# Prove that rank one projectors have the same partial trace iff they differ by a local unitary operation

In nielsen and chuang's QCQI book, there is a theorem called Unitary freedom in the ensemble for density matrices, which states that the sets $$|\psi_i\rangle$$ and $$|\phi_i\rangle$$ generate the same density matrix iff $$|\psi_i\rangle=\sum_iu_{ij}|\phi_j\rangle.$$ And I also found something interesting with respect to the reduced density matrix, which I stated as a theorem while I only prove a part of it and leave another part of it as the problem of the pose:

Theorem $$Tr_B|\Psi\rangle_{AB}\langle\Psi|=Tr_B|\Phi\rangle_{AB}\langle\Phi|$$ iff $$|\Psi\rangle_{AB}=I\otimes U|\Phi\rangle$$.

It's easy to see if $$|\Psi\rangle_{AB}=I\otimes U|\Phi\rangle$$, then the first part of the theorem is right, while I don't know how to prove the reverse of the theorem is true. (P.S. The theorem is inspired by the unitary freedom in the ensemble of density matrices, so it might not be true.)

• Are you familiar with the Schmidt decomposition? Sep 9, 2021 at 8:47
• @DaftWullie Yes, I am familiar with it. Sep 9, 2021 at 9:00

Consider the state $$|\Psi\rangle$$. This has a Schmidt decomposition $$|\Psi\rangle=U_A\otimes U_B\sum_i\lambda_i|ii\rangle.$$ Its reduced density matrix is $$\rho_A=U_A\left(\sum_i\lambda_i|i\rangle\langle i|\right)U_A^\dagger.$$
It must be that if $$|\Phi\rangle$$ has the same reduced density matrix, the density matrices have the same spectrum and hence $$|\Phi\rangle$$ has the same Schmidt coefficients. Hence, $$|\Phi\rangle=V_A\otimes V_B\sum_i\lambda_i|ii\rangle.$$
To start with (for simplicity), let's take the case where all the $$\lambda_i$$ are unique. It is also clear that $$U_A=V_A$$ so, in this case, we have the required result with $$U=U_BV_B^\dagger$$.
If there are repeated $$\lambda_i$$, the situation is a little more fiddly. For all distinct $$\lambda_i$$, it must be that $$U_A|i\rangle=V_A|i\rangle$$. But it could be that the way you've happened to choose the $$U_A$$ and $$V_A$$ is different, relying on a unitary freedom within the degenerate subspaces. However, we only have to prove that there exists one choice of $$V_A,V_B$$ such that the theorem holds. To do this, note that, within this degenerate subspace, what we effectively have (up to normalisation) is a maximally entangled state $$\lambda\sum_i|ii\rangle$$. But we know that $$U\otimes U^\star\sum_i|ii\rangle=\sum_i|ii\rangle.$$ Hence, if we decompose the choice $$V_A=U_AV_2$$, then we also have $$|\Phi\rangle=U_A\otimes(V_B\cdot V_2^\star)\sum_i\lambda_i|ii\rangle,$$ and hence you have the required result with $$U=U_B V_2^T V_B^\dagger$$.
• could I ask about a notation you've used here? You state the SD of $|\psi_{AB}\rangle=U_{A}\otimes U_{B} \sum_{i} \lambda_{i} |ii\rangle$. But isnt the SD just $\sum_{i} \lambda_{i} |ii\rangle$. Why the use of the two unitary operators? Sep 9, 2021 at 11:26
• @GaussStrife The statement of the Schmidt decomposition is that there exist bases $|i\rangle_A$ and $|i\rangle_B$ such that you can write it $\sum_i\lambda_i|i\rangle_A|i\rangle_B$. I'm just choosing to use a fixed basis (e.g. computational basis) $|i\rangle$ and convert into either of those specific bases using local unitaries. It makes comparisons between two states which could use different bases much more convenient. Sep 9, 2021 at 12:34