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!

  • $\begingroup$ What is $| \phi ^+ \rangle$? A Bell state? $\endgroup$
    – FDGod
    Jan 23 at 15:15
  • $\begingroup$ @FDGod: Yes, the maximum entagled state $\endgroup$ Jan 23 at 15:20
  • $\begingroup$ and what is $\rho_A$? any arbitrary state $\in H_A$? or $\text{Tr}_B\bigg(|\phi^+ \rangle \langle \phi ^+ |\bigg)$? $\endgroup$
    – FDGod
    Jan 23 at 15:38
  • $\begingroup$ Also, is $|\phi ^+\rangle $ specifically a Bell state, i.e. 2-qubit state, or is it a maximally entangled state of arbitrary dimension? $\endgroup$
    – FDGod
    Jan 23 at 15:42
  • $\begingroup$ @FDGod yes, it is meant to be a Bell state, so a 2-qubit state, maximally entangled. $$\lvert \phi^+ \rangle = 1/\sqrt{2} (\lvert 00 \rangle + \lvert 11\rangle)$$ $\endgroup$ Jan 23 at 18:21

1 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 $d=2$.

  • $\begingroup$ Thanks for your answer. The question arises from an coursera online lecture regarding an introduction about Quantum information/computing. The equality is introduced in a chapter about quantum channels (Kraus-Operators, Choi-Jamiolkowski isomorphism, ...). So I would assume, that indeed the notation corresponds to partial traces and $\rho_A$ indeed depicts a reduced state of state $\rho$ on system $A$. However I would also not exclude a mistake in the lecture script here as I have found others already. $\endgroup$ Jan 24 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.