0
$\begingroup$

If Alice and Bob share the state: $$\left| {{\psi _{AB}}} \right\rangle = \sin \theta \left| {10} \right\rangle + \cos \theta \left| {01} \right\rangle $$ then $\rho_{AB}$ can be obtained as: $${\rho _{AB}} = \left| {{\psi _{AB}}} \right\rangle \left\langle {{\psi _{AB}}} \right|.$$ Is there a way to get $\rho_{BA}$ instead?

$\endgroup$
3
  • 3
    $\begingroup$ Crossposted from Mathematics. I'm not sure why though, I already answered it on MSE. $\endgroup$
    – Rammus
    Jul 21 '21 at 11:34
  • $\begingroup$ what do you mean with "get". A quantum circuit sending one to the other? Also, what's your definition of $\rho_{BA}$ here? Is it the same state after swapping the spaces or something else? $\endgroup$
    – glS
    Jul 21 '21 at 13:04
  • $\begingroup$ @glS By $\rho_{BA}$ I mean the density matrix after permutation of the subsystems. I want to get the formula that enables me to calculate $\rho_{BA}$. $\endgroup$
    – Bekaso
    Jul 21 '21 at 14:55
2
$\begingroup$

$A$ and $B$ are labels for the Hilbert spaces in which each subsystem exists. There is no different physical content between $\mathcal{H}_A\otimes \mathcal{H}_B$ and $\mathcal{H}_B\otimes \mathcal{H}_A$, they are just different ways of bookkeeping.

As such, we can immediately trade all of the information about subspaces $A$ and $B$ and write $$|\psi_{BA}\rangle=\sin\theta |0\rangle_B\otimes|1\rangle_A+\cos\theta |1\rangle_B\otimes|0\rangle_A$$ and $$\rho_{BA}=|\psi_{BA}\rangle\langle \psi_{BA}|.$$

$\endgroup$
2
  • $\begingroup$ What about the coefficients $\sin \theta $ and $\cos \theta $? Won't they have any change? $\endgroup$
    – Bekaso
    Jul 22 '21 at 1:40
  • $\begingroup$ Nope, no change! We are simply changing the labels - I'll make it more explicit $\endgroup$ Jul 22 '21 at 1:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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