# How to prove that maximally entangled state remains maximally entangled under local unitaries?

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?

• are you asking why local unitary operations do not affect the entanglement?
– glS
Mar 25 at 20:03

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.