I am doing a homework problem on spin precession. Constant magnetic field $\vec{B}$ is in the z-direction. Initially, the electron is in an eigen state of $\vec{\sigma}.\vec{n}$, where $\vec{\sigma}$ is the Pauli matrix vector. $\vec{n} $ is some vector in the $x-z$ plane. I am asked to find the expectation value of $\hat{S_x}$ as a function of time. I initially found the eigen state of $\vec{\sigma}.\vec{n}$ and then found the unitary evolution operator (considering the magnetic field) and let it act on that eigen state (expressing the result as function of time). I then found the expectation value of $\hat{S_x}$ in this continuously evolving state. I found that the expactation value is independent of time and is equal to $\frac{\hbar}{2}$. Does this make sense? Can you give me some intuition for why this so. If this makes sense, can you tell me under what conditions, if any, will the expectation value be time dependent. I don't want to write down my derivation as it is extremely long.


One great way of understanding the time dependence of expectation values is by looking into the Ehrenfest Theorem (link here). This is \begin{equation} \frac{d}{d t} \langle A \rangle = \frac{1}{i \hbar} \langle \left[A , H\right] \rangle + \left\langle \frac{\partial A}{\partial t} \right\rangle \, . \end{equation} Let us now consider the case of a 2D hermitian Hamiltonian of the form \begin{equation} H= a_0 \, 1_{2\times 2} + \vec{a}\cdot\vec{\sigma} \, , \end{equation} with $\vec{a} = \left\{ a_1,a_2,a_3\right\} $ , $\vec{\sigma} = \left\{ \sigma_1,\sigma_2,\sigma_3\right\} $ and $a_0, \, a_1, \, a_2, \, a_3$ being real. Then, summing over repeated indices, we find \begin{equation} \frac{d}{dt} \langle \sigma_k \rangle = \frac{1}{i \hbar} \langle \left[ \sigma_k,H\right]\rangle = \frac{a_m}{i \hbar} \langle \left[ \sigma_k,\sigma_m\right]\rangle = \frac{a_m}{i \hbar} 2 i \epsilon_{kml} \langle \sigma_l \rangle = \frac{2}{\hbar} \epsilon_{kml} a_m \langle \sigma_l \rangle = \frac{2}{\hbar} \left( \vec{a} \times \langle \vec{\sigma} \rangle \right)_k \, , \end{equation} or \begin{equation} \frac{d}{dt} \langle \vec{\sigma} \rangle = \frac{2}{\hbar} \, \vec{a} \times \langle \vec{\sigma} \rangle \, . \end{equation} Here we see explicitly the influence of the vector $\vec{a}$. In particular, we directly find that $\vec{a} \cdot \frac{d}{dt} \langle \vec{\sigma} \rangle = 0 $. In your case $\vec{a}$ is time independent and proportional to $\vec{B} = B \, \hat{k}$, so the expectation value of $\sigma_3$ is constant in time.

But this is just half of the story. This expression clearly depends on the initial state. Let's pick $\vec{a} = a \, \hat{k}$ like in your problem, so we find \begin{eqnarray} \frac{d}{dt} \langle \sigma_1 \rangle &=& - \frac{2}{\hbar} \, a \langle \sigma_2 \, \rangle \\ \frac{d}{dt} \langle \sigma_2 \rangle &=& \frac{2}{\hbar} \, a \langle \sigma_1 \, \rangle \, . \end{eqnarray} For your initial state, an eigenstate of $\hat{n} \cdot \vec{\sigma}$ with $\hat{n}$ in the $x,z$ plane, you can see how the expectation value of $\sigma_1$ is constant in time while the expectation value of $\sigma_2$ is generically not. To get a more physical intuition you can also check for references on spin precession.

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    $\begingroup$ I should add, if you pick an arbitrary $\hat{n}$ in the $x,z$ plane, I can't see how OP gets specifically $\frac{1}{2} \hbar$ for the expectation value of $S_x$ when you use an eigenvector of $\hat{n} \cdot \vec{\sigma}$ . It seems to me it depends on the angle $\hat{n}$ forms with the $z$ axis and the corresponding eigenvalue. $\endgroup$ – secavara Jan 29 '18 at 18:44
  • $\begingroup$ Yes, I redid the problem and found there was a time dependence of the expectation value of $S_x$. Thanks for pointing out the Ehrenfest theorem. $\endgroup$ – IanDsouza Jan 30 '18 at 1:26
  • $\begingroup$ I am not sure if I misunderstood the problem but I found that the expectation value of $S_x$ is constant in time, as I point out in the answer. $\endgroup$ – secavara Jan 30 '18 at 8:54

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