# Prove that maximally entangled states $|\Phi\rangle$ satisfy the identity $(U\otimes I)|\Phi\rangle=(I\otimes U^T)|\Phi\rangle$

The definition of maximally entangled state is $$\begin{equation} \vert \Phi \rangle = \frac{1}{\sqrt{d}} \sum_i \vert i \rangle \vert i \rangle, \tag{1} \end{equation}$$ where $$d$$ is the dimension of the hilbert space. Then we have the following identity $$\begin{equation} (U\otimes I)\vert \Phi \rangle= (I\otimes U^T)\vert \Phi \rangle, \tag{2} \end{equation}$$ where $$U$$ denotes the unitary matrix.

Equation (2) is from the lower right corner at Page. 7 in Cross-Platform Verification in Quantum Networks.

My question is how to proof equation (2).

• This is also true for non-unitary operations, which don't need maximally entangled states but only states with maximal Schmidt number arxiv.org/abs/1906.07731 Jan 31 at 20:14

Start by writing $$U=\sum_{ij}U_{ji}|j\rangle\langle i|.$$ Now evaluate $$U\otimes I|\Phi\rangle=\frac{1}{\sqrt{d}}\sum_{ij}U_{ji}|j\rangle|i\rangle.$$ Next, realise that (again,if you don't see it,just write it out long-hand) $$\sum_iU_{ji}|i\rangle=U^T|j\rangle,$$ so this just becomes $$U\otimes I|\Phi\rangle=I\otimes U^T|\Phi\rangle.$$