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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?

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    $\begingroup$ Crossposted from Mathematics. I'm not sure why though, I already answered it on MSE. $\endgroup$ – Rammus Jul 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 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 at 14:55
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$A$ 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 $$\rho_{BA}=|\psi_{BA}\rangle\langle \psi_{BA}|.$$

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  • $\begingroup$ What about the coefficients $\sin \theta $ and $\cos \theta $? Won't they have any change? $\endgroup$ – Bekaso 2 days ago
  • $\begingroup$ Nope, no change! We are simply changing the labels - I'll make it more explicit $\endgroup$ – Quantum Mechanic 2 days ago

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