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.



SU(2) and entangled particles
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.


