# Expected trace distance between two types of random ensembles

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?

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

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 $$k$$: $$\bar{F}= \int d\psi d\phi_{1} d\phi_{2}|\langle \psi| (|\phi_{1}\otimes |\phi_{2})|^2.$$