We have a maximally entangled state $\phi$ of composite system $R$ and $Q$. Apply unitary $U$ to $\phi$ on the Q system. Now how to prove that $U \phi$ is also a maximally entangled state?

This statement was in this article by Nielsen.

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  • $\begingroup$ are you asking why local unitary operations do not affect the entanglement? $\endgroup$
    – glS
    Mar 25, 2023 at 20:03

1 Answer 1


TL;DR: Local unitaries leave entanglement unchanged.

Schmidt decomposition

Any pure bipartite quantum state $|\phi_{RQ}\rangle$ can be written as $$ |\phi_{RQ}\rangle=\sum_{i=1}^r\lambda_i|i_R\rangle|i_Q\rangle\tag1 $$ where the positive real numbers $\lambda_i$ are known as Schmidt coefficients, $r$ is known as the Schmidt rank (or Schmidt number) of $|\phi_{RQ}\rangle$, $|i_R\rangle$ is an orthonormal basis in the Hilbert space $\mathcal{H}_R$ of the subsystem $R$ and $|i_Q\rangle$ is an orthonormal basis in the Hilbert space $\mathcal{H}_Q$ of the subsystem $Q$.

Maximal entanglement

A maximally entangled state of $R$ and $Q$ is any pure bipartite state which maximizes von Neumann entropy across the partitioning of $RQ$ into $R$ and $Q$. We can restate it equivalently in terms of the parameters of the Schmidt decomposition $(1)$ by saying that the Schmidt coefficients are all equal $\lambda_i=\lambda$ for $i=1,\dots,r$ and Schmidt rank is maximum, i.e. $r=d$ where $d:=\min(\dim\mathcal{H}_R, \dim\mathcal{H}_Q)$.

Effect of local unitaries

Thus, we can write any maximally entangled state $|\phi_{RQ}\rangle$ as $$ |\phi_{RQ}\rangle=\frac{1}{\sqrt{d}}\sum_{i=1}^d|i_R\rangle|i_Q\rangle.\tag2 $$ If we then act with $U$ on subsystem $Q$, the state becomes $$ (I\otimes U)|\phi_{RQ}\rangle=\frac{1}{\sqrt{d}}\sum_{i=1}^d|i_R\rangle|i_Q'\rangle\tag3 $$ where $|i_Q'\rangle:=U|i_Q\rangle$. Clearly, the state $(3)$ is still maximally entangled. In general, entanglement depends on $r$ and $\lambda_i$, so local unitary transformations don't affect it.


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