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 $$e^{-\frac{\pi \sigma_x}{2}}|0\rangle = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} \begin{pmatrix} 1 \\0\\ \end{pmatrix} = \begin{pmatrix} 0 \\ -i \end{pmatrix}$$
Answer
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 $e^{-itH}\rho e^{itH}$.
No comments:
Post a Comment