Let $A,B$ be Hilbert spaces of dimension $d$. Let $\rho$ be some separable quantum state of the composite system $AB$. Given a maximally entangled state:

$$\vert\phi\rangle = \frac{1}{\sqrt{d}}\sum_{i=1}^d \vert i\rangle_A\vert i\rangle_B.\tag{1}$$

Can one say anything about the maximum and minimum possible value of the overlap:


where we maximize or minimize over all separable $\rho$?

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    $\begingroup$ +1 and welcome to our community! Thank you for contributing your question here, and we hope to see much more of you in the future !!! I've just edited it a bit to improve the formatting, and also to label the equations in case someone wants to refer to them later. $\endgroup$ Aug 23, 2021 at 17:58

1 Answer 1


The minimum overlap is zero and the maximum overlap is $\frac{1}{d}$.

The overlap is a linear function of $\rho$ and the set $S$ of separable states is convex, so the overlap is both minimized and maximized at extreme points. Extreme points of $S$ are the states of the form$^1$ $\rho = \overline\sigma\otimes\tau$. The reason we choose to define $\sigma$ as the complex conjugate of the first factor will become clear shortly. We can compute the overlap as

$$ \begin{align} \langle\phi|\rho|\phi\rangle &= \frac{1}{d}\sum_{i,j=1}^d\langle i|\overline\sigma|j\rangle \langle i|\tau|j\rangle \\ &= \frac{1}{d}\sum_{i,j=1}^d\overline\sigma_{ij}\tau_{ij} \\ &= \frac{1}{d}\mathrm{tr}(\sigma^\dagger\tau)\tag1. \end{align} $$

This immediately provides us with an example demonstrating that the minimum overlap is zero

$$ \rho_{min} = \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}\otimes\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix} = |0\rangle\langle 0|\otimes|1\rangle\langle 1|. $$

Now, $\mathrm{tr}(\sigma^\dagger\tau)$ is the Hilbert-Schmidt inner product of $\sigma$ and $\tau$. By Cauchy-Schwarz inequality, we have

$$ \mathrm{tr}(\sigma^\dagger\tau) \le \sqrt{\mathrm{tr}(\sigma^\dagger\sigma)\,\mathrm{tr}(\tau^\dagger\tau)} = \sqrt{\mathrm{tr}(\sigma^2)\,\mathrm{tr}(\tau^2)}\tag2 $$

i.e. $\mathrm{tr}(\sigma^\dagger\tau)$ is at most the square root of the product of the purities of $\sigma$ and $\tau$. The upper bound is achieved when $\sigma$ is a scalar multiple of $\tau$. However, the unit trace condition on density matrices means that the scalar must be $1$, so $\sigma = \tau$. Moreover, the upper bound in $(2)$ is greatest when the purity is highest, i.e. when $\sigma = \tau$ is a pure state. We conclude that the overlap $\langle\phi|\rho|\phi\rangle$ is maximized when

$$ \rho_{max} = \overline{|\psi\rangle\langle\psi|}\otimes|\psi\rangle\langle\psi|. $$

In this case, from $(1)$ and $(2)$ the overlap is $\frac{1}{d}$.

$^1$ In fact, we can say more. Extreme points of $S$ are of the form $\rho=|a\rangle\langle a|\otimes|b\rangle\langle b|$. The weaker form of $\rho$ is sufficient for our purposes, but note that both $\rho_{min}$ and $\rho_{max}$ do turn out to be pure product states as expected.


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