What linear map is needed for acting on a maximally entangled state?

Question

I was reading a textbook and I encountered this question. I was wondering why we don't consider $M^\dagger$ instead of $M^{T}$, so I didn't show this relation, could you please help me to show below relation?

Let $M: \mathcal{H}^{\tilde{A}} \rightarrow \mathcal{H}^{B}$ be a linear map and denote its transpose map by $M^{T}: \mathcal{H}^{\tilde{B}} \rightarrow \mathcal{H}^{A}$. Show that $$ I \otimes M\left|\phi_{+}^{A \tilde{A}}\right\rangle=M^{T} \otimes I\left|\phi_{+}^{\tilde{B} B}\right\rangle $$ where $\left|\phi_{+}^{\tilde{A} A}\right\rangle:=\sum_{y=1}^{|A|}|y y\rangle^{\tilde{A} A}$ and $\left|\phi_{+}^{\tilde{B} B}\right\rangle:=\sum_{y=1}^{|B|}|y y\rangle^{\tilde{B} B}$ and these are maximally entangled states.

Answer 1

Both states live in $ H_A \otimes H_B $. We will compute their inner product with the basis states $ |i \rangle_A | j \rangle_B $ and show that they are equal. Indeed:

• $$ \big(I \otimes M\big) |\phi_+^{AA} \rangle = \sum_{y=1}^{|A|} |y \rangle_A M|y\rangle_A \implies \\ \langle i|_A \langle j|_B \big(I \otimes M\big) |\phi_+^{AA} \rangle = \sum_{y=1}^{|A|}\langle i|y \rangle_A \langle j|_BM|y\rangle_A = \langle j|_BM|i\rangle_A = M_{ji} $$

• $$ \big(M^T \otimes I\big) |\phi_+^{BB} \rangle = \sum_{y=1}^{|B|} M^T|y \rangle_B |y\rangle_B \implies \\ \langle i|_A \langle j|_B \big(M^T \otimes I\big) |\phi_+^{BB} \rangle = \sum_{y=1}^{|B|}\langle i|_A M^T |y\rangle_B \langle j|y \rangle_B = \langle i|_A M^T |j\rangle_B = \big(M^T\big)_{ij} = M_{ji} $$