Spin Precession in a Magnetic Field

[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,

[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!

 

[Jhero]

Hi there,

It seems like you have a good understanding of the basics of spin-1/2 particles in a magnetic field. Your calculations for the expectation values of S_{x}, S_{y}, and S_{z} are correct and show the spin precession about the z-axis.

To answer your question about why the expectation value of S_{z} is zero, we first need to understand what the expectation value represents. The expectation value of an observable is the average value that we would expect to measure if we were to perform a large number of measurements on an ensemble of identical systems.

In the case of the spin-1/2 particle, the eigenstates of the Hamiltonian are the states with definite spin in the z-direction, |+\rangle and |-\rangle. These states have equal probabilities of being measured, so the expectation value of S_{z} is zero, as it represents the average value of the spin in the z-direction.

Now, let's consider the expectation values of the magnetic moment \mu. The expectation value of \mu_{z} is not necessarily zero because the magnetic moment is a vector quantity and can have a component in the z-direction even if the spin is precessing about the z-axis. This is why we see a non-zero value for \langle \mu_{z} \rangle in your calculations.

In summary, the expectation value of S_{z} is zero because the spin-1/2 particle can be in either of the two eigenstates with equal probabilities, while the expectation value of the magnetic moment \mu_{z} can be non-zero because it is a vector quantity that takes into account the spin precession. I hope this helps clarify your confusion. Keep up the good work!