehat{b}$), Schwinger spin angular momentum operators ${\widehat{S}}_{\xi }$ ($\xi =x,y,z$) are defined as

$$\begin{eqnarray}{\widehat{S}}_{x} & = & ({\widehat{b}}^{\dagger }\widehat{a}+{\widehat{a}}^{\dagger }\widehat{b})/2,\qquad {\widehat{S}}_{y}=({\widehat{b}}^{\dagger }\widehat{a}-{\widehat{a}}^{\dagger }\widehat{b})/2i,\\ {\widehat{S}}_{z} & = & ({\widehat{b}}^{\dagger }\widehat{b}-{\widehat{a}}^{\dagger }\widehat{a})/2\end{eqnarray} \tag{ 1 }$$

and which satisfy the commutation rules $[{\widehat{S}}_{\xi }$ $,{\widehat{S}}_{\mu }$ $]={\rm{i}}{\epsilon }_{\xi \mu \lambda }{\widehat{S}}_{\lambda }$ for angular momentum operators. For bosons the square of the angular momentum operators is given by ${\widehat{S}}_{x}^{2}+{\widehat{S}}_{y}^{2}\,+{\widehat{S}}_{z}^{2}=(\widehat{N}/2)(\widehat{N}/2+1)$, where $\widehat{N}=({\widehat{b}}^{\dagger }\widehat{b}+{\widehat{a}}^{\dagger }\widehat{a})$ is the boson total number operator, those for the separate modes being ${\widehat{n}}_{e}={\widehat{e}}^{\dagger }\widehat{e}$ ($\widehat{e}=\widehat{a}$ or $\widehat{b}$). The Schwinger spin operators are the second quantization form of symmetrized one body operators ${\widehat{S}}_{x}={\sum }_{i}(| {\phi }_{b}(i)\rangle \langle {\phi }_{a}(i)| +| {\phi }_{a}(i)\rangle \langle {\phi }_{b}(i)| )/2;$ ${\widehat{S}}_{y}={\sum }_{i}(| {\phi }_{b}(i)\rangle \langle {\phi }_{a}(i)| -| {\phi }_{a}(i)\rangle \langle {\phi }_{b}(i)| )/2i;$ ${\widehat{S}}_{z}\,={\sum }_{i}(| {\phi }_{b}(i)\rangle \langle {\phi }_{b}(i)| -| {\phi }_{a}(i)\rangle \langle {\phi }_{a}(i)| )/2,$ where the sum i is over the identical bosonic particles. In the case of the two mode EM field the spin angular momentum operators are related to the Stokes parameters.

As well as spin operators for the simple case of two modes we can also define spin operators in multimode cases involving two sub-systems A and B. For example, there may be two types of bosonic particle involved, each component distinguished from the other by having different hyperfine internal states $| A\rangle ,| B\rangle$. Each component may be associated with a complete orthonormal set of spatial mode functions ${\phi }_{{ai}}(r)$ and ${\phi }_{{bi}}(r)$, so there will be two sets of modes $| {\phi }_{{ai}}\rangle$, $| {\phi }_{{bi}}\rangle$, where in the $| {\bf{r}}\rangle$ representation we have $\langle {\bf{r}}\,| {\phi }_{{ai}}\rangle ={\phi }_{{ai}}(r)| A\rangle$ and $\langle {\bf{r}}\,| {\phi }_{{bi}}\rangle ={\phi }_{{bi}}(r)| B\rangle$. Mode orthogonality between A and B modes arises from $\langle A| B\rangle =0$ rather from the spatial mode functions being orthogonal. The multimode spin operators are defined in appendix A (see equations (193) and (196)). These satisfy the standard commutation rules for angular momentum operators.

If the density operator for the overall system is $\widehat{\rho }$ then expectation values of the three spin operators $\langle {\widehat{S}}_{\xi }\rangle ={\rm{Tr}}(\widehat{\rho }{\widehat{S}}_{\xi })$ ($\xi =x,y,z$) define the Bloch vector. Spin squeezing is related to the fluctuation operators ${\rm{\Delta }}{\widehat{S}}_{\xi }={\widehat{S}}_{\xi }-\langle {\widehat{S}}_{\xi }\rangle$, in terms of which a real, symmetric covariance matrix $C({\widehat{S}}_{\xi },{\widehat{S}}_{\mu })$ ($\xi ,\mu =x,y,z$) is defined [7, 15] via

$$\begin{eqnarray}C({\widehat{S}}_{\xi },{\widehat{S}}_{\mu }) & = & (\langle {\rm{\Delta }}{\widehat{S}}_{\xi }\,{\rm{\Delta }}{\widehat{S}}_{\mu }\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{\mu }\,{\rm{\Delta }}{\widehat{S}}_{\xi }\rangle )/2\\ & = & \langle {\widehat{S}}_{\xi }\,{\widehat{S}}_{\mu }+{\widehat{S}}_{\mu }\,{\widehat{S}}_{\xi }\rangle /2-\langle {\widehat{S}}_{\xi }\rangle \langle {\widehat{S}}_{\mu }\rangle \end{eqnarray} \tag{ 2 }$$

and whose diagonal elements $C({\widehat{S}}_{\xi },{\widehat{S}}_{\xi })=\langle {\rm{\Delta }}{\widehat{S}}_{\xi }^{2}\rangle$ gives the variance for the fluctuation operators. The covariance matrix is also positive definite. The variances for the spin operators satisfy the three Heisenberg uncertainty principle relations $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant \tfrac{1}{4}| \langle {\widehat{S}}_{z}\rangle {| }^{2};$ $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \geqslant \tfrac{1}{4}| \langle {\widehat{S}}_{x}\rangle {| }^{2};$ $\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \geqslant \tfrac{1}{4}| \langle {\widehat{S}}_{y}\rangle {| }^{2}$, and spin squeezing is defined via conditions such as $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ with $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \gt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle | ,$ for ${\widehat{S}}_{x}$ being squeezed compared to ${\widehat{S}}_{y}$ and so on. Spin squeezing in these components is relevant to tests for entanglement of the modes $\widehat{a}$ and $\widehat{b}$, as will be shown later. Spin squeezing in rotated components is also important, in particular in the so-called principal components for which the covariance matrix is diagonal.

Since the spin operators also satisfy Heisenberg uncertainty principle relationships

$$\begin{eqnarray}\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle & \geqslant & \displaystyle \frac{1}{4}| \langle {\widehat{S}}_{z}\rangle {| }^{2},\\ \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle & \geqslant & \displaystyle \frac{1}{4}| \langle {\widehat{S}}_{x}\rangle {| }^{2},\\ \langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle & \geqslant & \displaystyle \frac{1}{4}| \langle {\widehat{S}}_{y}\rangle {| }^{2}\end{eqnarray} \tag{ 5 }$$

spin squeezing will now be defined for the spin operators via conditions such as

$$\begin{eqnarray}\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle & \lt & \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \ {\rm{and}}\ \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \gt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | ,\\ \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle & \lt & \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{x}\rangle | \ {\rm{and}}\ \langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \gt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{x}\rangle | ,\\ \langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle & \lt & \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{y}\rangle | \ {\rm{and}}\ \langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \gt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{y}\rangle | \end{eqnarray} \tag{ 6 }$$

for ${\widehat{S}}_{x}$ being squeezed compared to ${\widehat{S}}_{y}$, and so on.

Note also that the Heisenberg uncertainty principle proof (based on $\langle ({\rm{\Delta }}{\widehat{S}}_{\alpha }+{\rm{i}}\lambda {\rm{\Delta }}{\widehat{S}}_{\beta })({\rm{\Delta }}{\widehat{S}}_{\alpha }-{\rm{i}}\lambda {\rm{\Delta }}{\widehat{S}}_{\beta })\rangle \geqslant 0$ for all real λ) also establishes the general result for all quantum states

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{\alpha }^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{\beta }^{2}\rangle \geqslant | \langle {\widehat{S}}_{\gamma }\rangle | ,\end{eqnarray} \tag{ 7 }$$

where α, β and γ are x, y and z in cyclic order.

Other criteria for spin squeezing are also used, for example in the article by Wineland et al [17], where the focus is on spin squeezing for one component compared to any perpendicular component, and is set out in appendix B. A further special case is that of planar squeezing [18] where the Bloch vector is in one plane and the focus is on the spin fluctuation in this plane being squeezed compared to that perpendicular to the plane. This case is also discussed in appendix B.

Since the two new mode spin operators defined in equation (3) satisfy the standard angular momentum operator commutation rules, the usual Heisenberg Uncertainty rules analogous to (5) apply, so that spin squeezing can also exist in the two mode cases involving the new spin operators ${\widehat{J}}_{x},{\widehat{J}}_{y}$ and ${\widehat{J}}_{z}$ as well. These uncertainty principle features also apply to multimode spin operators as defined in appendix B. The criteria for spin squeezing in the multimode case is of the same form as (6), but for completeness is included in appendix B.

It should be noted that finding spin squeezing for one principal spin operator ${\widehat{J}}_{y}$ with respect to another ${\widehat{J}}_{x}$ does not mean that there is spin squeezing for any of the old spin operators ${\widehat{S}}_{x},{\widehat{S}}_{y}$ and ${\widehat{S}}_{z}$ In the case of the relative phase eigenstate (see section 3.7) ${\widehat{J}}_{y}$ is squeezed with respect to ${\widehat{J}}_{x}$—however none of the old spin components are spin squeezed.

The new spin operators are also related to the original spin operators via a unitary rotation operator $\widehat{R}(\alpha ,\beta ,\gamma )$ parameterized in terms of Euler angles so that

$$\begin{eqnarray}&&{\widehat{J}}_{\xi }=\widehat{R}(\alpha ,\beta ,\gamma ){\widehat{S}}_{\xi }\,\widehat{R}{(\alpha ,\beta ,\gamma )}^{-1},\end{eqnarray} \tag{ 8 }$$

where

$$\begin{eqnarray}&&\widehat{R}(\alpha ,\beta ,\gamma )={\widehat{R}}_{z}(\alpha ){\widehat{R}}_{y}(\beta ){\widehat{R}}_{z}(\gamma )\end{eqnarray} \tag{ 9 }$$

with ${\widehat{R}}_{\xi }(\phi )=\exp ({\rm{i}}\phi {\widehat{S}}_{\xi })$ describing a rotation about the ξ axis anticlockwise through an angle ϕ. Details for the rotation operators and matrices are set out in [7]. Note that equation (8) specifies a rotation of the vector spin operator rather than a rotation of the axes, so ${\widehat{J}}_{\xi }$ ($\xi =x,y,z$) are the components of the rotated vector spin operator with respect to the original axes.

We can also see that the new spin operators are related to new mode operators $\widehat{c}$ and $\widehat{d}$ via

$$\begin{eqnarray}{\widehat{J}}_{x} & = & ({\widehat{d}}^{\dagger }\widehat{c}+{\widehat{c}}^{\dagger }\widehat{d})/2,\qquad {\widehat{J}}_{y}=({\widehat{d}}^{\dagger }\widehat{c}-{\widehat{c}}^{\dagger }\widehat{d})/2i,\\ {\widehat{J}}_{z} & = & ({\widehat{d}}^{\dagger }\widehat{d}-{\widehat{c}}^{\dagger }\widehat{c})/2,\end{eqnarray} \tag{ 10 }$$

where

$$\begin{eqnarray}\widehat{c} & = & \widehat{R}(\alpha ,\beta ,\gamma )\widehat{a}\,\widehat{R}{(\alpha ,\beta ,\gamma )}^{-1},\\ \widehat{d} & = & \widehat{R}(\alpha ,\beta ,\gamma )\widehat{b}\,\widehat{R}{(\alpha ,\beta ,\gamma )}^{-1}.\end{eqnarray} \tag{ 11 }$$

For the bosonic case a straight-forward calculation gives the new mode operators as

$$\begin{eqnarray}\widehat{c} & = & \exp \left(\displaystyle \frac{1}{2}{\rm{i}}\gamma \right)\left(\cos \left(\displaystyle \frac{\beta }{2}\right)\exp \left(\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)\widehat{a}\right.\\ & & +\sin \left(\displaystyle \frac{\beta }{2}\right)\exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)\widehat{b}),\\ \widehat{d} & = & \exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\gamma \right)\left(-\sin \left(\displaystyle \frac{\beta }{2}\right)\exp \left(\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)\widehat{a}\right.\\ & & \left.+\cos \left(\displaystyle \frac{\beta }{2}\right)\exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)\widehat{b}\right)\end{eqnarray} \tag{ 12 }$$

and it is easy to then check that $\widehat{c}$ and $\widehat{d}$ satisfy the expected non-zero bosonic commutation rules are $[\widehat{e},{\widehat{e}}^{\dagger }]=\widehat{1}$ ($\widehat{e}=\widehat{c}$ or $\widehat{d}$) and that the total boson number operator is $\widehat{N}=({\widehat{d}}^{\dagger }\widehat{d}+{\widehat{c}}^{\dagger }\widehat{c})$. As $\widehat{N}$ is invariant under unitary rotation operators it follows that ${\widehat{J}}_{x}^{2}+{\widehat{J}}_{y}^{2}+{\widehat{J}}_{z}^{2}=(\widehat{N}/2)(\widehat{N}/2+1)$.

The new mode operators correspond to new single particle states $| {\phi }_{c}\rangle$, $| {\phi }_{d}\rangle$ where

$$\begin{eqnarray}| {\phi }_{c}\rangle & = & \exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\gamma \right)\left(\cos \left(\displaystyle \frac{\beta }{2}\right)\exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)| {\phi }_{a}\rangle \right.\\ & & +\sin \left(\displaystyle \frac{\beta }{2}\right)\exp \left(\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)| {\phi }_{b}\rangle ),\\ | {\phi }_{d}\rangle & = & \exp \left(\displaystyle \frac{1}{2}{\rm{i}}\gamma \right)\left(-\sin \left(\displaystyle \frac{\beta }{2}\right)\exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)| {\phi }_{a}\rangle \right.\\ & & \left.+\cos \left(\displaystyle \frac{\beta }{2}\right)\exp \left(\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)| {\phi }_{b}\rangle \right)\end{eqnarray} \tag{ 13 }$$

These are two orthonormal quantum superpositions of the original single particle states $| {\phi }_{a}\rangle$, $| {\phi }_{b}\rangle$, and as such represent an alternative choice of modes that could be realized experimentally.

