# Can local projections on different parties give the same reduced state?

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?

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.

• can $A,B$ be arbitrary matrices here, or is it true only for projectors?
– glS
Jun 27, 2020 at 15:17
• @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. Jun 27, 2020 at 16:29

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 $$\sqrt 5 (I\otimes B)\lvert\phi\rangle = 2\lvert0\rangle\otimes(B_{11}\lvert0\rangle + B_{21} \lvert1\rangle) + \lvert1\rangle\otimes(B_{12} \lvert0\rangle + B_{22} \lvert1\rangle), \\ \sqrt 5 (A\otimes I)\lvert\phi\rangle = 2(B_{11}\lvert0\rangle + B_{12}/2 \lvert1\rangle)\otimes \lvert0\rangle + (2 B_{21} \lvert0\rangle + B_{22} \lvert1\rangle)\otimes \lvert1\rangle.$$