Changing the initial conditions

As we've been discussing, the angular momentum of physical objects is quantized: it can only be observed in integral units of $h/4\pi$. Additionally, spin angular momentum is unique as a physical property of particles. While a projection of a particle's spin along some axis can be measured, the total spin of the particle itself never slows down and never speeds up. These facts lead to a remarkable constraint on particles: we can only measure the direction of a particle's spin angular momentum along a few, quantized angles.

For particles with the smallest nonzero value of spin angular momentum, $h/4\pi$, particles like electrons, protons or even silver atoms, there are only two possible, observable angles: $0^{\circ}$ and $180^{\circ}$.

Last time we explored the first experimental verification of this physical fact. Today, we'll revisit the Stern-Gerlach experiment with one distinct alteration. We shall adjust the initial conditions of the silver atoms in the beam.

The Initial Conditions

In the Stern-Gerlach experiment, the beam of silver atoms was created thermally: atoms sublimated off a heated lump of silver. They bounced around a cavity, moving randomly until a few of the managed to exit through a pinhole. This spray of silver atoms was further restricted by a screen with a second pin hole, effectively focusing that spray into a beam.

We say the beam was create “thermally” to emphasize two points. First of course is the heat used to free them from their metallic crystal lattice. Second, because this process is essentially random, the direction and velocities of the newly freed, individual atoms are also random. The probability distribution that governs the collective properties of those atoms, like their average speed, depends on physical parameters like the temperature of the heated lump of silver.

The probability of finding a particle at a given speed in a gas at different temperatures. Otherwise known as the Maxwell Boltzmann distribution. Arbitrary units are used here just to emphasize the fact that as the temperature goes up, so too does the average velocity and the associated spread in those velocities. The spin direction is of course uniformly random.

Unlike the random distribution of positions and velocities, the spin of each silver atom$^{1}$ is fixed at $h/4\pi$. This is just a property of how the outermost electron in the silver atom is arranged. Nevertheless, while the magnitude of each silver atom's spin is the same, the direction - the axis along which it points - is entirely random.

It is important to note that the directions of the spins of these atoms are distributed randomly, but have not yet been observed. We can only observe their angles as $0^{\circ}$ or $180^{\circ}$. Those angles are made with respect to the observational apparatus - the direction of the Stern-Gerlach magnet. To understand the distinction between these two ideas, it is sometimes helpful to remember that we can position the Stern-Gerlach device along any angle we like$^{2}$.

As an exercise, you might reiterate in your own words why the random initial conditions causes the beam of silver atoms to split in two.

Changing the Initial Conditions

An easy augmentation of the Stern-Gerlach experiment is to expose the beam's source to a strong, uniform magnetic field. As we've previously discussed, the magnetic moments in the silver atoms will experience a torque, which causes them to align with this uniform magnetic field$^{3}$. As a result, the beam of silver atoms will all have spins pointed in that specific direction.

Let's imagine that we have a knob which can change the direction of that uniform magnetic field to anything we like. In particular, we can adjust it relative to the fixed, sharply pointed Stern-Gerlach magnetic field used in the detector. Unlike the original experiment, where the silver atoms are spinning entirely randomly, we now have precise control over the spins of the silver atoms as they approach the detector. Let's now explore what happens for a few different choices of angle.

The Augmented Stern-Gerlach Experiment

To be crystal clear, let us imagine the Stern-Gerlach apparatus from last time, with the detector magnet pointed in the $\widehat{x}$-direction. To the silver atoms we can then apply the additional, uniform magnetic field $\vec{B}$,

$$\vec{B} = B_{x}\widehat{x} + B_{y}\widehat{y} + B_{z}\widehat{z}.$$

We can then define the angle between them as the inner product as

\begin{equation}\label{defcos}\vec{B} \cdot \widehat{x} = B_{x} = |\vec{B}|\cos\varphi.\end{equation}

In other words, $\varphi$ is the angle $\vec{B}$ makes with respect to the $\widehat{x}$-axis, and therefore the detector.

In what follows we'll describe properties of this new, applied, uniform magnetic field $\vec{B}$ at various angles, and then discuss how that beam of particles behaves as it passes through the Stern-Gerlach device.

$\varphi$ at 90 degrees

Suppose we start with $\vec{B}$ purely in the $\widehat{y}$-direction.

$$\vec{B} = B \,\widehat{y}.$$

Clearly it's orthogonal to the detector, as

$$\vec{B}\cdot \widehat{x} = B\,\widehat{y}\cdot\widehat{x} = 0,$$

so in terms of \eqref{defcos}, $\varphi$ must equal 90 degrees.

At this angle, the Stern-Gerlach device appears to behave precisely as it did previously. The beam splits precisely in two.

$\varphi$ at 0 degrees

Now suppose we adjust the knob and tune $\vec{B}$ so that it's purely in the $\widehat{x}$-direction:

$$\vec{B} = B\,\widehat{x}.$$

In other words, it's exactly aligned with the detector:

$$\vec{B}\cdot \widehat{x} = B\cos\varphi = B.$$

This means that $\cos\varphi = 1$, which is true when $\varphi = 0$ degrees, as expected.

In this case, the beam of the Stern-Gerlach experiment does not split, but it is deflected. It is entirely deflected upwards.

