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Suppose I have a bipartite pure state $\vert\psi\rangle_{AB}$. By the Schmidt decomposition, we know that the reduced states $\rho_A$ and $\rho_B$ have the same eigenvalues. I am now interested in applying a projector on subsystem $B$, where I project onto some smaller subspace of $\mathcal{H}_B$. On the full state, this action is given by

$$\vert\psi\rangle\langle\psi\vert_{AB} \rightarrow \vert\omega\rangle\langle\omega\vert_{AB} := (I\otimes\Pi_B)\vert\psi\rangle\langle\psi\vert_{AB}(I\otimes\Pi_B)$$

This projected state $\vert\omega\rangle_{AB}$ is still pure and possibly subnormalized. Let its reduced states be $\sigma_A$ and $\sigma_B$, where $\sigma_B = \Pi_B\rho_B\Pi_B$. Since $\vert\omega\rangle\langle\omega\vert_{AB}$ is pure, the eigenvalues of $\sigma_A$ and $\sigma_B$ are identical.

Is it the case that there is an equivalent projector $\Sigma_A$ acting only on subsystem $A$ such that

$$\vert\psi\rangle\langle\psi\vert_{AB} \rightarrow \vert\omega'\rangle\langle\omega'\vert_{AB} := (\Sigma_A\otimes I)\vert\psi\rangle\langle\psi\vert_{AB}(\Sigma_A\otimes I)$$

where $\vert\omega'\rangle\langle\omega'\vert$ has the same reduced states as $\vert\omega\rangle\langle\omega\vert$?

In general, is a local projection acting on one subsystem of a bipartite pure state equivalent to another local projection acting on the other subsystem of the state?

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For pure states $|\phi\rangle$, it is always possible to write $$ (A\otimes I)|\phi\rangle = (I\otimes B)|\phi\rangle\ , $$ given that the Schmidt vectors of $|\phi\rangle$ span the full space on both parties.

This is an easy consequence of the fact that any state can be written as $$ |\phi\rangle=(F\otimes I)|\mu\rangle $$ for some $F$, with $|\mu\rangle$ the maximally entangled state, and that $$ (M\otimes I)|\mu\rangle = (I\otimes M^T)|\mu\rangle\ , $$ for arbitrary $M$.

If the Schmidt decomposition does not span the whole space, it is equally easy to find counterexamples: Just pick operators which take you out of the state spanned by the Schmidt decomposition.

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  • $\begingroup$ can $A,B$ be arbitrary matrices here, or is it true only for projectors? $\endgroup$ – glS Jun 27 at 15:17
  • $\begingroup$ @glS Arbitrary matrices. I think the only contraint will be that this only works if you restrict the local Hilbert spaces to the support of the corresponding RDMs. $\endgroup$ – Norbert Schuch Jun 27 at 16:29
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This is an elaboration of some of the results mentioned in the other answer.


All pure states can be written as $|\phi\rangle=(F\otimes I)|\mu\rangle$ with $|\mu\rangle$ maximally entangled and $F$ some matrix.

A way to see this is to think of the bipartite states as matrices. You can always do this by taking the typical expansion of a state as $|\phi\rangle=\sum_{ij}\phi_{ij} |i,j\rangle$, and denoting with $\phi\equiv (\phi_{ij})_{ij}$ the set of coefficients organised in a matrix. Note that in this notation, the maximally entangled state is, up to coefficients, equal to the identity: $\mu = \frac{1}{\sqrt N}I$, with $N$ the number of spanning vectors (i.e. the dimension of the space). We thus have

$$|\phi\rangle=(F\otimes I)|\mu\rangle \sim \newcommand{\on}[1]{\operatorname{#1}}\phi=\frac{1}{\sqrt N} F.$$ Finding $F$ is now trivial: $F=\sqrt N \phi$. More explicitly, $F_{ij}=\sqrt N\phi_{ij}\equiv \sqrt N \langle i,j|\phi\rangle$.


For all states $\lvert\phi\rangle$ and matrices $B$ that preserves the support of $|\phi\rangle$, there is some $A$ such that $$(A\otimes I)\lvert\phi\rangle=(I\otimes B)\lvert\phi\rangle.$$ We can immediately derive a necessary condition for this to be possible: the support of $\operatorname{tr}_1(\lvert\phi\rangle\!\langle\phi\rvert)$ must be invariant under $B$, and the support of $\operatorname{tr}_2(\lvert\phi\rangle\!\langle\phi\rvert)$ must be invariant under $A$.

One way to show the result is using the previous result about writing states as local operations on a maximally entangled state. More directly, we can get it noticing that again using the matrix notation used above, the condition reads $$A\phi=\phi B^T.$$ If $\phi$ is invertible, then $A=\phi B^T \phi^{-1}$ and we're done. Notice that $\phi$ is invertible if and only if it has full rank, i.e. if and only if the reduced states have full support.

More generally, we have $$A\phi\phi^+ = \phi B^T \phi^+,$$ where $\phi^+$ is the pseudo-inverse of $\phi$. Notice that $\phi\phi^+$ is the projector onto the range of $\phi$, which corresponds to the space covered by the reduced state $\operatorname{tr}_B(\lvert\phi\rangle\!\langle\phi\rvert)$. Replacing the matrices $A,B$ with their restrictions on the supports of the reduced states, we can restrict our attention to these, and reduce to the case of $\phi$ invertible.


As a concrete example, consider a three-dimensional space, and a state $$\sqrt5 |\phi\rangle= 2|0,+_{12}\rangle + |-_{23},3\rangle.$$ This is not maximally entangled, and looks complicated. We can however change coordinates, so that it becomes $\sqrt5\lvert\phi\rangle=2|00\rangle+\lvert11\rangle$. If we restrict our attention to the subspace spanned by these basis states, $\phi$ is invertible: $$\phi\equiv\frac{1}{\sqrt5}\begin{pmatrix}2 & 0 \\ 0 & 1\end{pmatrix}, \qquad\phi^{-1}\equiv \sqrt5\begin{pmatrix}1/2 & 0 \\ 0 & 1\end{pmatrix}.$$ The condition then reads $$A=\phi B^T \phi^{-1} = \begin{pmatrix}B_{11} & 2 B_{21} \\ B_{12}/2 & B_{22}\end{pmatrix}.$$

To see that this choice of $A$ does indeed work, observe that $$ \sqrt 5 (I\otimes B)\lvert\phi\rangle = 2\lvert0\rangle\otimes(B_{11}\lvert0\rangle + B_{21} \lvert1\rangle) + \lvert1\rangle\otimes(B_{12} \lvert0\rangle + B_{22} \lvert1\rangle), \\ \sqrt 5 (A\otimes I)\lvert\phi\rangle = 2(B_{11}\lvert0\rangle + B_{12}/2 \lvert1\rangle)\otimes \lvert0\rangle + (2 B_{21} \lvert0\rangle + B_{22} \lvert1\rangle)\otimes \lvert1\rangle. $$

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