Equations (12) can be inverted to give the old mode operators via

$$\begin{eqnarray}\widehat{a} & = & \exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)\left(\cos \left(\displaystyle \frac{\beta }{2}\right)\exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\gamma \right)\widehat{c}\right.\\ & & \left.-\sin \left(\displaystyle \frac{\beta }{2}\right)\exp \left(+\displaystyle \frac{1}{2}{\rm{i}}\gamma \right)\widehat{d}\right)\\ \widehat{b} & = & \exp \left(+\displaystyle \frac{1}{2}{\rm{i}}\alpha \right)\left(\sin \left(\displaystyle \frac{\beta }{2}\right)\exp \left(\displaystyle \frac{1}{2}{\rm{i}}\gamma \right)\widehat{c}\right.\\ & & \left.+\cos \left(\displaystyle \frac{\beta }{2}\right)\exp \left(-\displaystyle \frac{1}{2}{\rm{i}}\gamma \right)\widehat{d}\right)\end{eqnarray} \tag{ 14 }$$

Next, we find the mean values of the spin operators for the product state ${\widehat{\rho }}_{R}\,=\,$ ${\widehat{\rho }}_{R}^{A}\otimes {\widehat{\rho }}_{R}^{B}$

$$\begin{eqnarray}{\langle {\widehat{S}}_{x}\rangle }_{R} & = & \displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\rangle }_{R}{\langle \widehat{a}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\rangle }_{R}{\langle \widehat{b}\rangle }_{R})=0,\\ {\langle {\widehat{S}}_{y}\rangle }_{R} & = & \displaystyle \frac{1}{2i}({\langle {\widehat{b}}^{\dagger }\rangle }_{R}{\langle \widehat{a}\rangle }_{R}-{\langle {\widehat{a}}^{\dagger }\rangle }_{R}{\langle \widehat{b}\rangle }_{R})=0\end{eqnarray} \tag{ 22 }$$

and

$$\begin{eqnarray}&&{\langle {\widehat{S}}_{z}\rangle }_{R}=\displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}-{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})\end{eqnarray} \tag{ 23 }$$

for SSR compliant sub-system states using (16), and thus using (19) the overall mean values for the separable state is

$$\begin{eqnarray}&&\langle {\widehat{S}}_{x}\rangle =0,\qquad \langle {\widehat{S}}_{y}\rangle =0\end{eqnarray} \tag{ 24 }$$

and

$$\begin{eqnarray}&&\langle {\widehat{S}}_{z}\rangle =\displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}-{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})).\end{eqnarray} \tag{ 25 }$$

Hence if either $\langle {\widehat{S}}_{x}\rangle \ne 0$ or $\langle {\widehat{S}}_{y}\rangle \ne 0$ the state must be entangled. This may be called the Bloch vector test. This result will also have significance later.

Next we calculate ${\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle }_{R}$, ${\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle }_{R}$ and ${\langle {\widehat{S}}_{x}\rangle }_{R}$, ${\langle {\widehat{S}}_{y}\rangle }_{R}$, ${\langle {\widehat{S}}_{z}\rangle }_{R}$ for the case of the separable state (15) where ${\widehat{\rho }}_{R}\,=$ ${\widehat{\rho }}_{R}^{A}\otimes {\widehat{\rho }}_{R}^{B}$. From equation (1) we find that

$$\begin{eqnarray}{\widehat{S}}_{x}^{2} & = & \displaystyle \frac{1}{4}({({\widehat{b}}^{\dagger })}^{2}{(\widehat{a})}^{2}+{\widehat{b}}^{\dagger }\widehat{b}\widehat{a}{\widehat{a}}^{\dagger }+{\widehat{a}}^{\dagger }\widehat{a}\widehat{b}{\widehat{b}}^{\dagger }+{(\widehat{b})}^{2}{({\widehat{a}}^{\dagger })}^{2}),\\ {\widehat{S}}_{y}^{2} & = & -\displaystyle \frac{1}{4}({({\widehat{b}}^{\dagger })}^{2}{(\widehat{a})}^{2}-{\widehat{b}}^{\dagger }\widehat{b}\widehat{a}{\widehat{a}}^{\dagger }-{\widehat{a}}^{\dagger }\widehat{a}\widehat{b}{\widehat{b}}^{\dagger }+{(\widehat{b})}^{2}{({\widehat{a}}^{\dagger })}^{2})\end{eqnarray} \tag{ 26 }$$

so that on taking the trace with ${\widehat{\rho }}_{R}$ and using equations (16) we get after applying the commutation rules $[\widehat{e},{\widehat{e}}^{\dagger }]=\widehat{1}$ ($\widehat{e}=\widehat{a}$ or $\widehat{b}$)

$$\begin{eqnarray}{\langle {\widehat{S}}_{x}^{2}\rangle }_{R} & = & \displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}),\\ {\langle {\widehat{S}}_{y}^{2}\rangle }_{R} & = & \displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}).\end{eqnarray} \tag{ 27 }$$

Hence using (22) for ${\langle {\widehat{S}}_{x}\rangle }_{R}$ and ${\langle {\widehat{S}}_{y}\rangle }_{R}$ we see finally that the variances are

$$\begin{eqnarray}{\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle }_{R} & = & \displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}),\\ {\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle }_{R} & = & \displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})\end{eqnarray} \tag{ 28 }$$

and therefore from equation (20)

$$\begin{eqnarray}\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle & \geqslant & \displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})\\ & & +\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}),\\ \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle & \geqslant & \displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})\\ & & +\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}).\end{eqnarray} \tag{ 29 }$$

Now using (25) for $\langle {\widehat{S}}_{z}\rangle$ we see that

$$\begin{eqnarray}\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | & \leqslant & \displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{4}| ({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}-{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}))| \\ & \leqslant & \displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}))\end{eqnarray} \tag{ 30 }$$

and thus for any non-entangled state of modes $\widehat{a}$ and $\widehat{b}$

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle -\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \\ &&\quad \geqslant \displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})\\ &&\qquad -\displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}))\\ &&\quad \geqslant \displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})\\ &&\quad \geqslant 0.\end{eqnarray} \tag{ 31 }$$

Similar final steps show that $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle -\tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \geqslant 0$ for all non-entangled state of modes $\widehat{a}$ and $\widehat{b}$.

This shows that for the general non-entangled state with modes $\widehat{a}$ and $\widehat{b}$ as the sub-systems, the variances for two of the principal spin fluctuations $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle$ and $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle$ are both greater than $\tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$, and hence there is no spin squeezing for ${\widehat{S}}_{x}$ compared to ${\widehat{S}}_{y}$ (or vice versa). Note that as $| \langle {\widehat{S}}_{y}\rangle | =0$, the quantity $\sqrt{(| \langle {\widehat{S}}_{\perp 1}\rangle {| }^{2}+| \langle {\widehat{S}}_{\perp 2}\rangle {| }^{2})}$ is the same as $| \langle {\widehat{S}}_{z}\rangle |$, so the alternative criterion in appendix B equation (201) is the same as that in equation (6) which is used here.

It is easy to see from (24) that

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle -\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{y}\rangle | \geqslant 0,\qquad \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle -\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{x}\rangle | \geqslant 0\end{eqnarray} \tag{ 32 }$$

for any non-entangled state of modes $\widehat{a}$ and $\widehat{b}$. This completes the set of inequalities for the variances of ${\widehat{S}}_{x}$ and ${\widehat{S}}_{y}$. These last inequalities are of course trivially true and amount to no more than showing that the variances $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle$ and $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle$ are not negative.

For the other principal spin fluctuation we find that for separable states

$$\begin{eqnarray}{\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle }_{R} & = & \displaystyle \frac{1}{4}({\langle ({\widehat{b}}^{\dagger }\widehat{b}-{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})({\widehat{b}}^{\dagger }\widehat{b}-{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})\rangle }_{R}\\ & & +{\langle ({\widehat{a}}^{\dagger }\widehat{a}-{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})({\widehat{a}}^{\dagger }\widehat{a}-{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})\rangle }_{R}\end{eqnarray} \tag{ 33 }$$

so that using (20)

$$\begin{eqnarray}\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle & \geqslant & \displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{4}({\langle {({\widehat{b}}^{\dagger }\widehat{b}-{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})}^{2}\rangle }_{R}\\ & & +{\langle {({\widehat{a}}^{\dagger }\widehat{a}-{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})}^{2}\rangle }_{R}.\end{eqnarray} \tag{ 34 }$$

From equation (24) it follows that

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle -\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{x}\rangle | \,\\ &&\quad \geqslant \displaystyle \sum _{R}{P}_{R}\,\displaystyle \frac{1}{4}({\langle {({\widehat{b}}^{\dagger }\widehat{b}-{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})}^{2}\rangle }_{R}+{\langle {({\widehat{a}}^{\dagger }\widehat{a}-{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})}^{2}\rangle }_{R}\\ &&\quad \geqslant 0.\end{eqnarray} \tag{ 35 }$$

Similarly $\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle -\tfrac{1}{2}| \langle {\widehat{S}}_{y}\rangle | \,\geqslant 0$. Again, these results are trivial and just show that the variances are non-negative.

So overall, we have for the general non-entangled state of modes $\widehat{a}$ and $\widehat{b}$

$$\begin{eqnarray}\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle & \geqslant & \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \ {\rm{and}}\ \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | ,\\ \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle & \geqslant & \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{x}\rangle | \ {\rm{and}}\ \langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \geqslant \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{x}\rangle | ,\\ \langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle & \geqslant & \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{y}\rangle | \ {\rm{and}}\ \langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \geqslant \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{y}\rangle | .\end{eqnarray} \tag{ 36 }$$

The first result tells us that for any non-entangled state of modes $\widehat{a}$ and $\widehat{b}$ the spin operator ${\widehat{S}}_{x}$ is not squeezed compared to ${\widehat{S}}_{y}$ (or vice-versa). The same is also true for the other pairs of spin operators, as we will now see.

The key value of these results is the spin squeezing test for entanglement. We see that from the first inequality in (36) for separable states, that if for a quantum state we find that

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \qquad {\text{}}{\rm{or}}\quad \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \end{eqnarray} \tag{ 37 }$$

then the state must be entangled for modes $\widehat{a}$ and $\widehat{b}$. Thus we only need to have spin squeezing in either of ${\widehat{S}}_{x}$ with respect to ${\widehat{S}}_{y}$ or vice-versa to demonstrate entanglement. Note that one cannot have both $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ and $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ etc due to the Heisenberg uncertainty principle.

Because ${\langle {\widehat{S}}_{x}\rangle }_{\rho }={\langle {\widehat{S}}_{y}\rangle }_{\rho }=0$ the second and third results in (36) merely show that $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \geqslant 0$, $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant 0$ and $\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \geqslant 0\$ for SSR compliant non-entangled states, it may be thought that no conclusion follows regarding the spin squeezing involving ${\widehat{S}}_{z}$ for entangled states. This is not the case. If for a given state we find that

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{x}\rangle | \qquad {\text{}}{\rm{or}}\quad \langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{x}\rangle | \end{eqnarray} \tag{ 38 }$$

or

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{y}\rangle | \qquad {\text{}}{\rm{or}}\quad \langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{y}\rangle | \end{eqnarray} \tag{ 39 }$$

then the state must be entangled. This is because if any of these situations apply then at least one of ${\langle {\widehat{S}}_{x}\rangle }_{\rho }$ or ${\langle {\widehat{S}}_{y}\rangle }_{\rho }$ must be non-zero. But as we have seen from (24) both of the quantities are zero in non-entangled states. Thus we only need to have spin squeezing in either of ${\widehat{S}}_{z}$ with respect to ${\widehat{S}}_{y}$ or vice-versa or spin squeezing in either of ${\widehat{S}}_{z}$ with respect to ${\widehat{S}}_{x}$ or vice-versa to demonstrate entanglement.

Hence the general conclusion stated in [2], that spin squeezing in any spin operator ${\widehat{S}}_{x}$, ${\widehat{S}}_{y},$ ${\widehat{S}}_{z}$ shows that the state must be entangled for modes $\widehat{a}$ and $\widehat{b}$. The presence of spin squeezing is a conclusive test for entanglement. Note that the reverse is not true—there are many entangled states that are not spin squeezed. A notable example is the particular binomial state $| {\rm{\Phi }}\rangle ={({(\widehat{a}+\widehat{b})}^{\dagger }/\sqrt{2})}^{N}/\sqrt{N!}\ | 0\rangle$ for which ${\langle {\widehat{S}}_{x}\rangle }_{\rho }=N/2$, ${\langle {\widehat{S}}_{y}\rangle }_{\rho }={\langle {\widehat{S}}_{z}\rangle }_{\rho }=0$ and ${\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle }_{\rho }={\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle }_{\rho }\,=N/4$, ${\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle }_{\rho }=0$ (see [7]). The spin fluctuations in ${\widehat{S}}_{y}$ and ${\widehat{S}}_{z}$ correspond to the standard quantum limit.

This is a key result for two mode entanglement. All spin squeezed states are entangled. We emphasize again that the converse is not true. Not all entangled two mode states are spin squeezed. This important distinction is not always recognized—entanglement and spin squeezing are two distinct features of a two mode quantum state that do not always occur together.

For the two orthogonal spin operator components (199) (appendix B) in the xy plane ${\widehat{S}}_{\perp 1}$ and ${\widehat{S}}_{\perp 2}$ it is then straightforward to show that

$$\begin{eqnarray}&&{\rm{If}}{\text{}}\quad \langle {\rm{\Delta }}{\widehat{S}}_{\perp 1}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \end{eqnarray} \tag{ 40 }$$

or

$$\begin{eqnarray}&&{\rm{If}}{\text{}}\quad \langle {\rm{\Delta }}{\widehat{S}}_{\perp 2}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \end{eqnarray} \tag{ 41 }$$

that is, if ${\widehat{S}}_{\perp 1}$ is squeezed compared to ${\widehat{S}}_{\perp 2}$ or vice versa—then the state must be entangled. Spin squeezing in any of the spin operator component in the xy plane will demonstrate entanglement.

The derivation of the spin squeezing test for this two mode system of identical bosons was based on the requirements that the quantum state complied with both the symmetrization principle and the global superselection rule for total boson numbers, together with the subsystem states complying with the local particle number superselection rule in the case of. separable states. However, as discussed at length in paper I, there are some papers such as [4] in which the local particle number superselection rule is not applied to separable states—leading to the concept of 'separable but non-local' states (which are regarded here as entangled, not separable). Accordingly, such other approaches would result in different tests for what they define as entanglement. It is therefore of some interest to consider what interpretation could be placed on observing spin squeezing in any spin component, both from the point of view about entanglement presented here and from that in other papers such as [4]. This discussion is set out in appendix C. Figures 2 and 3 (within the supplementary material) therein depict the various types of quantum states involved—separable, 'separable but non-local', other entangled states, as well as indicating whch are spin squeezed.

Of the results for a non-entangled physical state for modes $\widehat{a}$ and $\widehat{b}$ we will later find it particularly important to consider the first of (36)

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \geqslant \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \quad {\rm{and}}\quad \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | .\end{eqnarray} \tag{ 42 }$$

This is because we can show that for any quantum state

$$\begin{eqnarray}| \langle {\widehat{S}}_{z}\rangle | & = & \left|\langle \displaystyle \frac{1}{2}({\widehat{n}}_{b}-{\widehat{n}}_{a})\rangle \right|\\ & \leqslant & \displaystyle \frac{1}{2}(| \langle {\widehat{n}}_{b}\rangle | \,+| \langle {\widehat{n}}_{a}\rangle | )=\displaystyle \frac{1}{2}\langle \widehat{N}\rangle \end{eqnarray} \tag{ 43 }$$

there is an inequality involving $| \langle {\widehat{S}}_{z}\rangle | \,$ and the mean number of bosons $\langle \widehat{N}\rangle$ in the two mode system. Note that there are some entangled states (see section 3.5) for which $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle$ and $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle$ are both greater than $\tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$, since all that has been proven is that for all non-entangled states we must have both $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \geqslant \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ and $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$.

Hence we may conclude that spin squeezing in any of the original spin fluctuations ${\widehat{S}}_{x}$, ${\widehat{S}}_{y}$ or ${\widehat{S}}_{z}$ requires the quantum state to be entangled for the modes $\widehat{a}$ and $\widehat{b}$ as the sub-systems. Similarly, we may conclude that spin squeezing in any of the principal spin fluctuations ${\widehat{J}}_{x}$, ${\widehat{J}}_{y}$ or ${\widehat{J}}_{z}$ requires the quantum state to be entangled for the modes $\widehat{c}$ and $\widehat{d}$ as the sub-systems, these modes being associated with the principal spin fluctuations via equation (10). Although finding spin squeezing tells us that the state is entangled, there are however no simple relationships between the measures of entanglement and those of spin squeezing, so the linkage is essentially a qualitative one. For general quantum states, measures of entanglement for the specific situation of two sub-systems (bipartite entanglement) are reviewed in [20].

