# Matrix representation of the symmetric subspace for two copies

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

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

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:

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

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.

• 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? Commented Apr 29 at 7:51

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