Larmor Precession and the Stern-Gerlach Experiment

These are two examples that show up in Griffiths, and the professor said they have a really deep physical meaning and are extremely important — basically “know these cold.” So I figured I’d cover both in one post.

In one sentence: this post is about how an electron behaves inside a magnetic field!!!!!!

Let’s go!!!!!

1. Larmor precession

OK so — say we have an up spin. Pretend the electron is actually spinning (even though, you know, it’s not really spinning).

Whatever the mechanism, there’s a magnetic moment associated with it ($\vec{\mu}$, the mu vector). Spin fast → $\vec{\mu}$ big. Spin slow → $\vec{\mu}$ small.

So $\vec{S}$ and $\vec{\mu}$ are proportional:

$$\overrightarrow{S} \propto \overrightarrow{\mu} \\ \text{We define the constant } \gamma \text{ that determines how fast/slow the spin is,} \\ \text{and call it the proportionality constant:} \\ \gamma \cdot \overrightarrow{S} = \overrightarrow{\mu}$$

That $\gamma$ is the gyromagnetic ratio.

We’ll dig into $\gamma$ properly later, but for now: $\gamma = -e/m$.

Which means — the proton’s mass is thousands of times larger than the electron’s, so the proton’s magnetic moment is basically negligible.

“We only need to worry about the electron!”

So let’s fo.cus on our electron.

Now here’s the scary thing we just stumbled onto:

“A hydrogen (electron) inside a magnetic field” feels a force.

The very fact that there’s a $\vec{\mu}$ means it feels a force inside a $\vec{B}$!!!

If we draw it out like this — does it make sense that there’s a force?!?!?!?

There’s a force~

Ohhh~~~ Then there must be energy too~?

Yep. From E&M: force is minus the gradient of energy.

$$\overrightarrow{\tau} = \nabla\!\left(\overrightarrow{\mu} \cdot \overrightarrow{B}\right) \\ H(\text{hamiltonian}) = -\overrightarrow{\mu} \cdot \overrightarrow{B} = -\gamma \cdot \overrightarrow{S} \cdot \overrightarrow{B}$$

OK, groundwork laid. All of that was just so we could talk about Larmor precession.

Say the external magnetic field is $B_0$ in the $z$ direction.

Then the Hamiltonian is

$$H = -\gamma \cdot \overrightarrow{S} \cdot \overrightarrow{B} = -\gamma B_0 S_z$$

In QM these aren’t numbers, they’re operators — so it’s totally fine, totally unsurprising, that they show up as matrices.

$$S_z = \frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix} \text{ — we had this already, so} \\ \therefore H = -\gamma B_0\frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$

The eigenvalues and eigenvectors of this $H$? Same as $S_z$, just scaled by a constant. That’s the only difference.

$$\lambda_1 = -\gamma B_0\frac{\hbar}{2} \quad \chi_+ = \begin{pmatrix}1\\0\end{pmatrix} \\ \lambda_2 = \gamma B_0\frac{\hbar}{2} \quad \chi_- = \begin{pmatrix}0\\1\end{pmatrix}$$

The whole reason we bothered getting $H$ — it’s so we can plug into Schrödinger.

The (time-dependent) Schrödinger equation is!!!

$$i\hbar\frac{\partial\psi}{\partial t} = H\psi$$

If I write $\psi$, chapters 1–3 come flooding back, so let me write it this way instead:

$$i\hbar\frac{\partial}{\partial t}(\text{state}) = H(\text{state})$$

We know what $H$ looks like as a matrix. Any state is a linear combination of $\chi_+$ and $\chi_-$.

In other words — we already know the eigenstates!!!!!

$$\chi_+$$

or

$$\chi_-$$

Those are it.

And remember: the reason the time-independent Schrödinger equation worked in the first place was because we had eigenstates, right?

In that case, to get the “state as a function of time,” we just multiply by a factor. That factor was

$$e^{-i\frac{E}{\hbar}t}$$

right~~~

OK now let’s nail down the precession.

What are we going to do? We’re going to measure $S_x$ on this state a bunch of times and average. In other words, we want the expectation value $\langle S_x \rangle$.

$$\langle S_x \rangle = \int\!\left(\text{state}^{\dagger}\, S_x\, \text{state}\right)dV$$

Writing it as an integral feels clunky. Let me just switch to Dirac notation.

