Spin Precession in a Magnetic Field

  • Thread starter Higgy
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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,

[tex]\vec{B}=B \hat{k}[/tex]

The Hamiltonian is

[tex]H=-\vec{\mu}\cdot\vec{B}=-\gamma B S_{z}[/tex]

where [itex]\gamma[/itex] is the gyromagnetic ratio.

Working in the [itex]S_{z},S^{2}[/itex] basis,

The eigenstates are

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

I can calculate the expectation values for the observables [itex]S_{x},S_{y},S_{z}[/itex], and I get,

[tex]\langle S_{x}\rangle=\frac{\hbar}{2} \cos\omega t[/tex]
[tex]\langle S_{y}\rangle=\frac{\hbar}{2} \sin\omega t[/tex]
[tex]\langle S_{z}\rangle=0[/tex]

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

Question
Why is [itex]\langle S_{z}\rangle=0[/itex]?

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

[tex]\vec{\mu}=\gamma \vec{S}[/tex]

are

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

where A, B, and C are some constants, and [itex]\delta[/itex] is a phase.

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

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