# Tracing out an observable vs integrating over unitaries

Let $$O$$ be an observable on a Hilbert space $$\mathcal{H}$$, and let $$B$$ be a subset of the spins composing $$\mathcal{H}$$, and let $$\bar{B}$$ be its complement. Now define

$$\displaystyle O_B = \frac{1}{\operatorname{Tr}_{\bar{B}}\mathbf{1}_{\bar{B}}} \operatorname{Tr}_{\bar{B}}(O) \otimes \mathbf{1}_{\bar{B}}.$$

Is this quantity equal to

$$\displaystyle \int d\mu(U) U O U^\dagger~?$$

The integral is taken over the set of unitary operators acting on $$\bar{B}$$ and $$\mu$$ is the Haar measure of $$U$$. If so, why is this the case?

Note: this question came up from trying to understand the following paper: http://arxiv.org/abs/quant-ph/0603121

It looks plausible to me, for two reasons:

1) The resulting expression must be invariant under conjugation by any unitary on $B$, because integration was by the Haar measure. So for product states the result of the integral has to be of the form $\rho_A \otimes \mathbf{1}_B$.

2) The integral is a linear superoperator. So what I said about product states can be extended to non-product states.

• Why do you write "looks plausible to me", rather than turning this into a proof? Commented Aug 22, 2019 at 9:27

One has

$$\int_{U(d)} d\mu(U) U_{i_1j_1} U^{\dagger}_{j_2i_2}= \frac1d \delta_{i_1,i_2}\delta_{j_1,j_2}$$

for Haar integration, from which it follows. (Source arXiv:math-ph/0402073)