We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space $\mathbb{C}^2$. We denote by $\left . \left | 0 \right \rangle\right .$ and $\left . \left | 1 \right \rangle\right .$ the standard basis elements $(1,0)^T$ and $(0,1)^T$. Let the Hamiltonian of the system in this basis be given by
$$
H=\omega\begin{pmatrix}
0 &1 \\
1 &0
\end{pmatrix}=\begin{pmatrix}
0 &-i \\
-i &0
\end{pmatrix}
$$
and assume that for $t=0$ the state of the system is just given by $\psi(t=0)=\left . \left | 0 \right \rangle\right .$. In the following, we also assume natural units in which $\hbar=1$.
I have found that $\psi(t)=\begin{pmatrix}
\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}\\
-\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}
\end{pmatrix}.$
And I have computed the expectation value of $\left \langle \sigma_z \right \rangle_{\psi(t)}=\left \langle \psi(t)\mid \sigma_z\psi(t) \right \rangle$ where $
\sigma_z=\begin{pmatrix}
1 &0 \\
0 &-1
\end{pmatrix}$ to 1.
Now I have to determine the time-evolved observable $\sigma_z=e^{iHt}\sigma_ze^{-iHt}$ (the Heisenberg evolution of $\sigma_z$) and I have to determine the resulting expectation value $\langle \sigma_z(t)\rangle_{\psi_0}=\langle\psi_0|\sigma_z(t)\psi_0\rangle$. But I'm not sure how to deal with it? Is the first one just a matrix product I have to solve? But what is $e^{iHt}$ then? I hope anyone can help me?
Best Answer
$e^{iHt}$ is $U^*(t)$, is the conjugate transpose of $U(t)$
(you can calculate $e^{iHt}$ to check this)With $\hat\sigma_z=e^{iHt}\sigma_ze^{-iHt}$ and $\mid\psi(t)\rangle=e^{-iHt}\mid\psi_0\rangle,\;\langle\psi(t)\mid=\langle\psi_0\mid e^{iHt}$
this identity follows,
$\left \langle \sigma_z \right \rangle_{\psi(t)}=\left \langle \psi(t)\mid \sigma_z \mid \psi(t) \right \rangle=$
$=\left \langle\psi_0\mid e^{iHt}\sigma_z e^{-iHt}\mid\psi_0 \right \rangle=$
$=\left \langle\psi_0\mid\hat\sigma_z\mid\psi_0 \right \rangle$
ADDED:
$U(t)=e^{-iHt}$ is the solution to the Schrodinger (differential) equation and it is the evolution operator (acting over $\mid\psi_0\rangle$ to give the state in any moment). The calculation of the exponential operator is not difficult, but some cumbersome. I tricked an online tool to do that. Follow the link to see how it's done.
Anyway, $U(t)=\begin{pmatrix} \frac{1}{2}e^{t}+\frac{1}{2}e^{-t}&-\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}\\ -\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}&\frac{1}{2}e^{t}+\frac{1}{2}e^{-t} \end{pmatrix}.$
EDIT:
The Hamiltonian is not self-adjoint, so $U^*\neq U^{-1}$. It's still true that $U^{-1}=e^{iHt}$. My response is valid in the case it was.
I only can speculate about what happened. Probably $H=\begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$ originally.