The state is

$$\text{state} : \chi = a\chi_+ + b\chi_-$$

(Normalization comes along for the ride: $|a|^2 + |b|^2 = 1$.)

I’ll just plug in arbitrary $a$, $b$ satisfying $a^2 + b^2 = 1$:

$$a = \cos\frac{\alpha}{2} \quad b = \sin\frac{\alpha}{2}$$

(Yeah, I’m assuming $a$, $b$ are real — but we don’t lose the essence, so onward.)

And we need to include the time-dependence too. So:

$$\langle S_x \rangle = \langle\chi(t)^{\dagger}|S_x\chi(t)\rangle \\ = \left(\left(\cos\frac{\alpha}{2}\right)e^{i\frac{E_1}{\hbar}t} \quad \left(\sin\frac{\alpha}{2}\right)e^{i\frac{E_2}{\hbar}t}\right) \left(\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\right) \begin{pmatrix}\left(\cos\frac{\alpha}{2}\right)e^{-i\frac{E_1}{\hbar}t}\\\left(\sin\frac{\alpha}{2}\right)e^{-i\frac{E_2}{\hbar}t}\end{pmatrix}$$

$E_1$, $E_2$ are the eigenvalues of $H$! So let’s swap them in:

$$E_1 = -\gamma B_0\frac{\hbar}{2}\\ E_2 = \gamma B_0\frac{\hbar}{2}$$$$= \left(\left(\cos\frac{\alpha}{2}\right)e^{-i\frac{\gamma B_0}{\hbar}\frac{\hbar}{2}t} \quad \left(\sin\frac{\alpha}{2}\right)e^{i\frac{\gamma B_0}{\hbar}\frac{\hbar}{2}t}\right) \left(\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}\right) \begin{pmatrix}\left(\cos\frac{\alpha}{2}\right)e^{i\frac{\gamma B_0}{\hbar}\frac{\hbar}{2}t}\\\left(\sin\frac{\alpha}{2}\right)e^{-i\frac{\gamma B_0}{\hbar}\frac{\hbar}{2}t}\end{pmatrix}$$

The calculation itself isn’t hard. Typesetting it in the equation editor, on the other hand, would be hell. So I’ll just drop the answer. (Anyone can do it — it’s just matrix multiplication from high school!!!!!)

$$\langle S_x \rangle = \frac{\hbar}{2}\sin\alpha\cos\!\left(\gamma B_0 t\right)$$

Same idea for $\langle S_y \rangle$:

$$\langle S_y \rangle = -\frac{\hbar}{2}\sin\alpha\sin\!\left(\gamma B_0 t\right)$$

And $\langle S_z \rangle$:

$$\langle S_z \rangle = \frac{\hbar}{2}\cos\alpha$$

Now you can probably guess why I picked $\cos(\alpha/2)$ and $\sin(\alpha/2)$ for $a$, $b$.

The bit I flagged in red up top — red was supposed to mean constant!! ^^

And if you squint at this:

Can you picture it?!?!?!?

$\langle \vec{S} \rangle$ is rotating.

Where?

“In the magnetic field~~!!~~!!!!!!”

At what speed?

“At angular velocity $\gamma B$!!”

$\omega = \gamma B$ (that’s why I colored it blue.)

This $\omega$ — the angular velocity of the precession — is called the Larmor frequency ^^^^^

2. Stern–Gerlach experiment

(Side note: Gerlach, the German physicist, got the Nobel Prize for this experiment and all that glory — but apparently he collaborated with the Nazis… and was shot after the war. (sigh) Grim.

Oh, and “Gerlach” is the English pronunciation — in actual German it’s more like “Gerloch-her~” haha.)

OK so — Larmor precession above was about an electron in a uniform magnetic field.

Stern–Gerlach is about an electron in a non-uniform magnetic field. Except — not hydrogen.

You do NOT put hydrogen in!! Do you know what happens if you run this experiment with hydrogen!!!!!!!! haha

There’s a reason the hydrogen bomb is a thing!!!!! Hydrogen sits in group 1 with the alkali metals and its reactivity is off the charts.

