show that $\mathrm{tr}_A \left[\rho_A \lvert \phi^+ \rangle_{AB} \langle \phi^+ \rvert\right]$ equals to $\rho_B^T$

Question

I have difficulties calculating with partial traces in terms of quantum operations.

For me it is not clear how to derive the equality stated in the question title for a quantum mechanical system whose state space is the tensor product $H_A \otimes H_B$ of Hilbert spaces with a mixed state described by the density matrix $\rho_{AB}$, partial traced out density matrices $\rho_A, \rho_B$ and $\lvert \phi^+ \rangle$ is a bell state :

$$ \mathrm{tr}_A \left[\rho_A \lvert \phi^+ \rangle_{AB} \langle \phi^+ \rvert\right] = \rho_B^T $$

I tried to apply the (to my knowledge) calculation rules $\rho_A = \mathrm{tr}_B\left[\rho_{AB}\right]$ and $\rho_{AB} = \rho_A \otimes \rho_B$. Together with $\lvert \phi^+ \rangle_{AB} \langle \phi^+ \rvert = \rho_{AB}$ I can come up with some transformations which lead up to points where I do not know how to continue, e.g.

$$ \mathrm{tr}_A \left[\rho_A \lvert \phi^+ \rangle_{AB} \langle \phi^+ \rvert\right] = \mathrm{tr}_A \left[\rho_A \, \rho_{AB} \right] = \mathrm{tr}_A \left[\mathrm{tr}_B \left[\rho_{AB}\right]\, \rho_{AB} \right] = \, \dots $$

Especially I do not know how to introduce $\rho_B$ at a given point. Thank you for pointing me towards the correct transformations and your help!

Answer 1

The equality is not true, at least the way that you seem to be interpreting the notation.

To see that it cannot possibly be true, consider $\rho_{AB}$ and $(I\otimes U)\rho_{AB}(I\otimes U^\dagger)$. Both states have the same reduced density matrix $\rho$ on system $A$. But they have different reduced density matrices on system $B$. So how could $\text{Tr}_A(\rho_A\otimes I |\phi^+\rangle\langle\phi^+|)$ possibly give two different answers?

I guess that what your notation is supposed to mean is that $\rho_A$ is a density matrix $\rho$ on subsystem $A$. $\rho_B$ would be the same density matrix $\rho$, but on subsystem $B$. (And we do not need to talk about any joint system and its partial traces.) If we interpret the notation this way, your claimed relation does hold (up to normalisation). \begin{align*} \text{Tr}(\rho\otimes I|\phi^+\rangle\langle\phi^+|)&=\sum_i(\langle i|\rho)\otimes I|\phi^+\rangle\langle\phi^+|(|i\rangle\otimes I) \\ &=\frac{1}{\sqrt{d}}\sum_i(\langle i|\rho)\otimes I|\phi^+\rangle_{AB}\langle i|_B \\ &=\frac{1}{\sqrt{d}}\sum_{i,j}(\langle i|\rho|j\rangle\langle j|)\otimes I|\phi^+\rangle_{AB}\langle i|_B \\ &=\frac{1}{d}\sum_{i,j}\langle i|\rho|j\rangle|j\rangle_{B}\langle i|_B \\ &=\frac{1}{d}\sum_{i,j}\langle j|\rho^T|i\rangle|j\rangle_{B}\langle i|_B \\ &=\rho_B^T/d \end{align*}

Note: I assumed that $$ |\phi^+\rangle=\frac{1}{\sqrt{d}}\sum_{i=0}^{d-1}|i\rangle, $$ while I see that there has subsequently been a comment restricting this to the specific case of $d=2$.