# How do get $\rho_{BA}$ if I have $\rho_{AB}$

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?

• Crossposted from Mathematics. I'm not sure why though, I already answered it on MSE. Jul 21 '21 at 11:34
• 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?
– glS
Jul 21 '21 at 13:04
• @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 $\rho_{BA}$. Jul 21 '21 at 14:55

$$A$$ 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 $$\rho_{BA}=|\psi_{BA}\rangle\langle \psi_{BA}|.$$
• What about the coefficients $\sin \theta$ and $\cos \theta$? Won't they have any change? Jul 22 '21 at 1:40