It is more instructive to consider spin squeezing in terms of new spin operators ${\widehat{J}}_{x}$, ${\widehat{J}}_{y}$, ${\widehat{J}}_{z}$ for which the covariance matrix is diagonal. The new spin operators are related to the original spin operators via

$$\begin{eqnarray}{\widehat{J}}_{x} & = & {\widehat{S}}_{z},\\ {\widehat{J}}_{y} & = & \sin {\theta }_{p}\,{\widehat{S}}_{x}+\cos {\theta }_{p}\,{\widehat{S}}_{y},\\ {\widehat{J}}_{z} & = & -\cos {\theta }_{p}\,{\widehat{S}}_{x}+\sin {\theta }_{p}\,{\widehat{S}}_{y}\end{eqnarray} \tag{ 67 }$$

corresponding to the transformation in equation (3) with Euler angles $\alpha =-\pi +{\theta }_{p}$, $\beta =-\pi /2$ and $\gamma =-\pi$.

The Bloch vector and covariance matrix for spin operators ${\widehat{J}}_{x}$, ${\widehat{J}}_{y}$, ${\widehat{J}}_{z}$ are given by (see equations (180) and (181) in [7]—note that the $C({\widehat{J}}_{y},{\widehat{J}}_{y})$ element is incorrect in equation (181), see second of [7])

$$\begin{eqnarray}&&\langle {\widehat{J}}_{x}\rangle =0,\qquad \langle {\widehat{J}}_{y}\rangle =0,\qquad \langle {\widehat{J}}_{z}\rangle =-N\,\displaystyle \frac{\pi }{8}\end{eqnarray} \tag{ 68 }$$

and

$$\begin{eqnarray}\left[\begin{array}{ccc}C({\widehat{J}}_{x},{\widehat{J}}_{x}) & C({\widehat{J}}_{x},{\widehat{J}}_{y}) & C({\widehat{J}}_{x},{\widehat{J}}_{z})\\ C({\widehat{J}}_{y},{\widehat{J}}_{x}) & C({\widehat{J}}_{y},{\widehat{J}}_{y}) & C({\widehat{J}}_{y},{\widehat{J}}_{z})\\ C({\widehat{J}}_{z},{\widehat{J}}_{y}) & C({\widehat{J}}_{z},{\widehat{J}}_{y}) & C({\widehat{J}}_{z},{\widehat{J}}_{z})\end{array}\right]\\ \quad \doteqdot \,\left[\begin{array}{lll}\tfrac{1}{12}{N}^{2} & 0 & 0\\ 0 & \tfrac{1}{4}+\tfrac{1}{8}\mathrm{ln}N & 0\\ 0 & 0 & \left(\tfrac{1}{6}-\tfrac{{\pi }^{2}}{64}\right){N}^{2}\end{array}\right]\qquad N\gg 1.\end{eqnarray} \tag{ 69 }$$

With $\langle {\rm{\Delta }}{\widehat{J}}_{x}^{2}\rangle =\tfrac{1}{12}{N}^{2}$, $\langle {\rm{\Delta }}{\widehat{J}}_{y}^{2}\rangle =\tfrac{1}{4}+\tfrac{1}{8}\mathrm{ln}N$ and $\langle {\rm{\Delta }}{\widehat{J}}_{z}^{2}\rangle \,=\left(\tfrac{1}{6}-\tfrac{{\pi }^{2}}{64}\right){N}^{2}$ and the only non-zero Bloch vector component being $\langle {\widehat{J}}_{z}\rangle =-N\,\tfrac{\pi }{8}$ it is easy to see that $\langle {\rm{\Delta }}{\widehat{J}}_{x}^{2}\rangle \langle {\rm{\Delta }}{\widehat{J}}_{y}^{2}\rangle \geqslant \tfrac{1}{4}| \langle {\widehat{J}}_{z}\rangle {| }^{2}$ as required by the Heisenberg Uncertainty Principle. The principal spin fluctuations in both ${\widehat{J}}_{x}$ and ${\widehat{J}}_{z}$ are comparable to the length of the Bloch vector and no spin squeezing occurs in either of these components. However, spin squeezing occurs in that ${\widehat{J}}_{y}$ is squeezed with respect to ${\widehat{J}}_{x}$, since $\langle {\rm{\Delta }}{\widehat{J}}_{y}^{2}\rangle$ only increases as $\tfrac{1}{8}\mathrm{ln}N$ while $\tfrac{1}{2}| \langle {\widehat{J}}_{z}\rangle |$ increases as $\tfrac{\pi }{16}N\,$ for large N. Hence the relative phase state satisfies the test in equation (37) to demonstrate entanglement for modes $\widehat{c}$, $\widehat{d}$. Here $\sqrt{\langle {\rm{\Delta }}{\widehat{J}}_{y}^{2}\rangle }/| \langle {\widehat{J}}_{z}\rangle | \sim \sqrt{\mathrm{ln}N}/N$ which indicates that the Heisenberg limit is being reached.

Note that none of the old spin components are spin squeezed. As shown in [7] $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle =\left(\tfrac{1}{6}-\tfrac{{\pi }^{2}}{64}\right){\cos }^{2}{\theta }_{p}\,{N}^{2}$, $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle =\left(\tfrac{1}{6}-\tfrac{{\pi }^{2}}{64}\right){\sin }^{2}{\theta }_{p}\,{N}^{2}$ and $\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle =\tfrac{1}{12}{N}^{2}$, along with $\langle {\widehat{S}}_{x}\rangle =N\,\tfrac{\pi }{8}\cos {\theta }_{p}\quad \langle {\widehat{S}}_{y}\rangle =-N\,\tfrac{\pi }{8}\sin {\theta }_{p}\quad \langle {\widehat{S}}_{z}\rangle =0$. All variances are of order ${N}^{2}$ while the non-zero means are only of order N. Hence spin squeezing in one of the principal spin operators does not imply spin squeezing in any of the original spin operators. This is relevant to spin squeezing tests for entanglement of the original modes.

To confirm that the relative phase state is in fact an entangled state for modes $\widehat{c}$, $\widehat{d}$ as well as for the original modes $\widehat{a}$, $\widehat{b}$, we note that the new mode operators $\widehat{c}$, $\widehat{d}$ are given in in equation (12) with Euler angles $\alpha =-\pi +{\theta }_{p}$, $\beta =-\pi /2$ and $\gamma =-\pi$. The old mode operators are given in equation (14) and with these Euler angles we have

$$\begin{eqnarray}\widehat{a} & = & -\exp \left(\displaystyle \frac{1}{2}{\rm{i}}{\theta }_{p}\right)\displaystyle \frac{1}{\sqrt{2}}(\widehat{c}-\widehat{d}),\\ \widehat{b} & = & -\exp \left(-\displaystyle \frac{1}{2}{\rm{i}}{\theta }_{p}\right)\displaystyle \frac{1}{\sqrt{2}}(\widehat{c}+\widehat{d}).\end{eqnarray} \tag{ 70 }$$

This enables us to write the phase state in terms of new mode operators $\widehat{c}$, $\widehat{d}$ as

$$\begin{eqnarray}\left|\displaystyle \frac{N}{2},{\theta }_{p}\right\rangle & = & \displaystyle \frac{1}{\sqrt{N+1}}{\left(\displaystyle \frac{-1}{\sqrt{2}}\right)}^{N}\displaystyle \sum _{k=-N/2}^{N/2}\displaystyle \sum _{r=-N/4+k/2}^{\,N/4-k/2}\displaystyle \sum _{s=-N/4-k/2}^{N/4+k/2}\\ & & \times \displaystyle \frac{1}{\sqrt{(N/2-k)!}}\displaystyle \frac{1}{\sqrt{(N/2+k)!}}{(-1)}^{N/4-k/2+r}\\ & & \times \displaystyle \frac{(N/2-k)!}{(N/4-k/2-r)!(N/4-k/2+r)!}\\ & & \times \displaystyle \frac{(N/2+k)!}{(N/4+k/2-s)!(N/4+k/2+s)!}\\ & & \times {({\widehat{c}}^{\dagger })}^{N/2-(r+s)}{({\widehat{d}}^{\dagger })}^{N/2+(r+s)}| 0\rangle .\end{eqnarray} \tag{ 71 }$$

We see that the expression does not depend explicitly on the relative phase ${\theta }_{p}$ when written in terms of the new mode un-normalized Fock states ${({\widehat{c}}^{\dagger })}^{N/2-(r+s)}{({\widehat{d}}^{\dagger })}^{N/2+(r+s)}| 0\rangle$. This pure state is a linear combination of product states of the form ${| N/2-m\rangle }_{c}\otimes {| N/2+m\rangle }_{d}$ for various m—each of which is an N boson state and an eigenstate for ${\widehat{J}}_{z}$ with eigenvalue m, and therefore is an entangled state for modes $\widehat{c}$, $\widehat{d}$ which is compatible with the global SSR. Note that there cannot just be a single term m involved, otherwise the variance for ${\widehat{J}}_{z}$ would be zero instead of $\left(\tfrac{1}{6}-\tfrac{{\pi }^{2}}{64}\right){N}^{2}$. We will return to the relative phase state again in section 4.1.

An entanglement test in which local particle number SSR compliance is ignored is presented in the paper by Hillery and Zubairy [22] entitled 'Entanglement conditions for two-mode states'. The paper actually dealt with EM field modes, and the density operators ${\widehat{\rho }}_{R}^{A}$, ${\widehat{\rho }}_{R}^{B}$ for photon modes allowed for coherences between states with differing photon numbers. A discussion of SSR for the case of photons is presented in paper I, in appendix L. Hence the conditions in equation (16) were not applied.

We will now derive the Hillery spin variance inequalities involving ${\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle }^{}$, ${\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle }^{}$ by applying a similar treatment to that in section 3.1, but now ignoring local particle number SSR compliance. It is found that for the original spin operators ${\widehat{S}}_{x}$, ${\widehat{S}}_{y}$, ${\widehat{S}}_{z}$ and modes $\widehat{a}$ and $\widehat{b}$

$$\begin{eqnarray}{\langle {\widehat{S}}_{x}^{2}\rangle }_{R} & = & \displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})\\ & & +\displaystyle \frac{1}{4}({\langle {({\widehat{b}}^{\dagger })}^{2}\rangle }_{R}{\langle {(\widehat{a})}^{2}\rangle }_{R}+{\langle {(\widehat{b})}^{2}\rangle }_{R}\langle {({\widehat{a}}^{\dagger })}^{2}\rangle ),\\ {\langle {\widehat{S}}_{y}^{2}\rangle }_{R} & = & \displaystyle \frac{1}{4}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})\\ & & -\displaystyle \frac{1}{4}({\langle {({\widehat{b}}^{\dagger })}^{2}\rangle }_{R}{\langle {(\widehat{a})}^{2}\rangle }_{R}+{\langle {(\widehat{b})}^{2}\rangle }_{R}\langle {({\widehat{a}}^{\dagger })}^{2}\rangle )\end{eqnarray} \tag{ 72 }$$

where terms such as ${\langle {({\widehat{b}}^{\dagger })}^{2}\rangle }_{R}$ and ${\langle {(\widehat{a})}^{2}\rangle }_{R}$ previously shown to be zero have been retained. Note that in [22] the operators ${\widehat{S}}_{x}$, ${\widehat{S}}_{y}$, ${\widehat{S}}_{z}$ constructed from the EM field mode operators as in equation (1)) would be related to Stokes parameters Hence

$$\begin{eqnarray}&&{\langle {\widehat{S}}_{x}^{2}\rangle }_{R}+{\langle {\widehat{S}}_{y}^{2}\rangle }_{R}\\ &&\quad =\displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})+({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}{\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})\\ &&\quad =\displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}+1)+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+1)),\end{eqnarray} \tag{ 73 }$$

where the terms ${\langle {({\widehat{b}}^{\dagger })}^{2}\rangle }_{R}$, ..., $\langle {({\widehat{a}}^{\dagger })}^{2}\rangle$ cancel out. This is the same as before.

However,

$$\begin{eqnarray}{\langle {\widehat{S}}_{x}\rangle }_{R} & = & \displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\rangle }_{R}{\langle \widehat{a}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\rangle }_{R}{\langle \widehat{b}\rangle }_{R}),\\ {\langle {\widehat{S}}_{y}\rangle }_{R} & = & \displaystyle \frac{1}{2{i}}({\langle {\widehat{b}}^{\dagger }\rangle }_{R}{\langle \widehat{a}\rangle }_{R}-{\langle {\widehat{a}}^{\dagger }\rangle }_{R}{\langle \widehat{b}\rangle }_{R})\end{eqnarray} \tag{ 74 }$$

is now non-zero, since the previously zero terms have again been retained. Hence

$$\begin{eqnarray}&&{\langle {\widehat{S}}_{x}\rangle }_{R}^{2}+{\langle {\widehat{S}}_{y}\rangle }_{R}^{2}={\langle {\widehat{b}}^{\dagger }\rangle }_{R}{\langle \widehat{b}\rangle }_{R}{\langle {\widehat{a}}^{\dagger }\rangle }_{R}{\langle \widehat{a}\rangle }_{R}\end{eqnarray} \tag{ 75 }$$

so that we now have

$$\begin{eqnarray}&&{\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle }_{R}+{\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle }_{R}\\ &&\quad =\displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}+1)+\displaystyle \frac{1}{2}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R})+1)\\ &&\qquad -{\langle {\widehat{b}}^{\dagger }\rangle }_{R}{\langle \widehat{b}\rangle }_{R}{\langle \widehat{a}\rangle }_{R}{\langle {\widehat{a}}^{\dagger }\rangle }_{R}{\langle \widehat{a}\rangle }_{R}\\ &&\quad =\displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})\\ &&\qquad +({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}-| {\langle \widehat{a}\rangle }_{R}{| }^{2}| {\langle {\widehat{b}}^{\dagger }\rangle }_{R}{| }^{2}).\end{eqnarray} \tag{ 76 }$$

But from the Schwarz inequality—which is based on $\langle ({\widehat{a}}^{\dagger }-\langle {\widehat{a}}^{\dagger }\rangle )(\widehat{a}-\langle \widehat{a}\rangle )\rangle \geqslant 0$ for any state

$$\begin{eqnarray}&&| {\langle \widehat{a}\rangle }_{R}{| }^{2}\leqslant {\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}\qquad | {\langle \widehat{b}\rangle }_{R}{| }^{2}\leqslant {\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}\end{eqnarray} \tag{ 77 }$$

so that

$$\begin{eqnarray}&&{\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle }_{R}+{\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle }_{R}\geqslant \displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})\end{eqnarray} \tag{ 78 }$$

and thus from equation (20) it follows that for a general non entangled state

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant \displaystyle \sum _{R}{P}_{R}\displaystyle \frac{1}{2}({\langle {\widehat{n}}_{b}\rangle }_{R}+{\langle {\widehat{n}}_{a}\rangle }_{R}).\end{eqnarray} \tag{ 79 }$$

However, half the expectation value of the number operator is

$$\begin{eqnarray}&&\displaystyle \frac{1}{2}\langle \widehat{N}\rangle =\displaystyle \frac{1}{2}\langle ({\widehat{n}}_{a}+{\widehat{n}}_{b})\rangle =\displaystyle \sum _{R}{P}_{R}\displaystyle \frac{1}{2}({\langle {\widehat{n}}_{b}\rangle }_{R}+{\langle {\widehat{n}}_{a}\rangle }_{R})\end{eqnarray} \tag{ 80 }$$

so for a non-entangled state

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant \displaystyle \frac{1}{2}\langle \widehat{N}\rangle .\end{eqnarray} \tag{ 81 }$$

This inequality for non-entangled states is given in [22] (see their equation (3)). The above proof was based on not invoking the SSR requirements for separable states that we apply in this paper.

