Consider the following interaction Hamiltonian $$H = \hbar \mu \sigma_{x} \otimes \sigma_x = \hbar \mu ( |01 \rangle \langle 1 0 | + |10\rangle\langle 01|)$$ acting on the joint states of qubits $\rho_{prim} \otimes \rho_{aux}$ for $t = \frac{\pi}{2 \mu}$. It is stated that if the primary and auxiliary systems (respectively $\rho_{prim}$ and $\rho_{aux}$) are in the state $|0\rangle$ then the interaction doesn't change the primary but if the primary is in state $|1\rangle$ and auxiliary in state $|0\rangle$ then the primary flips to $|0\rangle$.

For the first case my revised working is as follows: We have $$e^{-i\frac{\pi \sigma_x \otimes \sigma_x}{2}}[|0\rangle \langle0 |\otimes|0\rangle \langle0|]e^{i\frac{\pi \sigma_x \otimes \sigma_x}{2}}$$ where the state of the primary is $$e^{-\frac{\pi \sigma_x}{2}}|0\rangle = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \begin{pmatrix} 1 \\0\\ \end{pmatrix} = \begin{pmatrix} 0 \\ -i \end{pmatrix}$$

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    $\begingroup$ Why on earth should a Hamiltonian act on $\rho$ the way you describe it? $\endgroup$ – Norbert Schuch Jun 15 '18 at 15:34
  • $\begingroup$ @NorbertSchuch This is an example from Kurt Jacobs "Quantum Measurement Theory and Applications", see chapter 5 on quantum control theory. $\endgroup$ – John Doe Jun 15 '18 at 15:38
  • $\begingroup$ Yes, but I'm sure his Hamiltonian does not act on $\rho$ by left multiplication. This is not even hermitian! $\endgroup$ – Norbert Schuch Jun 15 '18 at 15:45
  • $\begingroup$ @NorbertSchuch It is Hermitian $\sigma_x \otimes \sigma_x = \sigma_x \sigma_x^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ $\endgroup$ – John Doe Jun 15 '18 at 16:15
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    $\begingroup$ With your edits, this is no longer a question, and in addition, the answer makes no sense any more whatsoever. You should roll back to the question you actually had and, if you wish, BELOW that post an edit where you explain the resolution yourself. (Or, if you wish, post an answer yourself.) $\endgroup$ – Norbert Schuch Jun 16 '18 at 18:01

Evolving a state $\rho$ according to an Hamiltonian $H$ does not work that way: $H\rho$ is not the evolved state (nor, in general, even a state at all).

The evolution with the Hamiltonian $H$ for time $t$ is described by the unitary operator $e^{-itH}$. To evolve a density matrix you have to compute $e^{-itH}\rho e^{itH}$.

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  • $\begingroup$ Not that this makes the original statement in the question correct. (Well, in some sense maybe, but not the way it is phrased.) $\endgroup$ – Norbert Schuch Jun 16 '18 at 17:59

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