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Consider two copies of an $n$ qubit Haar random state, given by:

\begin{equation} \rho = \mathbb{E}_{U \sim \mathsf{Haar}}\left[U |0^n\rangle \langle 0^n| U^{*}\otimes U |0^n\rangle \langle 0^n| U^{*}\right] = \frac{\Pi_{\text{symm}}}{2^n(2^n - 1)}, \end{equation}

where $\Pi_{\text{symm}}$ is the projector onto the symmetric subspace of appropriate dimensions and $\rho$ is over $2n$ qubits. For a particular $x \in \{0, 1\}^n$, I am trying to calculate the quantity:

\begin{equation} p_x = \mathsf{Tr}[|x\rangle \langle x| \otimes |x\rangle \langle x|~ \rho]. \end{equation}

Is there any nice expression of this quantity in terms of $x$? Moreover, is it true that for any choice of $x$,

$$0 \leq p_x \leq \frac{1}{2^n \cdot (2^n - 1)}?$$

I checked for $n=1$, when $\rho = \frac{I + \mathsf{SWAP}}{2}$, and it seemed to hold.

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    $\begingroup$ The way you phrased your question, why isn't the result simply $\langle x, x|\frac{\Pi_{\text{symm}}}{\binom{2^n+1}{2}}|x,x\rangle=\frac{1}{\binom{2^n+1}{2}}$? If the states are Haar-random, whywould this quantity even depend on $x$? Is there something I'm missing here? $\endgroup$
    – Tristan Nemoz
    Commented Apr 29 at 7:51

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TL;DR: No. As pointed out by @Tristan Nemoz in the comments, $p_x=\frac{1}{2^{n-1}(2^n+1)}$ independently of $x$, so for $n\geqslant 2$ the conjectured inequality fails.


The output probabilities $p$ of a Haar-distributed quantum state, like $U|0^n\rangle$, follow the Porter-Thomas distribution \begin{align} f(p)\,dp=(2^n - 1) (1 - p)^{2^n - 2}\,dp\tag1 \end{align} see for example this answer, so \begin{align} p_x&=\mathrm{tr}\left(|x\rangle\langle x|\otimes|x\rangle\langle x|\rho\right)\tag2\\ &=\mathrm{tr}\left(|x\rangle\langle x|\otimes|x\rangle\langle x|\,\mathbb{E}_{U\sim\text{Haar}}[U|0^n\rangle\langle 0^n|U^\dagger\otimes U|0^n\rangle\langle 0^n|U^\dagger]\right)\tag3\\ &=\mathbb{E}_{U\sim\text{Haar}}\left[\left(\mathrm{tr} (|x\rangle\langle x|U|0^n\rangle\langle 0^n|U^\dagger)\right)^2\right]\tag4\\ &=\mathbb{E}_{U\sim\text{Haar}}[|\langle x|U|0^n\rangle|^4]\tag5\\ &=\int_0^1 p^2 f(p) dp \tag6\\ &=(2^n-1)\int_0^1 p^2 (1 - p)^{2^n - 2} dp \tag7\\ &=(2^n-1)\frac{2\,\Gamma(2^n-1)}{\Gamma(2^n+2)}\tag8\\ &=(2^n-1)\frac{2}{(2^n-1)\,2^n\,(2^n+1)}\tag9\\ &=\frac{1}{2^{n-1}\,(2^n+1)}={2^n+1 \choose 2}^{-1}.\tag{10} \end{align} For $n=1$, we have \begin{align} \frac{1}{2^{n-1}\,(2^n+1)}<\frac{1}{2^n\,(2^n-1)}\tag{11} \end{align} but for $n\geqslant 2$ \begin{align} \frac{1}{2^{n-1}\,(2^n+1)}>\frac{1}{2^n\,(2^n-1)}\tag{12} \end{align} which disproves the inequality conjectured in the question.

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    $\begingroup$ Is there any reason to choose a random $x$? Seems like the same calculation would hold for any $x$. $\endgroup$
    – BlackHat18
    Commented Apr 28 at 16:13

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