The Hillery spin variance test can also be applied in multi-mode situations, where the spin operators are defined as in appendix A and there are three cases involving different choices of sub-systems (see appendix D). The Hillery spin variance test applies for Cases 1 and 2, but not for Case 3. The derivation of the Hillary spin variance test for the multi-mode situation is presented in appendix E.

However, it is interesting that the inequality (81) can be more readily derived from the definition of entangled states used in the present paper—which is based on local particle number SSR compliance for separable states. We would then find that ${\langle {\widehat{S}}_{x}\rangle }_{R}={\langle {\widehat{S}}_{y}\rangle }_{R}=0$ and hence

$$\begin{eqnarray}{\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle }_{R}+{\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle }_{R} & = & \displaystyle \frac{1}{2}({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}+{\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})\\ & & +({\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R})\end{eqnarray} \tag{ 82 }$$

instead of equation (76). Since the term ${\langle {\widehat{b}}^{\dagger }\widehat{b}\rangle }_{R}({\langle {\widehat{a}}^{\dagger }\widehat{a}\rangle }_{R}$ is always positive we find after applying equation (20) that

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant \displaystyle \frac{1}{2}\langle \widehat{N}\rangle \end{eqnarray} \tag{ 83 }$$

which is the same as in equation (81). Hence, finding that $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt \tfrac{1}{2}\langle \widehat{N}\rangle$ would show that the state was entangled, irrespective of whether or not entanglement is defined in terms of non-physical nonentangled states.

Thus, the Hillery spin variance entanglement test [22] is that if

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt \displaystyle \frac{1}{2}\langle \widehat{N}\rangle \end{eqnarray} \tag{ 84 }$$

then the state is an entangled state of modes $\widehat{a}$ and $\widehat{b}$. This test is still used in recent papers, for example [23, 24] which deal with the entanglement of sub-systems each consisting of single modes $\widehat{a}$, $\widehat{b}$ for a double well situation (in these papers ${\widehat{S}}_{x}\to {\widehat{J}}_{{AB}}^{X}$, ${\widehat{S}}_{y}\to -{\widehat{J}}_{{AB}}^{Y}$, ${\widehat{S}}_{z}\to -{\widehat{J}}_{{AB}}^{Z}$).

Previously we had found for a general non-entangled state that is based on physically valid density operators ${\widehat{\rho }}_{R}^{A}$, ${\widehat{\rho }}_{R}^{B}$

$$\begin{eqnarray}\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle -\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | & \geqslant & 0,\\ \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle -\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | & \geqslant & 0\end{eqnarray} \tag{ 85 }$$

so that the sum of the variances satisfies the inequality

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant | \langle {\widehat{S}}_{z}\rangle | .\end{eqnarray} \tag{ 86 }$$

This is another valid inequality required for a non-entangled state as defined in the present paper. It follows that if only physical states ${\widehat{\rho }}_{R}^{A}$, ${\widehat{\rho }}_{R}^{B}$ are allowed, the related entanglement test involving $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle$ would be

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt | \langle {\widehat{S}}_{z}\rangle | .\end{eqnarray} \tag{ 87 }$$

For any quantum state we have

$$\begin{eqnarray}&&| \langle {\widehat{S}}_{z}\rangle | =\displaystyle \frac{1}{2}| (\langle {\widehat{n}}_{b}\rangle -\langle {\widehat{n}}_{a}\rangle )| \,\leqslant \displaystyle \frac{1}{2}(\langle {\widehat{n}}_{b}\rangle +\langle {\widehat{n}}_{a}\rangle )=\displaystyle \frac{1}{2}\langle \widehat{N}\rangle \end{eqnarray} \tag{ 88 }$$

which means that it is now required that $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle$ be less than a quantity that is smaller than in the criterion in (81).

However, it should be noted (see (7)) that all states, entangled or otherwise, satisfy the inequality

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant | \langle {\widehat{S}}_{z}\rangle | \end{eqnarray} \tag{ 89 }$$

so the inequality in (86)—though true, is of no use in establishing whether a state is entangled in the terms of the meaning of entanglement in the present paper. There are no quantum states, entangled or otherwise that satisfy the proposed entanglement test given in equation (87). This general result was stated by Hillery et al [22]. To show this we have

$$\begin{eqnarray}&&\langle {({\rm{\Delta }}{\widehat{S}}_{x}-{\rm{i}}\lambda {\rm{\Delta }}{\widehat{S}}_{y})}^{\dagger }({\rm{\Delta }}{\widehat{S}}_{x}+{\rm{i}}\lambda {\rm{\Delta }}{\widehat{S}}_{y})\rangle \geqslant 0,\end{eqnarray} \tag{ 90 }$$

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\lambda \langle {\widehat{S}}_{z}\rangle +{\lambda }^{2}\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant 0\end{eqnarray} \tag{ 91 }$$

for all real λ. The condition that this function of λ is never negative gives the Heisenberg uncertainty principle $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant \tfrac{1}{4}| \langle {\widehat{S}}_{z}\rangle {| }^{2}$ and (89) follows from taking $\lambda =1$ and $\lambda =-1$. Even spin squeezed states with $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ still have $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \geqslant | \langle {\widehat{S}}_{z}\rangle |$, so it is never found that $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt | \langle {\widehat{S}}_{z}\rangle |$ and hence this latter inequality cannot used as a test for entanglement.

Fortunately—as we have seen, showing that spin squeezing occurs via either $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ or $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ is sufficient to establish that the state is an entangled state for modes $\widehat{a},\widehat{b}$, with analogous results if principle spin operators are considered. Applying the Hillery et al entanglement test in equation (84) involving $\tfrac{1}{2}\langle \widehat{N}\rangle$ is also a valid entanglement test, but is usually less stringent than the spin squeezing test involving either $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ or $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$. For the Hillery et al entanglement test to be satisfied at least one of $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle$ or $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle$ is required to be less than $\tfrac{1}{2}\langle \widehat{N}\rangle$, whereas for the spin squeezing test to apply at least one of $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle$ or $\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle$ must be less than $\tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$. The quantity $\tfrac{1}{2}| \langle {\widehat{S}}_{z}\rangle |$ is likely to be smaller than $\tfrac{1}{2}\langle \widehat{N}\rangle$—for example the Bloch vector may lie close to the xy plane, so a greater degree of reduction in spin fluctuation is needed to satisfy the spin squeezing test for entanglement.

However, this is not always the case as the example of the relative phase state discussed in section 3.7 shows. The results in the current section can easily be modified to apply to new spin operators ${\widehat{J}}_{x}$, ${\widehat{J}}_{y}$, ${\widehat{J}}_{z}$, with entanglement being considered for new modes $\widehat{c}$ and $\widehat{d}.$ The Hillery et al [22] entanglement test then becomes

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{J}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{J}}_{y}^{2}\rangle \lt \displaystyle \frac{1}{2}\langle \widehat{N}\rangle .\end{eqnarray} \tag{ 92 }$$

In the case of the relative phase eigenstate we have from equation (69) that $\langle {\rm{\Delta }}{\widehat{J}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{J}}_{y}^{2}\rangle =\tfrac{1}{12}{N}^{2}+\tfrac{1}{4}+\tfrac{1}{8}\mathrm{ln}N\,\approx \tfrac{1}{12}{N}^{2}$ for large N. This clearly exceeds $\tfrac{1}{2}\langle \widehat{N}\rangle =\tfrac{1}{2}N$, so the Hillery et al [22] test for entanglement fails. On the other hand, as we have seen in section 3.7 $\langle {\rm{\Delta }}{\widehat{J}}_{y}^{2}\rangle \lt \tfrac{1}{2}| \langle {\widehat{J}}_{z}\rangle | \approx \tfrac{\pi }{16}N$, so the spin squeezing test is satisfied for this entangled state of modes $\widehat{c}$ and $\widehat{d}$.

There are numerous choices for defining spin operators, but the most useful would be the local spin operators for each well [23] defined by

$$\begin{eqnarray}{\widehat{S}}_{x}^{1} & = & ({\widehat{b}}_{1}^{\dagger }{\widehat{a}}_{1}+{\widehat{a}}_{1}^{\dagger }{\widehat{b}}_{1})/2,\qquad {\widehat{S}}_{y}^{1}=({\widehat{b}}_{1}^{\dagger }{\widehat{a}}_{1}-{\widehat{a}}_{1}^{\dagger }{\widehat{b}}_{1})/2{i},\\ {\widehat{S}}_{z}^{1} & = & ({\widehat{b}}_{1}^{\dagger }{\widehat{b}}_{1}-{\widehat{a}}_{1}^{\dagger }{\widehat{a}}_{1})/2,\\ {\widehat{S}}_{x}^{2} & = & ({\widehat{b}}_{2}^{\dagger }{\widehat{a}}_{2}+{\widehat{a}}_{2}^{\dagger }{\widehat{b}}_{2})/2,\qquad {\widehat{S}}_{y}^{2}=({\widehat{b}}_{2}^{\dagger }{\widehat{a}}_{2}-{\widehat{a}}_{2}^{\dagger }{\widehat{b}}_{2})/2{i},\\ {\widehat{S}}_{z}^{2} & = & ({\widehat{b}}_{2}^{\dagger }{\widehat{b}}_{2}-{\widehat{a}}_{2}^{\dagger }{\widehat{a}}_{2})/2.\end{eqnarray} \tag{ 94 }$$

These satisfy the usual angular momentum commutation rules and those or the different wells commute. The squares of the local vector spin operators are related to the total number operators ${\widehat{N}}_{1}={\widehat{b}}_{1}^{\dagger }{\widehat{b}}_{1}+{\widehat{a}}_{1}^{\dagger }{\widehat{a}}_{1}$ and ${\widehat{N}}_{2}={\widehat{b}}_{2}^{\dagger }{\widehat{b}}_{2}+{\widehat{a}}_{2}^{\dagger }{\widehat{a}}_{2}$ as ${\sum }_{\alpha }{({\widehat{S}}_{\alpha }^{1})}^{2}\,=$ (${\widehat{N}}_{1}/2)({\widehat{N}}_{1}/2+1)$ and ${\sum }_{\alpha }{({\widehat{S}}_{\alpha }^{2})}^{2}\,=({\widehat{N}}_{2}/2)$ $({\widehat{N}}_{2}/2+1)$. The total spin operators are

$$\begin{eqnarray}&&{\widehat{S}}_{\alpha }={\widehat{S}}_{\alpha }^{1}+{\widehat{S}}_{\alpha }^{2},\qquad \alpha =x,y,z\end{eqnarray} \tag{ 95 }$$

and these satisfy the usual angular momentum commutation rules. Hence there may be cases of spin squeezing, but these do not in general provide entanglement tests.

For the local spin operators we have in general

$$\begin{eqnarray}{\langle {\widehat{S}}_{\alpha }^{1}\rangle }_{{ab}(1)} & = & {\rm{Tr}}({\widehat{\rho }}_{R}^{{ab}(1)}\,{\widehat{S}}_{\alpha }^{1})\ne 0,\\ {\langle {\widehat{S}}_{\alpha }^{2}\rangle }_{{ab}(2)} & = & {\rm{Tr}}({\widehat{\rho }}_{R}^{{ab}(2)}\,{\widehat{S}}_{\alpha }^{2})\ne 0,\qquad \alpha =x,y,z\end{eqnarray} \tag{ 96 }$$

based on (93), and applying (19) we see that in general $\langle {\widehat{S}}_{\alpha }\rangle \ne 0$ for separable states. Thus the Bloch vector test for entanglement does not apply.

Furthermore, there is no spin squeezing test either. Following a similar approach as in section 3 we can obtain the following inequalities for separable states of the sub-systems 1 and 2

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle -\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \,\\ &&\quad \geqslant \,\displaystyle \sum _{R}{P}_{R}(\langle {({\rm{\Delta }}{\widehat{S}}_{x}^{1})}^{2}{\rangle }_{R}-\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}^{1}{\rangle }_{R}| )\\ &&\qquad +\displaystyle \sum _{R}{P}_{R}(\langle {({\rm{\Delta }}\widehat{S}{\text{}\text{}}_{x}^{2})}^{2}{\rangle }_{R}-\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}^{2}{\rangle }_{R}| ),\end{eqnarray} \tag{ 97 }$$

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle -\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}\rangle | \,\\ &&\quad \geqslant \,\displaystyle \sum _{R}{P}_{R}(\langle {({\rm{\Delta }}{\widehat{S}}_{y}^{1})}^{2}{\rangle }_{R}-\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}^{1}{\rangle }_{R}| )\\ &&\qquad +\displaystyle \sum _{R}{P}_{R}(\langle {({\rm{\Delta }}{\widehat{S}}_{y}^{2})}^{2}{\rangle }_{R}-\displaystyle \frac{1}{2}| \langle {\widehat{S}}_{z}^{2}{\rangle }_{R}| ).\end{eqnarray} \tag{ 98 }$$

Similar inequalities can be obtained for other pairs of spin operators. In neither case can we state that the right sides are always non-negative, For example, each ${\widehat{\rho }}_{R}^{{ab}(1)}$ may be a spin squeezed state for ${\widehat{S}}_{x}^{1}$ versus ${\widehat{S}}_{y}^{1}$ and each ${\widehat{\rho }}_{R}^{{ab}(2)}$ may be a spin squeezed state for ${\widehat{S}}_{x}^{2}$ versus ${\widehat{S}}_{y}^{2}$. In this case the right side of the first inequality is a negative quantity, so we cannot conclude that the total ${\widehat{S}}_{x}$ is not squeezed versus ${\widehat{S}}_{y}$ for all separable states. As the ${\widehat{\rho }}_{R}^{{ab}(1)}$ and ${\widehat{\rho }}_{R}^{{ab}(2)}$ can be chosen independently we see that separable states for the sub-systems 1 and 2 may be spin squeezed, so the presence of spin squeezing in a total spin operator is not a test for bipartite entanglement in this four mode system. This does not of course preclude tests for bipartite entanglement involving spin operators, as we will now see.

