# Can spin operator expectation value be time-independent while commutator with Hamiltonian is non-zero?

Considering the following (magnetic field) hamiltonian: $$\hat{H}=-\gamma B_z \hat{S}_z$$ ($$\gamma$$ is a constant). Suppose an electron is in an eigenstate of $$S_x$$, and we ask ourselves the question whether the expected value of $$S_z$$ changes with time. This is not the case, as $$[\hat{H}, \hat{S}_z]=0$$.

Now suppose we are in an eigenstate of $$S_z$$, does the $$S_x$$ expectation value change with time? Logically I would say it does because of the answer to the previous question. However, $$[\hat{H}, \hat{S}_x]\neq 0$$ indicating time dependence. So, will the expectation value of $$S_x$$ change with time or not?

• But... look at the m=1/2 eigenstate of $S_z$. It gives $\langle S_x\rangle=0$. – Cosmas Zachos Oct 8 '20 at 21:05

## 2 Answers

The spin is undergoing Larmor precession about the magnetic field, so yes the $$x$$ and $$y$$ components will change while the component parallel to the field remains fixed..

• ...unless both x and y expectation values vanish. – Cosmas Zachos Oct 8 '20 at 21:09
• Alright, so x and y components ofcourse change (i.e. are time dependent), but nevertheless the expectation values for x and y are constant ($=0$) (makes sense as the precession creates a perfect circle). I guess I should not be too quick to make assumptions solely on the Ehrenfest theorem. – 6thsense Oct 8 '20 at 21:16
• @CosmosZachos. Only if $J=0$. – mike stone Oct 8 '20 at 23:31
• What do you mean by "expectation of $x$ $y$" do you mean $S_x$, $S_y$? – mike stone Oct 8 '20 at 23:32
• Yes, consider J=1/2, and $\langle S_x\rangle=\langle S_y\rangle=0$. – Cosmas Zachos Oct 9 '20 at 0:17

The Ehrenfest theorem uses the expected value of the commutator. i.e. for some operator $$\hat{A}$$ the Ehrenfest theorem says $$\frac{d}{dt} \langle\hat{A}\rangle = \frac{1}{i\hbar} \langle[\hat{A}, \hat{H}]\rangle + \langle\frac{\partial\hat{A}}{\partial t}\rangle.$$

In your case while $$[\hat{S}_x, \hat{H}] \neq 0,$$ its expectation value will still be zero thus the expectation value of $$\hat{S}_x$$ will not depend on time. Should be straightforward to show that $$\langle[\hat{S}_x, \hat{S}_z]\rangle = 0$$ with the spin commutator relations.

Edit: For your first point, if the only Hamiltonian is the one you've given then the particle won't be in an eigenstate of $$\hat{S}_x$$. If it was in such an eigenstate before the magnetic field was turned on the wavefunction will collapse into one of the possible $$\hat{S}_z$$ eigenstates once the field is turned on.