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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

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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.

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  • $\begingroup$ Why do you write "looks plausible to me", rather than turning this into a proof? $\endgroup$ – Norbert Schuch Aug 22 at 9:27

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