SU(2) and entangled particles [closed]

Question

We have two particles $A$ and $B$ in a maximally entangled state $|\Psi\rangle \in \cal{H}_A \times \cal{H}_B$ $$ \left|\Psi\right\rangle = \frac{1}{\sqrt{2}} ( \left| 0 \right\rangle_A\otimes \left| 0 \right\rangle_B + \left| 1 \right\rangle_A\otimes \left| 1 \right\rangle_B ), $$ where $_A\left\langle i | j \right\rangle_A = {_B\left\langle i | j \right\rangle_B} = \delta_{ij}$ and $U^T$ is the transpose of the matrix $U$.

It is claimed that if $U \in SU(\cal{H}_A)$, then $$ U \otimes I \left|\Psi\right\rangle = I \otimes U^T \left|\Psi\right\rangle $$

How can we show this to be true? I am trying to make sense of a "quantum game theory" paper, but it assumes a high comfort level with the mathematical apparatus of quantum physics.

Answer 1

This is presumably simply an issue of notation - \begin{eqnarray} U\otimes I \left|\Psi\right\rangle = \frac{1}{\sqrt{2}} \left( (U_{11} \left| 0 \right\rangle_A + U_{21} \left| 1 \right\rangle_A ) \otimes \left| 0 \right\rangle_B \\ \hspace{3cm} + ( U_{12} \left| 0 \right\rangle_A + U_{22} \left| 1 \right\rangle_A ) \otimes \left| 1 \right\rangle_B \right) \end{eqnarray} and \begin{eqnarray} I\otimes U^T \left|\Psi\right\rangle = \frac{1}{\sqrt{2}} \left( \left| 0 \right\rangle_A \otimes (U_{11} \left| 0 \right\rangle_B + U_{12} \left| 1 \right\rangle_B ) \\ \hspace{3cm} + \ \left| 1 \right\rangle_A \otimes ( U_{21} \left| 0 \right\rangle_B + U_{22} \left| 1 \right\rangle_B ) \right) \end{eqnarray} They're indeed the same.