# Spin Precession in a Magnetic Field

1. Dec 4, 2008

### Higgy

I'm having a mental block on this and I was hoping that someone could help me.

A little background
Consider a spin-1/2 particle in static, homogeneous magnetic field,

$$\vec{B}=B \hat{k}$$

The Hamiltonian is

$$H=-\vec{\mu}\cdot\vec{B}=-\gamma B S_{z}$$

where $\gamma$ is the gyromagnetic ratio.

Working in the $S_{z},S^{2}$ basis,

The eigenstates are

$$|S_{z} + \rangle = |s,m=+1/2\rangle = |+\rangle$$
$$|S_{z} - \rangle = |s,m=-1/2\rangle = |-\rangle$$

I can calculate the expectation values for the observables $S_{x},S_{y},S_{z}$, and I get,

$$\langle S_{x}\rangle=\frac{\hbar}{2} \cos\omega t$$
$$\langle S_{y}\rangle=\frac{\hbar}{2} \sin\omega t$$
$$\langle S_{z}\rangle=0$$

This is to be interpreted as spin precession about the z-axis.

Question
Why is $\langle S_{z}\rangle=0$?

I found the above mostly in Sakurai, but in another text, it is shown that the expectation values of the magnetic moment $\mu$,

$$\vec{\mu}=\gamma \vec{S}$$

are

$$\langle \mu_x \rangle= \gamma \hbar A \cos (\omega t + \delta)$$
$$\langle \mu_y \rangle= - \gamma \hbar A \sin (\omega t + \delta)$$
$$\langle \mu_z \rangle= \gamma \hbar C$$

where A, B, and C are some constants, and $\delta$ is a phase.

If the expectation value of $S_{z}$ is zero, wouldn't the same be true for $\mu_z$? This is where I'm confused!