So — Stern and Gerlach wanted to see what a non-uniform field does to the electron, but “load it with hydrogen” means “you die,” so what do you use?

You use silver. Ag.

Silver shows up in experiments all the time!!!

Why~~~!!!??

$$Ag = [\text{Kr}]\, 4d^{10}5s^1$$

Because of this config, orbital angular momentum is 0 (the $s$-shell means $l=0$),

and the only angular momentum left is spin!!!

Plus it’s, you know, not going to kill you.

So it’s a go-to experimental material!!

OK so Stern and Gerlach shove silver into a non-uniform magnetic field. How’d they build the non-uniform field?

(I drew something that kind of screams Lord of the Rings.)

And they fire a stream of silver atoms through it — pew pew pew.

The result, up front:

It splits into up and down.

Now let’s sail our way to that conclusion.

“Huh — looks like these guys are feeling a force????”

Force is minus the gradient of energy, right?!?!?!?!

$$\overrightarrow{F} = -\overrightarrow{\nabla}H \quad \text{and } H \text{, as we had above, is the torque-energy from } \overrightarrow{\mu} \text{ and } \overrightarrow{B}, \\ = -\overrightarrow{\nabla}\!\left(-\overrightarrow{\mu}\cdot\overrightarrow{B}\right) = \overrightarrow{\nabla}\!\left(\overrightarrow{\mu}\cdot\overrightarrow{B}\right)$$

Now we need a non-uniform $\vec{B}$. Let’s think about it like this.

If we had normal magnets like these,

$$B = B_0\hat{k}$$

would’ve been fine. But —

Since the geometry looks like this, think of the field as the old flat $B_0$ plus a term that scales with $z$:

$$\overrightarrow{B} = \left(B_0 + \alpha z\right)\hat{k}$$

But Maxwell says $\nabla \cdot \vec{B} = 0$ everywhere, always — so to keep that satisfied,

$$\overrightarrow{B} = \left(B_0 + \alpha z\right)\hat{k} - \alpha x\hat{i}$$

let’s patch it like that. Treating the visible “horizontal” as the $x$-axis, I added an $x$-component to make Maxwell happy.

Alright, so now:

$$\overrightarrow{F} = \overrightarrow{\nabla}\!\left(\overrightarrow{\mu}\cdot\overrightarrow{B}\right)$$

Let’s just crank it out!!! heh heh heh

$$= \overrightarrow{\nabla}\left\{\left(\mu_x,\; \mu_y,\; \mu_z\right)\cdot\left(-\alpha x,\; 0,\; B_0+\alpha z\right)\right\} \\ = \overrightarrow{\nabla}\left\{-\mu_x\alpha x + \mu_z\left(B_0+\alpha z\right)\right\} \\ = \left(\frac{\partial}{\partial x}\hat{i} + \frac{\partial}{\partial y}\hat{j} + \frac{\partial}{\partial z}\hat{k}\right)\left\{-\mu_x\alpha x + \mu_z\left(B_0+\alpha z\right)\right\} \\ = -\alpha\mu_x\hat{i} + \alpha\mu_z\hat{k} \\ = -\alpha\gamma S_x\hat{i} + \alpha\gamma S_z\hat{k}$$

That red term in front — by Larmor precession, it averages to zero. So we just chuck it:

$$\text{Net force} = \alpha\gamma S_z\hat{k} \\ \to \text{since the eigenvalue of } S_z \text{ is } \pm\frac{\hbar}{2} \\ \sum F = \pm\frac{\alpha\gamma\hbar}{2}\hat{k}$$

That $\pm$ — that’s up spin vs. down spin, right?????

Ahhh — so Stern–Gerlach was the experiment that first let us tell up-spin electrons and down-spin electrons apart!!!!!!

That’s why it’s such a big deal.

Honestly, separating up spin from down spin — that’s basically a revolution, no?????

I mean, who would’ve thought you could just… sort them, collect the up-spin ones over here and the down-spin ones over there?! Wild.

Of course these days, with modern tech, nobody’s doing it by flinging atoms through giant magnets anymore — they do it with semiconductors. Semiconductors!

Modern tech is absurdly precise. Apparently this kind of splitting is trivial now… How cool is that. heh heh heh

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Originally written in Korean on my Naver blog (2015-12). Translated to English for gdpark.blog.