# Does taking partial trace commute with local operators?

Suppose we have a big system with two subsystems $$H=A\otimes B$$. For a unitary $$U$$ in the Hilbert space $$A$$ and a state $$\rho$$ in the Hilbert space $$H$$. Is the following statement true? $$\text{Tr}_B [U\otimes \mathbb{1}_A \rho (U\otimes \mathbb{1}_A)^\dagger] = U \text{Tr}_B(\rho) U^\dagger$$

Physically it feels quite convincing because someone doing experiments in his lab $$A$$ should not be concerned about the word outside ($$B$$). But mathematically, this statements seems nontrivial to me.

• The physical intuition you expressed seems misplaced to me because it can be true only if there exists a wavefunction for subsystem $B$. A wavefunction $\rho$ in full Hilbert space $H$ doesn't necessarily imply a wavefunction for $B$ as $\rho$ can be an entangled state. – Feynmans Out for Grumpy Cat Mar 7 at 21:50
• @NorbertSchuch I tried with random numerical examples and it seems to be true. – taper Mar 8 at 9:31
• @DvijMankad Thanks for you comment. I think here $\rho$ means (reduced) density matrix, which always exists. – taper Mar 8 at 9:44
• Did you try to use the formula for the partial trace? – Norbert Schuch Mar 8 at 14:10

Using Einstein's summation convention for repeated indices, $$[(U\otimes\mathbb 1)\rho(U^\dagger\otimes \mathbb 1)]_{ijkl} = U_{im} \delta_{jn}\rho_{mnpq} U^*_{kp}\delta_{lq} =U_{im}U^*_{kp}\rho_{mjpl}$$
$$(\operatorname{Tr}_B[(U\otimes\mathbb 1)\rho(U^\dagger\otimes \mathbb 1)])_{ik} =[(U\otimes\mathbb 1)\rho(U^\dagger\otimes \mathbb 1)]_{ijkj} = U_{im}U^*_{kp}\rho_{mjpj} = U_{im}U^*_{kp}\operatorname{Tr}_B(\rho)_{mp} = (U\operatorname{Tr}_B(\rho)U^\dagger)_{ik}$$