# What are min and max overlaps of a maximally entangled state with a separable state?

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:

$$\langle\phi\vert\rho\vert\phi\rangle,\tag{2}$$

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

• +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. Aug 23 at 17:58

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.