In section 2.8 of paper 1 it was shown that $| \langle {\widehat{{\rm{\Omega }}}}_{A}^{\dagger }{\widehat{{\rm{\Omega }}}}_{B}\rangle {| }^{2}\leqslant \langle {\widehat{{\rm{\Omega }}}}_{A}^{\dagger }\,{\widehat{{\rm{\Omega }}}}_{A}\,{\widehat{{\rm{\Omega }}}}_{B}^{\dagger }\,{\widehat{{\rm{\Omega }}}}_{B}\rangle$ for a non-entangled state of general sub-systems A and B, so with ${\widehat{{\rm{\Omega }}}}_{A}\to {\widehat{S}}_{-}^{1}={\widehat{S}}_{x}^{1}-{\rm{i}}{\widehat{S}}_{y}^{1}$ and ${\widehat{{\rm{\Omega }}}}_{B}\to {\widehat{S}}_{-}^{2}={\widehat{S}}_{x}^{2}-{\rm{i}}{\widehat{S}}_{y}^{2}={({\widehat{S}}_{+}^{2})}^{\dagger }$ to give

$$\begin{eqnarray}&&| \langle {\widehat{S}}_{+}^{1}\,{\widehat{S}}_{-}^{2}\rangle {| }^{2}\leqslant \langle {\widehat{S}}_{+}^{1}\,{\widehat{S}}_{-}^{1}\,{\widehat{S}}_{+}^{2}\,{\widehat{S}}_{-}^{2}\,\rangle \end{eqnarray} \tag{ 99 }$$

for a non-entangled state of sub-systems 1 and 2. For the non-entangled state of these two sub-systems we have

$$\begin{eqnarray}&&\langle {\widehat{S}}_{+}^{1}\,{\widehat{S}}_{-}^{2}\rangle =\displaystyle \sum _{R}{P}_{R}{\langle {\widehat{S}}_{+}^{1}\rangle }_{{ab}(1)}^{R}{\langle {\widehat{S}}_{-}^{2}\rangle }_{{ab}(2)}^{R}\end{eqnarray} \tag{ 100 }$$

which in general is non-zero from equation (96).

Hence a valid entanglement test involving spin operators for sub-systems 1 and 2—each consisting of two modes localized in each well exists, so if

$$\begin{eqnarray}&&| \langle {\widehat{S}}_{+}^{1}\,{\widehat{S}}_{-}^{2}\rangle {| }^{2}\gt \langle {\widehat{S}}_{+}^{1}\,{\widehat{S}}_{-}^{1}\,{\widehat{S}}_{+}^{2}\,{\widehat{S}}_{-}^{2}\,\rangle \end{eqnarray} \tag{ 101 }$$

then the two sub-systems are entangled. A similar conclusion is stated in [23], where the criterion was predicted to be satisfied for four mode two well BEC systems. This test for entanglement involves the local spin operators, though it is not then the same as spin squeezing criteria. It is referred to there as spin entanglement. Other similar tests may be obtained via different choices of ${\widehat{{\rm{\Omega }}}}_{A}$ and ${\widehat{{\rm{\Omega }}}}_{B}$.

In a paper entitled 'many-particle entanglement with Bose–Einstein condensates' Sørensen et al [14] consider the implications for spin squeezing for non-entangled states of the form

$$\begin{eqnarray}&&\widehat{\rho }=\displaystyle \sum _{R}{P}_{R}\,{\widehat{\rho }}_{R}^{1}\otimes {\widehat{\rho }}_{R}^{2}\otimes {\widehat{\rho }}_{R}^{3}\otimes ...,\end{eqnarray} \tag{ 105 }$$

where ${\widehat{\rho }}_{R}^{i}$ is a density operator for particle i. As discussed previously, a density operator of this general form is not consistent with the symmetrization principle—having separate density operators ${\widehat{\rho }}_{R}^{i}$ for specific particles i in an identical particle system (such as for a BEC) is not compatible with the indistinguishability of such particles. It is modes that are distinguishable, not identical particles, so the basis for applying their results to systems of identical bosons is flawed. However, they derive an inequality for the spin variance $\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle$

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \geqslant \displaystyle \frac{1}{N}({\langle {\widehat{S}}_{x}\rangle }^{2}+{\langle {\widehat{S}}_{y}\rangle }^{2})\end{eqnarray} \tag{ 106 }$$

that applies in the case of non-entangled states. Key steps in their derivation are stated in the appendix to [14], but as the justification of these steps is not obvious for completeness the full derivation is given in appendix G of the present paper. This inequality (106) establishes that if

$$\begin{eqnarray}&&{\xi }^{2}=\displaystyle \frac{\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle }{({\langle {\widehat{S}}_{x}\rangle }^{2}+{\langle {\widehat{S}}_{y}\rangle }^{2})}\lt \displaystyle \frac{1}{N}\end{eqnarray} \tag{ 107 }$$

then the state is entangled, so that if the condition for spin squeezing analogous to that in appendix B equation (201) is satisfied, then entanglement is required if spin squeezing for ${\widehat{S}}_{z}$ to occur. Spin squeezing is then a test for entanglement in terms of their definition of an entangled state.

If the Bloch vector is close to the Bloch sphere, for example with $\langle {\widehat{S}}_{x}\rangle =0$ and $\langle {\widehat{S}}_{y}\rangle =N/2$ then the condition (107) is equivalent to

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \lt \displaystyle \frac{1}{2}| \langle {\widehat{S}}_{y}\rangle | \end{eqnarray} \tag{ 108 }$$

which is the condition for squeezing in ${\widehat{S}}_{z}$ compared to ${\widehat{S}}_{x}$. Spin squeezing is then a test for entanglement in terms of their definition of an entangled state. Note that the condition (108) requires the Bloch vector to be in the xy plane and close to the Bloch sphere of radius $N/2$. By comparison with appendix B (201) we see that the Sørensen spin squeezing test is that if there is squeezing in ${\widehat{S}}_{z}$ with respect to any spin component in the xy plane and the Bloch vector is close to the Bloch sphere, then the state is entangled.

As explained above, the proof of Sørensen et al really applies only when the individual spins are distinguishable. It is possible however to modify the work of Sørensen et al [14] to apply to a system of identical bosons in accordance with the symmetrization and SSRs if the index i is re-interpreted as specifying different modes, for example modes localized on optical lattice sites $i=1,2,..,n$ or distinct free space momentum states listed $i=1,2,..,n$. On each lattice site or for each momentum state there would be two modes $a,b$—for example associated with two different internal states—so the sub-system density operator ${\widehat{\rho }}_{R}^{i}$ then applies to the two modes on site i. However the proof of (106) requires the ${\widehat{\rho }}_{R}^{i}$ to be restricted to the case where there is exactly one identical boson on each site or in a momentum state. Such a localization process in position or momentum space has the effect of enabling the identical bosons to be treated as if they are distinguishable. Details are given in appendix H. A similar modification has been carried out by Hyllus et al [27].

However, as we have seen in section 3.2 it does in fact turn out for two mode systems of identical bosons that showing that ${\widehat{S}}_{z}$ is spin squeezed compared to ${\widehat{S}}_{x}$ or ${\widehat{S}}_{y}$ is sufficient to prove that the quantum state is entangled. There are no restrictions either on the mean number of bosons occupying each mode. The proof is based on applying the requirement of local particle number SSR compliance to the separable states in the present case of massive bosons and treating modes (not particles) as sub-systems. In appendix D we have also shown that the same result applies to multi-mode situations in cases where the sub-systems are all single modes (Case 2) or where there are two sub-systems each containing all modes for a single component (Case 1). So the spin squeezing test is still valid for many particle BEC, though the justification is not as in the proof of Sørensen et al [14] (which was derived for systems of distinguishable particles), with each sub-system being a single two state particle.

For a non-entangled state based on SSR compliant ${\widehat{\rho }}_{R}^{A},{\widehat{\rho }}_{R}^{B}$ for modes $\widehat{a}$ and $\widehat{b}$ where the SSR is satisfied we have with ${\widehat{{\rm{\Omega }}}}_{A}={(\widehat{a})}^{m}$ and ${\widehat{{\rm{\Omega }}}}_{B}={(\widehat{b})}^{n}$

$$\begin{eqnarray}\langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle & = & \displaystyle \sum _{R}{P}_{R}{\langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle }_{R}\\ & = & \displaystyle \sum _{R}{P}_{R}{\langle {(\widehat{a})}^{m}\rangle }_{R}{\langle {({\widehat{b}}^{\dagger })}^{n}\rangle }_{R}=0\end{eqnarray} \tag{ 119 }$$

since from equations analogous to (16) ${\langle {(\widehat{a})}^{m}\rangle }_{R}\,={\langle {({\widehat{b}}^{\dagger })}^{n}\rangle }_{R}=0$. Hence for a SSR compliant non-entangled state as defined in the present paper the inequality becomes

$$\begin{eqnarray}&&0\leqslant \langle {({\widehat{a}}^{\dagger })}^{m}{(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}{(\widehat{b})}^{n}\rangle \end{eqnarray} \tag{ 120 }$$

which is trivially true and applies for any state, entangled or not.

Since $\langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle$ is zero for non-entangled states it follows that it is merely necessary to show that this quantity is non-zero to establish that the state is entangled. Hence an entanglement test [2] in the case of sub-systems consisting of single modes $\widehat{a}$ and $\widehat{b}$ becomes

$$\begin{eqnarray}&&| \langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle {| }^{2}\gt 0\end{eqnarray} \tag{ 121 }$$

for a non-entangled state based on SSR compliant ${\widehat{\rho }}_{R}^{A},{\widehat{\rho }}_{R}^{B}$. Note that for globally compliant states $\langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle =0$ unless n = m, so only that case is of interest. This is a useful weak correlation test for entanglement in terms of the definition of entanglement in the present paper. A related but different test is that of Hillery et al [22]—discussed in the next subsection.

For the case where $n=m=1$ the weak correlation test is

$$\begin{eqnarray}&&| \langle (\widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}\gt 0\end{eqnarray} \tag{ 122 }$$

which is equivalent to $\langle {\widehat{S}}_{x}\rangle \ne 0$ and/or $\langle {\widehat{S}}_{y}\rangle \ne 0$, the Bloch vector test.

In a later paper entitled 'detecting entanglement with non-Hermitian operators' Hillery et al [32] apply other inequalities for determining entanglement derived in the earlier paper [22] but now also to systems of massive identical bosons, while still retaining density operators ${\widehat{\rho }}_{R}^{A}$, ${\widehat{\rho }}_{R}^{B}$ that contain coherences between states with differing boson numbers. In particular, for a non-entangled state the following family of inequalities—originally derived in [22], is invoked.

$$\begin{eqnarray}&&| \langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle {| }^{2}\leqslant \langle {({\widehat{a}}^{\dagger })}^{m}{(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}{(\widehat{b})}^{n}\rangle .\end{eqnarray} \tag{ 123 }$$

This is just a special case of (118) with ${\widehat{{\rm{\Omega }}}}_{A}={(\widehat{a})}^{m}$ and ${\widehat{{\rm{\Omega }}}}_{B}={(\widehat{b})}^{n}$. Thus if

$$\begin{eqnarray}&&| \langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle {| }^{2}\gt \langle {({\widehat{a}}^{\dagger })}^{m}{(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}{(\widehat{b})}^{n}\rangle \end{eqnarray} \tag{ 124 }$$

then the state is entangled. The Hillery et al [22] entanglement test (124) is a valid test for entanglement and is actually a more stringent test than merely showing that $| \langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle {| }^{2}\gt 0$, since the quantity $| \langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle {| }^{2}$ is now required to be larger. In a paper by He et al [23] (see section 5.3) the Hillery et al [22] entanglement test $| \langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle {| }^{2}\gt \langle {({\widehat{a}}^{\dagger })}^{m}{(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}{(\widehat{b})}^{n}\rangle$ is applied for the case where A and B each consist of one mode localized in each well of a double well potential. This test while applicable could be replaced by the more easily satisfied test $| \langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle {| }^{2}\gt 0$ (see (121)). However, as will be seen below in section 5.3, the Hillery et al [22] entanglement criterion is needed if the sub-systems each consist of pairs of modes, as treated in [23, 25].

Note that if $n\ne m$ the left side is zero for states that are globally SSR compliant. In this case we can always substitute for two mode systems

$$\begin{eqnarray}{(\widehat{a}{\widehat{b}}^{\dagger })}^{n} & = & {({\widehat{S}}_{x}-{\rm{i}}{\widehat{S}}_{y})}^{n},\\ {({\widehat{a}}^{\dagger })}^{n}{(\widehat{a})}^{n} & = & {P}_{n}{({\widehat{a}}^{\dagger }\widehat{a})}^{}={P}_{n}{\left(\displaystyle \frac{\widehat{N}}{2}-{\widehat{S}}_{z}\right)}^{},\\ {({\widehat{b}}^{\dagger })}^{n}{(\widehat{b})}^{n} & = & {P}_{n}{({\widehat{b}}^{\dagger }\widehat{b})}^{}={P}_{n}{\left(\displaystyle \frac{\widehat{N}}{2}+{\widehat{S}}_{z}\right)}^{}\end{eqnarray} \tag{ 125 }$$

(where P_n is a polynomial of order n) to write both the Hillery and the weak correlation test in terms of spin operators.

A particular case for $n=m=1$ is the test $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}\gt \langle {\widehat{n}}_{a}\,{\widehat{n}}_{b}\rangle$ for an entangled state. To put this result in context, for a general quantum state and any operator $\widehat{{\rm{\Omega }}}$ we have $\langle {\widehat{{\rm{\Omega }}}}^{\dagger }\rangle ={\langle \widehat{{\rm{\Omega }}}\rangle }^{* }$ and $\langle ({\widehat{{\rm{\Omega }}}}^{\dagger }-\langle {\widehat{{\rm{\Omega }}}}^{\dagger }\rangle )(\widehat{{\rm{\Omega }}}-\langle \widehat{{\rm{\Omega }}}\rangle )\rangle \geqslant 0$, hence leading to the Schwarz inequality $| \langle \widehat{{\rm{\Omega }}}\rangle {| }^{2}\,=| \langle {\widehat{{\rm{\Omega }}}}^{\dagger }\rangle {| }^{2}\,\leqslant \langle {\widehat{{\rm{\Omega }}}}^{\dagger }\,\widehat{{\rm{\Omega }}}\rangle$. Taking $\widehat{{\rm{\Omega }}}=\widehat{a}\,{\widehat{b}}^{\dagger }$ leads to the inequality $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}\leqslant \langle {\widehat{n}}_{a}({\widehat{n}}_{b}+1)\rangle$, while choosing $\widehat{{\rm{\Omega }}}=\widehat{b}\,{\widehat{a}}^{\dagger }$ leads to the inequality $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}\leqslant \langle ({\widehat{n}}_{a}+1){\widehat{n}}_{b}\rangle$ for all quantum states. In both cases the right side of the inequality is greater than $\langle {\widehat{n}}_{a}\,{\widehat{n}}_{b}\rangle$, so if it was found that $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}\gt \langle {\widehat{n}}_{a}\,{\widehat{n}}_{b}\rangle$ (though of course still $\leqslant \,\langle {\widehat{n}}_{a}({\widehat{n}}_{b}+1)\rangle$ and $\leqslant \,\langle ({\widehat{n}}_{a}+1){\widehat{n}}_{b}\rangle$) then it could be concluded that the state was entangled. However, as we will see the left side $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}$ actually works out to be zero if physical states for ${\widehat{\rho }}_{R}^{A}$, ${\widehat{\rho }}_{R}^{B}$ are involved in defining non-entangled states, so that for a non-entangled state defined as in the present paper the true inequality replacing $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}\leqslant \langle {\widehat{n}}_{a}\,{\widehat{n}}_{b}\rangle$ is just $0\leqslant \langle {\widehat{n}}_{a}\,{\widehat{n}}_{b}\rangle$, which is trivially true for any quantum state.

