Collision probability of reduced density matrix of a Haar random state

Question

Consider a Haar random state on $n$ qubits. Trace out one qubit from the state and consider the reduced density matrix $\rho$. What is the value of \begin{equation} \mathbb{E}\left[\text{Tr}\left(|{0^{\otimes n-1}}\rangle \langle 0^{\otimes n-1}| \otimes |{0^{\otimes n-1}}\rangle \langle 0^{\otimes n-1}|~\rho \otimes \rho\right )\right], \end{equation} where the expectation is taken over the Haar random state.

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I think it should be close to being $\frac{2}{2^{n-1}}$. Is there any reference/proof that shows this a bit rigorously?

Answer 1

Note that the quantity that you want to compute is $$ \newcommand\ket[1]{\left|#1\right\rangle}\newcommand\bra[1]{\left\langle#1\right|} \begin{align} \mathbb{E}\left[\ket{0^{n-1}}\!\bra{0^{n-1}}\otimes\ket{0^{n-1}}\!\bra{0^{n-1}}\rho\otimes\rho\right]&=\mathbb{E}\left[\bra{0^{n-1},0^{n-1}}\rho\otimes\rho\ket{0^{n-1},0^{n-1}}\right]\\ &= \mathbb{E}\left[\left(\bra{0^{n-1}}\rho\ket{0^{n-1}}\right)^2\right]\,. \end{align} $$ Let us denote $\ket{\psi}$ the Haar-random state that is sampled on $n$ qubits. For an index $i=\left(i_1,\cdots,i_n\right)\in\{0, 1\}^n$, we note $\psi_i$ the coefficient of $\ket{i_1,\cdots,i_n}$ in $\ket{\psi}$. We then have $$\begin{align} \ket{\psi}\!\bra{\psi} ={}& \sum_{i,j\in\{0, 1\}^n}\psi_i\overline{\psi_j}\ket{i}\!\bra{j}\\ ={}&\sum_{i,j\in\{0, 1\}^{n-1}}\psi_{i,0}\overline{\psi_{j,0}}\ket{i,0}\!\bra{j,0}+{}&\\ &\sum_{i,j\in\{0, 1\}^{n-1}}\psi_{i,0}\overline{\psi_{j,1}}\ket{i,0}\!\bra{j,1}+{}\\ &\sum_{i,j\in\{0, 1\}^{n-1}}\psi_{i,1}\overline{\psi_{j,0}}\ket{i,1}\!\bra{j,0}+{}\\ &\sum_{i,j\in\{0, 1\}^{n-1}}\psi_{i,1}\overline{\psi_{j,1}}\ket{i,1}\!\bra{j,1}\,. \end{align}$$ Tracing over the last qubit (of course, the reasoning is the same for any qubit) yields $$\rho = \sum_{i,j\in\{0, 1\}^{n-1}}\psi_{i,0}\overline{\psi_{j,0}}\ket{i}\!\bra{j}+\sum_{i,j\in\{0, 1\}^{n-1}}\psi_{i,1}\overline{\psi_{j,1}}\ket{i}\!\bra{j}\,.$$ We thus have $$\bra{0^{n-1}}\rho\ket{0^{n-1}}=\left|\psi_{0^{n}}\right|^2+\left|\psi_{0^{n-1}, 1}\right|^2\,.$$ Since I'm still not up-to-date on Weingarten calculus despite how useful it is, we'll do it the old way: writing $\psi_i$ as $\frac{X_i+\mathrm{i}Y_i}{\sqrt{\sum_jX_j^2+Y_j^2}}$, with the $\left\{X_j\right\}_j$ and the $\left\{Y_j\right\}_j$ being i.i.d. according to a $\mathcal{N}(0, 1)$ law. In particular, we have $$\bra{0^{n-1}}\rho\ket{0^{n-1}}=\frac{X_{0^n}^2+Y_{0^n}^2+X_{0^{n-1}, 1}^2+Y_{0^{n-1}, 1}^2}{\sum_jX_j^2+Y_j^2}\,.$$ In particular, we can write this quantity as $\frac{A}{A+B}$, with $A\sim\chi^2(4)=\Gamma\left(2,2\right)$ and $B\sim\chi^2\left(2^{n+1}-4\right)=\Gamma\left(2^n-2,2\right)$ being independent. As such it follows a $\beta\left(2, 2^n-2\right)$ law.

Now that we have the distribution of the random variable, we know everything we want to know about it. In particular, its second moment is $$\frac{6}{2^n\left(2^n+1\right)}=\boxed{\frac{3}{2^{n-1}\left(2^n+1\right)}}\,.$$

Answer 2

Weingarten calculus tells us that$^1$ $$ \mathbb E\big[\, \lvert\psi\rangle\langle\psi\rvert \otimes \lvert\psi\rangle\langle\psi\rvert\,\big] = \frac{1}{D(D+1)} \sum (\delta_{ij}\delta_{k\ell} + \delta_{i\ell}\delta_{kj})\, \lvert i,k\rangle\langle j,\ell\rvert\ , $$ where $D$ is the dimension of the space (i.e. here $D=2^n$).

(Intuitively, the first term is the identity and the second a swap, which are indeed the only two permutations, and Weingarten calculus always gives weighted permutations).

If we now trace $m$ qubits (we have to trace them in both components of the tensor product!), we obtain $$ \frac{1}{D(D+1)} \sum (2^{2m}\delta_{ij}\delta_{k\ell} + 2^{m}\delta_{i\ell}\delta_{kj})\, \lvert i,k\rangle\langle j,\ell\rvert\ , $$ where I reuse the same indices $i,j,k,\ell$ for the remaining qubits.

(In a graphical picture, tracing the identity on two copies gives two closed loops, thus $2^{2m}$, and tracing the swap gives one closed loop, thus $2^{m}$.)

Projecting this onto $\lvert0\cdots 0\rangle$ then given the desired quantity, which is $$ \frac{2^{2m}+2^{m}}{2^{n}(2^{n}+1)}\ , $$ or for $m=1$, as asked in the question $$ \frac{6}{2^{n}(2^{n}+1)}\ . $$

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$^1$ This can be easily derived from the formulas on the Wikipedia page under "Explicit expressions for the integrals in the first cases", 2nd equation, by setting $j=\ell=n=q=0$. Or see e.g. (B1) in arXiv:2404.19023 if you just want to copy the formula, but note that there $D^4$ is the dimension.)