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Consider a Haar random state on $n$ qubits, and denote it by $|\psi\rangle$. Now consider the following state

$$|\phi\rangle = \frac{1}{\sqrt{k}} \sum_{i=1}^{k} |\phi_{1, i} \rangle \otimes |\phi_{2, i} \rangle,$$

where each $|\phi_{i,k}\rangle$, for each choice of $k$ is a Haar random state over $n/2$ qubits.

What is the expected trace distance between these two ensembles (denoted by $|\psi\rangle$ and $|\phi\rangle$)?

It clearly is very large when $k$ is $1$, but my hope is that it decreases with increasing $k$. How large of a $k$ suffices?

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    $\begingroup$ Should the summand be $\left|\phi_{1,i}\right\rangle\otimes\left|\phi_{2,i}\right\rangle$? $\endgroup$
    – Tristan Nemoz
    Commented Jan 23 at 10:14
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    $\begingroup$ And thinking about it, $|\phi\rangle$ isn't even a well-defined quantum state, is it? Why would its norm be equal to $1$? $\endgroup$
    – Tristan Nemoz
    Commented Jan 23 at 20:29

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Assume each of these states to be independently distributed. The question is too challenging as stated, so let us look at the expected fidelity between the two states, where trace distance is the square root of unity minus the fidelity (but averaging the trace distance will be less than the trace distance of the average, by Jensen's inequality and concavity).

We can express the fidelity using any starting state $|0\rangle$ as $$F=|\langle\psi|\phi\rangle|^2=|\langle 0|U^\dagger \sum_i V_i\otimes W_i|0\rangle|^2/k=\sum_{i,j=1}^k\frac{\langle 0|U^\dagger V_i\otimes W_i|0\rangle\langle 0| V_j^\dagger\otimes W_j^\dagger U|0\rangle}{k}.$$ Integrating this over all $dU$, $\{dV_i\}$, and $\{dW_i\}$ using Haar measures for the special unitary group of the appropriate dimension actually gives some simplifications, because all of the terms with $i\neq j$ vanish. Specifically, consider something like, for $i\neq j$, $$\int V_i |0\rangle\langle 0| V_j^\dagger dV_i=(\int V_i dV_i)|0\rangle\langle 0| V_j^\dagger $$ and note that $\int V_i dV_i=-\int V_i dV_i=0$ because the Haar measure is invariant under multiplication by a global phase. We are thus left with the averaged fidelity $$\bar{F}=\int F dU \prod_{i=1}^k dV_i dW_i=\sum_{i=1}^k \frac{\int dU dV_i dW_i|\langle 0|U^\dagger V_i\otimes W_i|0\rangle|^2}{k}=\sum_{i=1}^k \frac{\int d\psi d\phi_{1,i} d\phi_{2,i}|\langle \psi| (|\phi_{1,i}\otimes |\phi_{2,i})|^2}{k}.$$ This is just the average over $k$ of the results you would have gotten for $k=1$, which are all the same, so the result is unchanged with changing $k$: $$\bar{F}= \int d\psi d\phi_{1} d\phi_{2}|\langle \psi| (|\phi_{1}\otimes |\phi_{2})|^2.$$

Now I did just make this all up on the spot, so perhaps there is an error, but it seems conclusive. Of course one must deal with the state not actually being normalized in this scenario, and doing Haar averages with normalization constants will be much more challenging.

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