For the case where $n=m=1$ we can write the test (124) in terms of spin operators using $\widehat{a}\,{\widehat{b}}^{\dagger }={\widehat{S}}_{x}-{\rm{i}}{\widehat{S}}_{y}$ as

$$\begin{eqnarray}&&{\langle {\widehat{S}}_{x}\rangle }_{\rho }^{2}+{\langle {\widehat{S}}_{y}\rangle }_{\rho }^{2}\gt \displaystyle \frac{1}{4}{\langle {\widehat{N}}^{2}\rangle }_{\rho }-{\langle {\widehat{S}}_{z}^{2}\rangle }_{\rho }\end{eqnarray} \tag{ 126 }$$

which when combined with the general result ${\widehat{S}}_{x}^{2}+{\widehat{S}}_{y}^{2}\,+{\widehat{S}}_{z}^{2}=(\widehat{N}/2)(\widehat{N}/2+1)$ leads to the test

$$\begin{eqnarray}&&{\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle }_{\rho }+{\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle }_{\rho }\lt \displaystyle \frac{1}{2}{\langle \widehat{N}\rangle }_{\rho }.\end{eqnarray} \tag{ 127 }$$

This is the same as the Hillery spin variance test (84), so the Hillery first order correlation test does not add a further test for demonstrating non-SSR compliant entanglement. The Hillery correlation test for n = 2 leads to complex conditions involving higher powers of spin operators.

As an example of applying these tests consider the mixed two mode coherent states described in appendix K, whose density operator for the two mode $\widehat{a}$, $\widehat{b}$ system is given in equation (326). We can now examine the Hillery et al [32] entanglement test in equation (124) and the entanglement test in equation (121) for the case where $m=n=1$. It is straight-forward to show that

$$\begin{eqnarray}| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2} & = & | \alpha {| }^{4},\\ \langle ({\widehat{a}}^{\dagger }\widehat{a})({\widehat{b}}^{\dagger }\widehat{b})\rangle & = & | \alpha {| }^{4}\end{eqnarray} \tag{ 128 }$$

so that $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}=\langle ({\widehat{a}}^{\dagger }\widehat{a})({\widehat{b}}^{\dagger }\widehat{b})\rangle$. A non-entangled state defined in terms of the SSR requirement for the separate modes satisfies $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}=0$, while for a non-entangled state in which the SSR requirement for separate modes is not specifically required merely satisfies $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}\,\leqslant \langle ({\widehat{a}}^{\dagger }\widehat{a})({\widehat{b}}^{\dagger }\widehat{b})\rangle$. Hence the test for entanglement of modes A, B in the present paper $| \langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle {| }^{2}\gt 0$ is satisfied, while the Hillery et al [32] test $| \langle \widehat{a}{\widehat{b}}^{\dagger }\rangle {| }^{2}\gt \langle ({\widehat{a}}^{\dagger }\widehat{a})({\widehat{b}}^{\dagger }\widehat{b})\rangle$ is not.

In terms of the definition of non-entangled states in the present paper, the mixture of two mode coherent states given in equation (326) is not a separable state, but is an entangled state. As discussed in paper 1 (see section 3.4.3) this is because a coherent state gives rise to a non-zero coherence ($\langle \widehat{a}\rangle \ne 0$) and thus cannot represent a physical state for the SSR compliant states involving identical massive bosons (as in BECs). However, in terms of the definition of non-entangled states in other papers such as those of Hillery et al [22, 32] the mixture of two mode coherent states would be a non-entangled state. It is thus a useful state for providing an example of the different outcomes of definitions where the local SSR is applied or not.

A further example of applying correlation tests is provided by the NOON state defined in (59) where here we consider modes A, B. All matrix elements of the form $\langle {({\widehat{a}}^{\dagger })}^{m}{(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}{(\widehat{b})}^{n}\rangle$ are zero for all $m,n$ because both terms contain one mode with zero bosons. Matrix elements of the form $\langle {(\widehat{a})}^{m}{({\widehat{b}}^{\dagger })}^{n}\rangle$ are all zero unless $m=n=N$ and in this case

$$\begin{eqnarray}\langle {(\widehat{a})}^{N}{({\widehat{b}}^{\dagger })}^{N}\rangle & = & \langle {({\widehat{S}}_{x}-{\rm{i}}{\widehat{S}}_{y})}^{N}\rangle \\ & = & \cos \theta \,\sin \theta \langle N,0| {({\widehat{S}}_{-})}^{N}| 0,N\rangle \\ & = & \cos \theta \,\sin \theta \,\sqrt{N}\sqrt{2N-2}\\ & & \times \sqrt{3N-6}\sqrt{4N-12}...\sqrt{N}\end{eqnarray} \tag{ 129 }$$

which is non-zero in general. Hence $| \langle {(\widehat{a})}^{N}{({\widehat{b}}^{\dagger })}^{N}\rangle {| }^{2}\gt 0$ as required for both the weak and strong correlation tests, confirming that the NOON state is entangled. Carrying out this entanglement test experimentally for large N would involve measuring expectation values of high powers of the spin operators ${\widehat{S}}_{x}$ and ${\widehat{S}}_{y}$, which is difficult at present.

A further inequality aimed at providing a signature for entanglement is set out in the papers by Duan et al [33], Toth et al [34]. Duan et al [33] considered a general situation where the system consisted of two distinguishable sub-systems A and B, for which position and momentum Hermitian operators ${\widehat{x}}_{A},{\widehat{p}}_{A}$ and ${\widehat{x}}_{B},{\widehat{p}}_{B}$ were involved that satisfied the standard commutation rules $[{\widehat{x}}_{A},{\widehat{p}}_{A}]=[{\widehat{x}}_{B},{\widehat{p}}_{B}]=i$ in units where ${\hslash }=1$. These sub-systems were quite general and could be two distinguishable quantum particles A and B, but other situations can also be treated. An inequality was obtained for a two sub-system non-entangled state involving the variances for the commuting observables ${\widehat{x}}_{A}+{\widehat{x}}_{B}$ and ${\widehat{p}}_{A}-{\widehat{p}}_{B}$

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle \geqslant 2\end{eqnarray} \tag{ 132 }$$

which could be used to establish a variance test for entangled states of the A and B sub-systems, so that if

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle \lt 2\end{eqnarray} \tag{ 133 }$$

then the sub-systems are entangled. For the case of distinguishable particles such states are possible—consider for example any simultaneous eigenstate of the commuting observables ${\widehat{x}}_{A}+{\widehat{x}}_{B}$ and ${\widehat{p}}_{A}-{\widehat{p}}_{B}$. For such a state $\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle$ and $\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle$ are both zero, so the simultaneous eigenstates are entangled states of particles A, B. For simplicity we only set out the case for which a = 1 in [33]. The proof given in [33] considered separable states of the general form as in equation (15) for two sub-systems but where ${\widehat{\rho }}_{R}^{A}$ and ${\widehat{\rho }}_{R}^{B}$ are possible states for sub-systems A, B. Consequently, a first quantization case involving one particle states could be involved, where SSRs were not relevant. As explained in the introduction, the two distinguishable quantum particles are each equivalent to a whole set of single particle states (momentum eigenstates, harmonic oscillator states, ..) that each quantum particle can occupy, and because both ${\widehat{\rho }}_{R}^{A}$ and ${\widehat{\rho }}_{R}^{B}$ represent states for one particle we have $[{\widehat{n}}_{A},{\widehat{\rho }}_{R}^{A}]=[{\widehat{n}}_{B},{\widehat{\rho }}_{R}^{B}]=0$. Because $\widehat{\rho }$ represent a state for the two particles $[{\widehat{n}}_{A}+{\widehat{n}}_{B},\widehat{\rho }]=0$, the SSR are still true, though irrelevant in the case of distinguishable quantum particles A and B.

Another inequality that can be established is

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{x}}_{A}-{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}+{\widehat{p}}_{B})}^{2}\rangle \geqslant 2\end{eqnarray} \tag{ 134 }$$

which could also be used to establish a variance test for entangled states of the A and B sub-systems, so that if

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{x}}_{A}-{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}+{\widehat{p}}_{B})}^{2}\rangle \lt 2\end{eqnarray} \tag{ 135 }$$

then the sub-systems are entangled.

However, we can also consider cases of systems of identical bosons with two modes A, B rather than two distinguishable quantum particles A and B. In this case both the sub-systems may involve arbitrary numbers of particles, so it is of interest to see what implications follow from the physical sub-system states ${\widehat{\rho }}_{R}^{A}$ and ${\widehat{\rho }}_{R}^{B}$ now being required to satisfy the local particle number SSR, and all quantum states $\widehat{\rho }$ satisfying the global particle number SSR. It is well-known that in two mode boson systems quadrature operators can be defined via

$$\begin{eqnarray}{\widehat{x}}_{A} & = & \displaystyle \frac{1}{\sqrt{2}}(\widehat{a}+{\widehat{a}}^{\dagger }),\qquad {\widehat{p}}_{A}=\displaystyle \frac{1}{\sqrt{2}{i}}(\widehat{a}-{\widehat{a}}^{\dagger }),\\ {\widehat{x}}_{B} & = & \displaystyle \frac{1}{\sqrt{2}}(\widehat{b}+{\widehat{b}}^{\dagger }),\qquad {\widehat{p}}_{B}=\displaystyle \frac{1}{\sqrt{2}{i}}(\widehat{b}-{\widehat{b}}^{\dagger })\end{eqnarray} \tag{ 136 }$$

which have the same commutation rules as the position and momentum operators for distinguishable particles. Thus $[{\widehat{x}}_{A},{\widehat{p}}_{A}]=[{\widehat{x}}_{B},{\widehat{p}}_{B}]=i$ as for cases where A, B were distinguishable particles.

Since the proof of equation (132) in [33] did not involve invoking the SSR, then if the inequality in equation (133) is satisfied, then the state would be an entangled state for modes A, B as well as for distinguishable particles A, B. The situation would then be similar to that for the Hillery et al [22, 32] tests—the SSR compliant sub-system states are just a particular case of the set of all sub-system states. However, in regard to spin squeezing and correlation tests for entanglement, new tests were found when the SSR were explicitly considered and it is possible that this could occur here. This turns out not to be the case.

As we will see, the inequality (132) is replaced by an equation that is satisfied by all quantum states for two mode systems of identical bosons where the global particle number SSR applies. This equation is the same irrespective of whether the state is separable or entangled. To see this we evaluate $\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle$ for states that are global SSR compliant.

Firstly,

$$\begin{eqnarray}&&\langle ({\widehat{x}}_{A}+{\widehat{x}}_{B})\rangle =\langle ({\widehat{p}}_{A}-{\widehat{p}}_{B})\rangle =0\end{eqnarray} \tag{ 137 }$$

since $\langle \widehat{a}\rangle =\langle \widehat{b}\rangle =\langle {\widehat{a}}^{\dagger }\rangle =\langle {\widehat{b}}^{\dagger }\rangle =0$ for SSR compliant states.

Secondly,

$$\begin{eqnarray*}&&\langle {({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle =\displaystyle \frac{1}{2}\left(\begin{array}{c}\langle \widehat{a}\,{\widehat{a}}^{\dagger }\rangle +\langle \widehat{a}\,{\widehat{b}}^{\dagger }\rangle +\langle \widehat{b}\,{\widehat{a}}^{\dagger }\rangle +\langle \widehat{b}\,{\widehat{b}}^{\dagger }\rangle \\ +\langle {\widehat{a}}^{\dagger }\,\widehat{a}\rangle +\langle {\widehat{a}}^{\dagger }\,\widehat{b}\rangle +\langle {\widehat{b}}^{\dagger }\,\widehat{a}\rangle +\langle {\widehat{b}}^{\dagger }\,\widehat{b}\rangle \end{array}\right)\end{eqnarray*}$$

using $\langle {\widehat{a}}^{2}\rangle =\langle {({\widehat{a}}^{\dagger })}^{2}\rangle =\langle {\widehat{b}}^{2}\rangle =\langle {({\widehat{b}}^{\dagger })}^{2}\rangle =\langle \widehat{a}\,\widehat{b}\rangle =\langle {\widehat{a}}^{\dagger }\,{\widehat{b}}^{\dagger }\rangle =0$ for global SSR compliant states. Hence using the commutation rules, introducing the number operator $\widehat{N}$ and the spin operator ${\widehat{S}}_{x}$ and using (137) we find that

$$\begin{eqnarray}\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle & = & \langle {({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle \\ & = & 1+\langle {\widehat{a}}^{\dagger }\,\widehat{a}\rangle +\langle {\widehat{b}}^{\dagger }\,\widehat{b}\rangle +\langle {\widehat{b}}^{\dagger }\,\widehat{a}\rangle +\langle {\widehat{a}}^{\dagger }\,\widehat{b}\rangle \\ & = & 1+\langle \widehat{N}\rangle +2\langle {\widehat{S}}_{x}\rangle .\end{eqnarray} \tag{ 138 }$$

Similarly

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle =1+\langle \widehat{N}\rangle -2\langle {\widehat{S}}_{x}\rangle .\end{eqnarray} \tag{ 139 }$$

Thus we have for all globally SSR compliant states

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle =2+2\langle \widehat{N}\rangle .\end{eqnarray} \tag{ 140 }$$

Since $\langle \widehat{N}\rangle \geqslant 0$ for all quantum states we see that the Duan et al inequality (132) for separable states is still satisfied, but because (140) applies for all states irrespective of whether or not they are separable, we see that there is no quadrature variance entanglement test of the form

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle \lt 2+2\langle \widehat{N}\rangle \end{eqnarray} \tag{ 141 }$$

for the case of two mode systems of identical massive bosons. The situation is similar to the non-existent test $\langle {\rm{\Delta }}{\widehat{S}}_{x}^{2}\rangle +\langle {\rm{\Delta }}{\widehat{S}}_{y}^{2}\rangle \lt | \langle {\widehat{S}}_{z}\rangle | \,$ in section 4.1.3. The situation contrasts that in section 4.3, where a test $\langle {\rm{\Delta }}{({\widehat{S}}_{x}^{1}+{\widehat{S}}_{x}^{2})}^{2}\rangle \,+\langle {\rm{\Delta }}{({\widehat{S}}_{y}^{1}-{\widehat{S}}_{y}^{2})}^{2}\rangle \lt | \langle {\widehat{S}}_{z}\rangle |$ establishes entanglement between two sub-systems (1 and 2)—but in this case each consisting of two modes.

We can also show for all globally SSR compliant states that

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{x}}_{A}-{\widehat{x}}_{B})}^{2}\rangle =1+\langle \widehat{N}\rangle -2\langle {\widehat{S}}_{x}\rangle ,\end{eqnarray} \tag{ 142 }$$

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{p}}_{A}+{\widehat{p}}_{B})}^{2}\rangle =1+\langle \widehat{N}\rangle +2\langle {\widehat{S}}_{x}\rangle \end{eqnarray} \tag{ 143 }$$

and hence

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{({\widehat{x}}_{A}-{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}+{\widehat{p}}_{B})}^{2}\rangle =2+2\langle \widehat{N}\rangle \end{eqnarray} \tag{ 144 }$$

but again no entanglement test results.

The universal result (140) for the quadrature variance sum may seem paradoxical in view of the operators $({\widehat{x}}_{A}+{\widehat{x}}_{B})$ and $({\widehat{p}}_{A}-{\widehat{p}}_{B})$ commuting. Mathematically, this would imply that they would then have a complete set of simultaneous eigenvectors $| {X}_{A,B},{P}_{A,B}\rangle$ such that $({\widehat{x}}_{A}+{\widehat{x}}_{B})| {X}_{A,B},{P}_{A,B}\rangle \,={X}_{A,B}| {X}_{A,B},{P}_{A,B}\rangle$ and $({\widehat{p}}_{A}-{\widehat{p}}_{B})| {X}_{A,B},{P}_{A,B}\rangle ={P}_{A,B}| {X}_{A,B},{P}_{A,B}\rangle$. For these eigenstates $\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle \,=\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle =0$ which contradicts (140) for such states. However, no such eigenstates exist that are globally SSR compliant. For SSR compliant states $| {X}_{A,B},{P}_{A,B}\rangle$ must be an eigenstate of $\widehat{N}$ and for eigenvalue N we see that $({\widehat{x}}_{A}+{\widehat{x}}_{B})$ ${| {X}_{A,B},{P}_{A,B}\rangle }_{N}$ is a linear combination of eigenstates of $\widehat{N}$ with eigenvalues N ± 1. Hence $({\widehat{x}}_{A}+{\widehat{x}}_{B})$ ${| {X}_{A,B},{P}_{A,B}\rangle }_{N}\ne {X}_{A,B}{| {X}_{A,B},{P}_{A,B}\rangle }_{N}$ so simultaneous eigenstates that are SSR compliant do not exist and there is no paradox. As pointed out above, this issue does not arise for the case of two distinguishable particles where the operators ${\widehat{x}}_{A},{\widehat{x}}_{B},{\widehat{p}}_{A}$ and ${\widehat{p}}_{B}$ are not related to mode annihilation and creation operators—as in the present case.

We can also derive inequalities for separable states involving ${\widehat{x}}_{A},{\widehat{p}}_{A}$ and ${\widehat{x}}_{B},{\widehat{p}}_{B}$ based on the approach in section 4.3. Starting with appendix F equation (276) we choose ${\widehat{{\rm{\Omega }}}}_{A}=\text{}{\widehat{x}}_{\text{}\text{}\text{}\text{}A}$, ${\widehat{{\rm{\Omega }}}}_{B}={\widehat{x}}_{\text{}\text{}B}$, ${\widehat{{\rm{\Lambda }}}}_{A}={\widehat{p}}_{A}$ and ${\widehat{{\rm{\Lambda }}}}_{B}={\widehat{p}}_{B}$. Here ${\widehat{{\rm{\Theta }}}}_{A}={\widehat{1}}_{A}$ and ${\widehat{{\rm{\Theta }}}}_{B}={\widehat{1}}_{B}$ For separable states we have from (276)

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{(\alpha {\widehat{x}}_{A}+\beta {\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{(\alpha {\widehat{p}}_{A}-\beta {\widehat{p}}_{B})}^{2}\rangle \geqslant {\alpha }^{2}+{\beta }^{2}.\end{eqnarray} \tag{ 145 }$$

With the choice of ${\alpha }^{2}={\beta }^{2}=1$ we then find the following inequalities for separable states

$$\begin{eqnarray}\langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle & \geqslant & 2,\\ \langle {\rm{\Delta }}{({\widehat{x}}_{A}-{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}+{\widehat{p}}_{B})}^{2}\rangle & \geqslant & 2,\\ \langle {\rm{\Delta }}{({\widehat{x}}_{A}+{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}+{\widehat{p}}_{B})}^{2}\rangle & \geqslant & 2,\\ \langle {\rm{\Delta }}{({\widehat{x}}_{A}-{\widehat{x}}_{B})}^{2}\rangle +\langle {\rm{\Delta }}{({\widehat{p}}_{A}-{\widehat{p}}_{B})}^{2}\rangle & \geqslant & 2\end{eqnarray} \tag{ 146 }$$

depending on the choice of α and β. With $\alpha =\beta =1$ the first result is obtained and is the same as in (132). This result is consistent with (140). However using (138), (143), (142) and (139) we have for global SSR compliant states—separable and non-separable that the left sides of the last set of inequalities are respectively (a) $2+2\langle \widehat{N}\rangle$, (b) $2+2\langle \widehat{N}\rangle$, (c) $2+2\langle \widehat{N}\rangle$ $+\,4\langle {\widehat{S}}_{x}\rangle$ and (d) $2+2\langle \widehat{N}\rangle +4\langle {\widehat{S}}_{x}\rangle$. The implications for the first two equalities have been discussed above. In the case of the (+, +) and (−, −) cases, we note that for states with eigenvalue N for $\widehat{N}$ the eigenvalues for ${\widehat{S}}_{x}$ lie in the range $-N/2$ to $+N/2$ and hence $\langle \widehat{N}\rangle \pm 2\langle {\widehat{S}}_{x}\rangle$ is always $\geqslant \,0$. Thus (146) will apply for both separable and entangled states. Hence for global SSR compliant states none of (146) lead to an entanglement test.

On the other hand if neither the sub-system nor the overall system states are required to be SSR compliant—though they may be— we find that for separable states

$$\begin{eqnarray}&&{\langle {\rm{\Delta }}{({\widehat{x}}_{A}\pm {\widehat{x}}_{B})}^{2}\rangle }_{\rho }+{\langle {\rm{\Delta }}{({\widehat{p}}_{A}\mp {\widehat{p}}_{B})}^{2}\rangle }_{\rho }\geqslant 2+2{\langle \widehat{N}\rangle }_{\rho }\\ &&\quad +2(\langle \widehat{a}\,\widehat{b}\rangle +\langle {\widehat{a}}^{\dagger }\,{\widehat{b}}^{\dagger }\rangle )-2| {\langle \widehat{a}\rangle }_{\rho }+{\langle {\widehat{b}}^{\dagger }\rangle }_{\rho }{| }^{2}\end{eqnarray} \tag{ 147 }$$

so entanglement based on ignoring local particle number SSR in the separable states is now shown if

$$\begin{eqnarray}&&{\langle {\rm{\Delta }}{({\widehat{x}}_{A}\pm {\widehat{x}}_{B})}^{2}\rangle }_{\rho }+{\langle {\rm{\Delta }}{({\widehat{p}}_{A}\mp {\widehat{p}}_{B})}^{2}\rangle }_{\rho }\lt 2+2{\langle \widehat{N}\rangle }_{\rho }\\ &&\quad +2({\langle \widehat{a}\widehat{b}\rangle }_{\rho }+{\langle {\widehat{a}}^{\dagger }{\widehat{b}}^{\dagger }\rangle }_{\rho })-2| {\langle \widehat{a}\rangle }_{\rho }+{\langle {\widehat{b}}^{\dagger }\rangle }_{\rho }{| }^{2}.\end{eqnarray} \tag{ 148 }$$

However, even if local particle number SSR compliance is ignored for the sub-system states (as in [4]), global particle number SSR compliance is still required for the overall quantum state. This applies to both the separable states and to states that are being tested for entanglement. In this case the quantities ${\langle \widehat{a}\widehat{b}\rangle }_{\rho }$, ${\langle {\widehat{a}}^{\dagger }{\widehat{b}}^{\dagger }\rangle }_{\rho }$, ${\langle \widehat{a}\rangle }_{\rho }$, ${\langle {\widehat{a}}^{\dagger }\rangle }_{\rho }$, ${\langle \widehat{b}\rangle }_{\rho }$ and ${\langle {\widehat{b}}^{\dagger }\rangle }_{\rho }$ are all zero, so the entanglement test in (148) would become the same as the hypothetical entanglement test (141).

For the sceptic (see appendix C) who wishes to completely disregard the SSR (both locally and globally) and proposes to use tests based on quadrature variances such as (148) to establish entanglement, the challenge will be to find a way of measuring the allegedly non-zero quantities ${\langle \widehat{a}\widehat{b}\rangle }_{\rho }$, ${\langle {\widehat{b}}^{\dagger }\rangle }_{\rho }$. This would require some sort of system with a well-defined phase reference. Such a measurement is not possible with the beam splitter interferometer discussed in this paper, and the lack of such a detector system would preclude establishing SSR neglected entanglement for systems of identical bosons. Essentially the same problem arises in testing whether states that are non-SSR compliant exist in single mode systems of massive bosons.

As mentioned previously, the result in equation (132) was established in [33] without requiring the sub-system states ${\widehat{\rho }}_{R}^{A}$, ${\widehat{\rho }}_{R}^{B}$ to be compliant with the local particle number SSR or the density operator $\widehat{\rho }$ for the state being tested to comply with the global particle number SSR, as would be the case for physical sub-system and system states of identical bosons. However, in [33] it was pointed out that two mode squeezed vacuum states of the form $| {\rm{\Phi }}\rangle \,=\exp (-r({\widehat{a}}^{\dagger }{\widehat{b}}^{\dagger }-\widehat{a}\widehat{b}))| 0\rangle$ satisfy the entanglement test. However, such stand alone two-mode states are not allowed quantum states for massive identical boson systems. as they are not compliant with the global particle number SSR. To create states with correlated pairs of bosons in modes a and b processes such as the dissociation of a bosonic molecular BEC in a mode M into pair of bosonic atoms in modes a and b can indeed occur, but would involve interaction Hamiltonians such as $\widehat{V}=\kappa ({\widehat{a}}^{\dagger }{\widehat{b}}^{\dagger }\widehat{M}+{\widehat{M}}^{\dagger }\widehat{a}\widehat{b})$. The state produced would be an entangled state of the atoms plus molecules which would be compliant with the global total quanta number SSR—taking into account the boson particle content of the molecule via $\widehat{N}=2{\widehat{n}}_{M}+{\widehat{n}}_{a}+{\widehat{n}}_{b}$. It would not be a state of the form $| {\rm{\Phi }}\rangle =\exp (-r({\widehat{a}}^{\dagger }{\widehat{b}}^{\dagger }-\widehat{a}\widehat{b}))| 0\rangle$.

In the beam splitter case (for state tomography in the xy plane) we choose $2s=\pi /2$ and ϕ (variable) giving

$$\begin{eqnarray}&&{\widehat{M}}_{H}\left(\displaystyle \frac{\pi }{2},\phi \right)=\sin \phi \,{\widehat{S}}_{x}+\cos \phi {\widehat{S}}_{y}\end{eqnarray} \tag{ 172 }$$

and we find that for the output state of the BS interferometer the mean value and variance of $\widehat{M}$ are given by

$$\begin{eqnarray}&&\langle \widehat{M}\rangle =\sin \phi {\langle {\widehat{S}}_{x}\rangle }_{\rho }+\cos \phi {\langle {\widehat{S}}_{y}\rangle }_{\rho },\end{eqnarray} \tag{ 173 }$$

$$\begin{eqnarray}\langle {\rm{\Delta }}{\widehat{M}}^{2}\rangle & = & \displaystyle \frac{1}{2}(1-\cos 2\phi )C({\widehat{S}}_{x},{\widehat{S}}_{x})\\ & & +\displaystyle \frac{1}{2}(1+\cos 2\phi )C({\widehat{S}}_{y},{\widehat{S}}_{y})+\sin 2\phi \,C({\widehat{S}}_{x},{\widehat{S}}_{y})\end{eqnarray} \tag{ 174 }$$

showing the mean value for the measurable $\widehat{M}$ depends sinusoidally on the phase ϕ and the mean values of the spin operators ${\widehat{S}}_{x},{\widehat{S}}_{y}$. The variance for the measurable depends sinusoidally on $2\phi$ and on the covariance matrix of the same spin operators. Both the means and covariances of the spin operators ${\widehat{S}}_{x},{\widehat{S}}_{y}$ now depend on the input state $\widehat{\rho }$ for the interferometer rather than the output state ${\widehat{\rho }}^{\#}$.

For state tomography in the yz plane we obtain the means and covariances of the spin operators ${\widehat{S}}_{y},{\widehat{S}}_{z}$. To do this we choose $2s=\theta$ (variable) and $\phi =0$ so that

$$\begin{eqnarray}&&{\widehat{M}}_{H}(\theta ,0)=\sin \theta \,{\widehat{S}}_{y}+\cos \theta \,{\widehat{S}}_{z}\end{eqnarray} \tag{ 175 }$$

and find that for the output state the mean value and variance of $\widehat{M}$ are given by

$$\begin{eqnarray}&&\langle \widehat{M}\rangle =\sin \theta {\langle {\widehat{S}}_{y}\rangle }_{\rho }+\cos \theta {\langle {\widehat{S}}_{z}\rangle }_{\rho },\end{eqnarray} \tag{ 176 }$$

$$\begin{eqnarray}\langle {\rm{\Delta }}{\widehat{M}}^{2}\rangle & = & \displaystyle \frac{1}{2}(1-\cos 2\theta )C({\widehat{S}}_{y},{\widehat{S}}_{y})\\ & & +\displaystyle \frac{1}{2}(1+\cos 2\theta )C({\widehat{S}}_{z},{\widehat{S}}_{z})+\sin 2\theta \,C({\widehat{S}}_{y},{\widehat{S}}_{z}).\end{eqnarray} \tag{ 177 }$$

A single measurement does not of course determine the mean value $\langle \widehat{M}\rangle$. An average over a large number of independent repetitions of the measurement is needed to accurately determine $\langle \widehat{M}\rangle$ which can then be compared to theoretical predictions. This is a well-known practical issue for the experimenter that we need not dwell on here. A brief account of the issues involved is included in appendix N.

In this case we choose $2s=\theta =\pi$ and ϕ (arbitrary) giving

$$\begin{eqnarray}&&{\widehat{M}}_{H}(\pi ,\phi )=-{\widehat{S}}_{z}\end{eqnarray} \tag{ 178 }$$

and for the output state the mean value and variance of $\widehat{M}$ are given by

$$\begin{eqnarray}&&\langle \widehat{M}\rangle =-{\langle {\widehat{S}}_{z}\rangle }_{\rho },\end{eqnarray} \tag{ 179 }$$

$$\begin{eqnarray}&&\langle {\rm{\Delta }}{\widehat{M}}^{2}\rangle =\langle {\rm{\Delta }}{\widehat{S}}_{z}^{2}\rangle \end{eqnarray} \tag{ 180 }$$

so the phase changer measures the negative of the population difference. Effectively the phase changer interchanges the modes $\widehat{a}\to \widehat{b}$ and $\widehat{b}\to \widehat{a}$ and this is its role rather than being directly involved in a measurement. Phase changers are often included in complex interferometers at the midpoint of free evolution regions to cancel out unwanted causes of phase change.

Another useful choice of measurable is the square of the population difference

$$\begin{eqnarray}&&{\widehat{M}}_{2}={\left(\displaystyle \frac{1}{2}({\widehat{b}}^{\dagger }\widehat{b}-{\widehat{a}}^{\dagger }\widehat{a})\right)}^{2}={\widehat{S}}_{z}^{2}.\end{eqnarray} \tag{ 181 }$$

For the beam splitter case with $2s=\pi /2$ and ϕ (variable), we can easily show (see appendix M) that the mean value of ${\widehat{M}}_{2}$ for the output state is given by

$$\begin{eqnarray}\langle {\widehat{M}}_{2}\rangle & = & {\sin }^{2}\phi \langle {({\widehat{S}}_{x})}^{2}\rangle +{\cos }^{2}\phi \langle {({\widehat{S}}_{y})}^{2}\rangle \\ & & +\sin \phi \,\cos \phi \langle ({\widehat{S}}_{x}{\widehat{S}}_{y}+{\widehat{S}}_{y}{\widehat{S}}_{x})\rangle \end{eqnarray} \tag{ 182 }$$

showing that the mean for the new observable ${\widehat{M}}_{2}$ is a sinusoidal function of the BS interferometer variable ϕ with coefficients that depend on the means of ${\widehat{S}}_{x}^{2}$, ${\widehat{S}}_{y}^{2}$ and ${\widehat{S}}_{x}{\widehat{S}}_{y}+{\widehat{S}}_{y}{\widehat{S}}_{x}$.

Choosing special cases for the interferometer variable yields the following useful results

$$\begin{eqnarray}&&{\langle {\widehat{M}}_{2}\rangle }_{\phi =0}={\langle {({\widehat{S}}_{y})}^{2}\rangle }_{\rho },\qquad {\langle {\widehat{M}}_{2}\rangle }_{\phi =\pi /2}={\langle {({\widehat{S}}_{x})}^{2}\rangle }_{\rho },\\ &&\quad {\langle {\widehat{M}}_{2}\rangle }_{\phi =\pi /4}-{\langle {\widehat{M}}_{2}\rangle }_{\phi =-\pi /4}={\langle ({\widehat{S}}_{x}{\widehat{S}}_{y}+{\widehat{S}}_{y}{\widehat{S}}_{x})\rangle }_{\rho }.\end{eqnarray} \tag{ 183 }$$

Hence all three quantities $\langle {({\widehat{S}}_{x})}^{2}\rangle$ , $\langle {({\widehat{S}}_{y})}^{2}\rangle$ and $\langle ({\widehat{S}}_{x}{\widehat{S}}_{y}+{\widehat{S}}_{y}{\widehat{S}}_{x})\rangle$ can be measured. We note that just measuring $\langle {\widehat{M}}_{2}\rangle$ does not add to the results obtained by measuring the mean and variance of the original measurable $\widehat{M}$, since $\langle {\widehat{M}}_{2}\rangle \,=\langle {\rm{\Delta }}{\widehat{M}}^{2}\rangle +{\langle \widehat{M}\rangle }^{2}$. The variance $\langle {\rm{\Delta }}{\widehat{M}}_{2}^{2}\rangle$ does of course depend on higher moments, for example with $\phi =0$ $\langle {\rm{\Delta }}{\widehat{M}}_{2}^{2}\rangle =\langle {\rm{\Delta }}{({\widehat{S}}_{y}^{2})}^{2}\rangle$ and $\phi =\pi /2$ $\langle {\rm{\Delta }}{\widehat{M}}_{2}^{2}\rangle =\langle {\rm{\Delta }}{({\widehat{S}}_{x}^{2})}^{2}\rangle$, so these also could be measured.

To demonstrate spin squeezing in ${\widehat{S}}_{x}$ with respect to ${\widehat{S}}_{y}$ we need to measure the variances ${\langle {\rm{\Delta }}{\widehat{S}}_{x}{}^{2}\rangle }_{\rho }$ and ${\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }$ and the mean ${\langle {\widehat{S}}_{z}\rangle }_{\rho }$ and show that

$$\begin{eqnarray}&&{\langle {\rm{\Delta }}{\widehat{S}}_{x}{}^{2}\rangle }_{\rho }\lt \displaystyle \frac{1}{2}| {\langle {\widehat{S}}_{z}\rangle }_{\rho }| \qquad {\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }\gt \displaystyle \frac{1}{2}| {\langle {\widehat{S}}_{z}\rangle }_{\rho }| \end{eqnarray} \tag{ 184 }$$

As we have seen, the variances in ${\widehat{S}}_{y}$, ${\widehat{S}}_{x}$ are obtained by measuring the fluctuation in the measurable $\widehat{M}$ after applying the interferometer to the state $\widehat{\rho }$, with the interferometer phase set to $\phi =0$ or $\phi =\pi /2$ for the two cases respectively. The mean ${\langle {\widehat{S}}_{z}\rangle }_{\rho }$ is obtained by a direct measurement of the measurable $\widehat{M}$ without applying the interferometer to the state $\widehat{\rho }$.

As shown in section 3 (see [2]) squeezing in ${\widehat{S}}_{x}$ compared to ${\widehat{S}}_{y}$ or vice versa is sufficient to show that the state is entangled. However, as the interferometer measures the variance for the state $\widehat{\rho }$ in the quantity

$$\begin{eqnarray}&&{\widehat{M}}_{H}(\pi /2,\phi )=\sin \phi \,{\widehat{S}}_{x}+\cos \phi \,{\widehat{S}}_{y}={\widehat{S}}_{x}^{\#}\left(\displaystyle \frac{3\pi }{2}+\phi \right)\end{eqnarray} \tag{ 185 }$$

corresponding to the x component of the spin vector operator $(\underrightarrow{\widehat{S}})$ after it has been rotated about the z axis through an angle $\tfrac{3\pi }{2}+\phi$, it is desirable to extend the entanglement test to consider the squeezing of ${\widehat{S}}_{x}^{\#}\left(\tfrac{3\pi }{2}+\phi \right)$ with respect to the corresponding y component ${\widehat{S}}_{y}^{\#}\left(\tfrac{3\pi }{2}+\phi \right)$—and vice versa, where

$$\begin{eqnarray}&&{\widehat{S}}_{y}^{\#}\left(\displaystyle \frac{3\pi }{2}+\phi \right)=-\cos \phi \,{\widehat{S}}_{x}+\sin \phi \,{\widehat{S}}_{y}.\end{eqnarray} \tag{ 186 }$$

The variance in ${\widehat{S}}_{y}^{\#}\left(\tfrac{3\pi }{2}+\phi \right)$ can be obtained by changing the interferometer phase to $\phi +\tfrac{\pi }{2}$. Clearly $\left[{\widehat{S}}_{x}^{\#}\left(\tfrac{3\pi }{2}+\phi \right),{\widehat{S}}_{y}^{\#}\left(\tfrac{3\pi }{2}+\phi \right)\right]={i}\,{\widehat{S}}_{z}$, as before.

The question is—does squeezing in either ${\widehat{S}}_{x}^{\#}\left(\tfrac{3\pi }{2}+\phi \right)$ or ${\widehat{S}}_{y}^{\#}\left(\tfrac{3\pi }{2}+\phi \right)$ demonstrate entanglement of the modes $\widehat{a}$ and $\widehat{b}$? The answer is that it does. The proof is set out in section M.4 of appendix M.

The question remaining is whether the mean value ${\langle {\widehat{S}}_{z}\rangle }_{\rho }$ can be measured accurately enough to apply the test for entanglement. With an infinite number of repeated measurements this is always possible, since then both the variances ${\langle {\rm{\Delta }}{\widehat{S}}_{x}^{\#}{\left(\tfrac{3\pi }{2}+\phi \right)}^{2}\rangle }_{\rho }$ and the mean ${\langle {\widehat{S}}_{z}\rangle }_{\rho }$. would become error free. For a finite number of measurements R the measurement of ${\langle {\widehat{S}}_{z}\rangle }_{\rho }$ requires a consideration of the variance in ${\widehat{S}}_{z}$. For general entangled states general considerations indicate that this mean will be of order N and the variance will be at worst of order N², Hence the variance ${\langle {\rm{\Delta }}{\langle \widehat{{S}_{z}}\rangle }^{2}\rangle }_{{\rm{sample}}}$ in determining the mean $\langle {\widehat{S}}_{z}\rangle$ for R repetitions of the measurement would be $\sim {N}^{2}/R$, giving a fluctuation of $\sim N/\sqrt{R}$. For this to be small compared to $\sim N$ we merely require $R\gg 1$, which is not unexpected. This result indicates that the application of the spin squeezing test via interferometric measurement of both the variances ${\langle {\rm{\Delta }}{\widehat{S}}_{x}^{\#}{\left(\tfrac{3\pi }{2}+\phi \right)}^{2}\rangle }_{\rho }$ and the mean ${\langle {\widehat{S}}_{z}\rangle }_{\rho }$ looks feasible.

To demonstrate spin squeezing in ${\widehat{S}}_{z}$ with respect to ${\widehat{S}}_{y}$ we need to measure the variances ${\langle {\rm{\Delta }}{\widehat{S}}_{z}{}^{2}\rangle }_{\rho }$ and ${\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }$ and the mean ${\langle {\widehat{S}}_{x}\rangle }_{\rho }$ and show that

$$\begin{eqnarray}&&{\langle {\rm{\Delta }}{\widehat{S}}_{z}{}^{2}\rangle }_{\rho }\lt \displaystyle \frac{1}{2}| {\langle {\widehat{S}}_{x}\rangle }_{\rho }| ,\qquad {\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }\gt \displaystyle \frac{1}{2}| {\langle {\widehat{S}}_{x}\rangle }_{\rho }| .\end{eqnarray} \tag{ 187 }$$

As we have seen, the variances in ${\widehat{S}}_{z}$, ${\widehat{S}}_{y}$ are obtained by measuring the fluctuation in the measurable $\widehat{M}$ after applying the interferometer to the state $\widehat{\rho }$, with the interferometer phase set to $\phi =0$ and the pulse area $2s=\theta$ made variable. From equation (177) we see that choosing $\theta =0$ gives ${\langle {\rm{\Delta }}{\widehat{S}}_{z}{}^{2}\rangle }_{\rho }$ and choosing $\theta =\tfrac{\pi }{2}$ gives ${\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }$. From equation (173) the mean ${\langle {\widehat{S}}_{x}\rangle }_{\rho }$ is obtained by a measurement of the mean in the measurable $\widehat{M}$ after applying the interferometer to the state $\widehat{\rho }$, with the interferometer phase set to $\phi =\pi /2$ and the pulse area $2s=\pi /2$.

Unless stated otherwise, we again focus on correlation tests for SSR compliant entanglement. For the beam splitter case and by choosing the phase $\phi =0$ we see that $\langle \widehat{M}\rangle ={\langle {\widehat{S}}_{y}\rangle }_{\rho }$ and by choosing the phase $\phi =\pi /2$ we see that $\langle \widehat{M}\rangle ={\langle {\widehat{S}}_{x}\rangle }_{\rho }$. The simplest form of the correlation test with $n=m=1$ requires

$$\begin{eqnarray}&&{\langle {\widehat{S}}_{x}\rangle }_{\rho }\ne 0,\qquad {\langle {\widehat{S}}_{y}\rangle }_{\rho }\ne 0\end{eqnarray} \tag{ 188 }$$

for establishing that the state is entangled. For the separable state with ${\widehat{M}}_{H}=\sin \phi \,{\widehat{S}}_{x}+\cos \phi \,{\widehat{S}}_{y}={\widehat{S}}_{x}^{\#}\left(\tfrac{3\pi }{2}+\phi \right)$

$$\begin{eqnarray}&&\langle \widehat{M}\rangle ={\langle {\widehat{M}}_{H}\rangle }_{\rho }=0\end{eqnarray} \tag{ 189 }$$

so that the mean value of the measurable is zero and independent of the beam splitter phase ϕ for all ϕ. Finding any non-zero value for ${\langle {\widehat{M}}_{H}\rangle }_{\rho }$ would then show that the state $\widehat{\rho }$ is entangled. More importantly from the general result, ${\langle {\widehat{M}}_{H}\rangle }_{\rho }$ would show a sinusoidal dependence on the interferometer phase ϕ, so the appearance of such a dependence would be an indication that the state was entangled.

The question remaining is whether the mean values ${\langle {\widehat{S}}_{x,y}\rangle }_{\rho }$ can be measured accurately enough to apply the test for entanglement. With an infinite number of repeated measurements this is always possible, since then both the variances ${\langle {\rm{\Delta }}{\widehat{S}}_{x,y}^{2}\rangle }_{\rho }$ and the means ${\langle {\widehat{S}}_{x,y}\rangle }_{\rho }$. would become error free. For a finite number of measurements the measurement of ${\langle {\widehat{S}}_{x,y}\rangle }_{\rho }$ requires a consideration of the variances in ${\widehat{S}}_{x,y}$ (or ${\widehat{M}}_{H}$ to cover both cases). For a general entangled state we can assume that ${\langle {\widehat{M}}_{H}\rangle }_{\rho }\sim N/2$ and the variance will be at worst of order N², Hence the variance ${\langle {\rm{\Delta }}{\langle {\widehat{M}}_{H}\rangle }^{2}\rangle }_{{\rm{sample}}}$ in determining the mean $\langle {\widehat{S}}_{x,y}\rangle$ for R repetitions of the measurement would be $\sim {N}^{2}/R$, giving a fluctuation of $\sim N/\sqrt{R}$. For this to be small compared to $\sim N$ we merely require $R\gg 1$, which is not unexpected. This result indicates that the application of the simple correlation test via interferometric measurement of ${\langle {\widehat{S}}_{x}\rangle }_{\rho }$ and ${\langle {\widehat{S}}_{y}\rangle }_{\rho }$ looks feasible.

For the second order form of the correlation test with $n=m=2$ requires

$$\begin{eqnarray}{\langle {\rm{\Delta }}{\widehat{S}}_{x}{}^{2}\rangle }_{\rho }+{\langle {\widehat{S}}_{x}\rangle }_{\rho }^{2}-{\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }-{\langle {\widehat{S}}_{y}\rangle }_{\rho }^{2} & \ne & 0\\ (C({\widehat{S}}_{x},{\widehat{S}}_{y})+{\langle {\widehat{S}}_{x}\rangle }_{\rho }{\langle {\widehat{S}}_{y}\rangle }_{\rho }) & \ne & 0.\end{eqnarray} \tag{ 190 }$$

We have already shown using equations (173) and (174) that the variances ${\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }$ and ${\langle {\rm{\Delta }}{\widehat{S}}_{x}{}^{2}\rangle }_{\rho }$ and the means ${\langle {\widehat{S}}_{y}\rangle }_{\rho }$ and ${\langle {\widehat{S}}_{x}\rangle }_{\rho }$ can be obtained via the BS interferometer from the mean $\langle \widehat{M}\rangle$ and the variance $\langle {\rm{\Delta }}{\widehat{M}}^{2}\rangle$ for the choices of $\phi =0$ and $\phi =\pi /2$. To obtain the covariance matrix element $C({\widehat{S}}_{x},{\widehat{S}}_{y})$ we see that if we choose $\phi =\pi /4$ then $\langle {\rm{\Delta }}{\widehat{M}}^{2}\rangle \,=\tfrac{1}{2}{\langle {\rm{\Delta }}{\widehat{S}}_{x}{}^{2}\rangle }_{\rho }+\tfrac{1}{2}{\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }+\,C({\widehat{S}}_{x},{\widehat{S}}_{y})$, from which the covariance can be measured. Thus the second order correlation test can be applied.

Alternately, if the measurement quantity for the BS interferometer is the square of the population difference then we see from (183) that the mean value of ${\widehat{M}}_{2}$ for certain choices of the BS variable ϕ measures ${\langle {\rm{\Delta }}{\widehat{S}}_{x}{}^{2}\rangle }_{\rho }+{\langle {\widehat{S}}_{x}\rangle }_{\rho }^{2}\,={\langle {\widehat{S}}_{x}{}^{2}\rangle }_{\rho }$, ${\langle {\rm{\Delta }}{\widehat{S}}_{y}{}^{2}\rangle }_{\rho }+{\langle {\widehat{S}}_{y}\rangle }_{\rho }^{2}={\langle {\widehat{S}}_{y}{}^{2}\rangle }_{\rho }$ and $(C({\widehat{S}}_{x},{\widehat{S}}_{y})+{\langle {\widehat{S}}_{x}\rangle }_{\rho }{\langle {\widehat{S}}_{y}\rangle }_{\rho })\,={\langle ({\widehat{S}}_{x}{\widehat{S}}_{y}+{\widehat{S}}_{y}{\widehat{S}}_{x})\rangle }_{\rho }$. The second order correlation test is that if

$$\begin{eqnarray}{\langle {\widehat{S}}_{x}{}^{2}\rangle }_{\rho } & \ne & {\langle {\widehat{S}}_{y}{}^{2}\rangle }_{\rho }\\ {\langle ({\widehat{S}}_{x}{\widehat{S}}_{y}+{\widehat{S}}_{y}{\widehat{S}}_{x})\rangle }_{\rho } & \ne & 0\end{eqnarray} \tag{ 191 }$$

then the state is entangled.