$\varphi$ at 180 degrees

Carrying on, we now adjust the knob and tune $\vec{B}$ so that it's still purely in the $\widehat{x}$-direction, but in the opposite orientation:

$$\vec{B} = -B\,\widehat{x}.$$

In other words, it's antialigned with the detector:

$$\vec{B}\cdot \widehat{x} = B\cos\varphi = -B.$$

This means that $\cos\varphi = -1$, which is true when $\varphi = 180$ degrees, as expected.

In this case, the beam of the Stern-Gerlach experiment again does not split, but it is again deflected. This time, it is entirely deflected downwards.

$\varphi$ at 90 degrees, different ways

There are a few more options for us to consider. Before continuing, it's worth pointing out that there is more than one way for $\vec{B}$ to be orthogonal to the $\widehat{x}$-direction. We could turn our knob to set

$$\vec{B} = B\,\widehat{z},$$

so that

$$\vec{B}\cdot \widehat{x} = B\cos\varphi = 0.$$

Again, $\varphi$ will be 90 degrees and the beam, once passed through the Stern-Gerlach apparatus will split evenly in two.

Since this is true for both $\vec{B} = B\,\widehat{y}$ and $\vec{B} = B\,\widehat{z}$, it's not hard to see that it's true also for

$$\vec{B} = B_{y} \widehat{y} + B_{z}\widehat{z}.$$

In other words, for any uniform magnetic field pointed along any direction in the $yz$-plane, $\varphi$ = 90 degrees and the beam, passed through the Stern-Gerlach apparatus, will likewise split in two.

$\varphi$ at 45 degrees

Finally, let's turn those knobs to rotate the uniform magnetic field to a slightly different direction, perhaps one that mixes the $\widehat{x}$- and $\widehat{z}$-directions:

$$\vec{B} = B_{x}\widehat{x} + B_{z}\widehat{z}.$$

To keep things simple, let's suppose that $B_{x}$ and $B_{z}$ has exactly the same value, $B_{\diamond}$:

$$\vec{B} = B_{\diamond}\widehat{x} + B_{\diamond}\widehat{z}.$$

The magnitude of $\vec{B}$ is then

$$|\vec{B}| = \sqrt{B_{\diamond}^{2}+ B_{\diamond}^{2}} = B,$$

so that

$$B = \sqrt{2}B_{\diamond}.$$

We therefore have

$$\vec{B}\cdot \widehat{x} = B_{\diamond}.$$

From \eqref{defcos}, we can write this in terms of the angle $\varphi$:

$$B\cos\varphi = \frac{1}{\sqrt{2}}B,$$

or

$$\cos\varphi = \frac{1}{\sqrt{2}}.$$

So $\varphi$ must equal 45 degrees.

When we pass this beam through the Stern-Gerlach device, with it's magnet aligned in $\widehat{x}$-direction, the beam does split, but it doesn't split evenly. Looking at the screen, we see that far more silver atoms have been deflected “up” than “down”.

Evidently, the likelihood of the spin being measured in the “up” orientation is much higher than likelihood of being “down”.

A Preview of the Mathematics

We shall revisit this mathematical framework many times in the next few lectures, but for now I just want to give you the nuts and bolts to see how it works.

The spin of a silver atom is a vector, $\vec{S}$, which we assume to be aligned with the imposed, uniform magnetic field $\vec{B}$. Like $\vec{B}$ can write $\vec{S}$ as a vector in three-dimensions,

$$\vec{S} = S_{x}\widehat{x}+S_{y}\widehat{y}+S_{z}\widehat{z}.$$

This is the unobserved, spin angular momentum presented from the atom's point of view. From the detector's point of view, the observed spin angular momentum is either up or down. We sometimes represent these as “ket” state vectors$^{4}$:

$$| \uparrow\; \rangle \quad \mathrm{or }\quad|\downarrow\;\rangle.$$

These are not the familiar three-dimensional vectors from the world around us, these are just abstract vectors that we'll learn a lot more about soon. Suffice it to say, a silver atom in the state $|\uparrow\;\rangle$ will be and will have been deflected up by the Stern-Gerlach magnet. Similarly, $|\downarrow\;\rangle$ will be and will have been deflected down.

Before observation, from the detector's point of view, the physical state of a generic silver atom will be a combination of up and down,

$$|\mathrm{Ag}{\kern 0.5pt}\rangle = \alpha |\uparrow \;\rangle + \beta | \downarrow\;\rangle,$$

where $\alpha$ and $\beta$ are two constants that parametrize the physical state $|\mathrm{Ag}{\kern 0.5pt}\rangle$.

Mathematically, the probability finding the quantum state $|\mathrm{Ag}{\kern 0.5pt}\rangle$ deflected up, that is in the state $|\uparrow\;\rangle$ is $|\alpha|^{2}$. The probability of finding $|\mathrm{Ag}{\kern 0.5pt}\rangle$ deflected down is $|\beta|^{2}$. Because these are the only two options available to us, we have

\begin{equation}\label{ab}|\alpha|^{2} + |\beta|^{2} = 1.\end{equation}

Clearly $\alpha$ and $\beta$ will depend on $\vec{S}$. Over the next few episodes, we'll learn how to precisely relate the vectors $\vec{S}$ and $|\mathrm{Ag}{\kern 0.5pt}\rangle$. For our purposes today, let's just guess and check values for $\alpha$ and $\beta$.

Computing the Likelihoods

Clearly $\alpha$ and $\beta$ will depend on the angle $\varphi$, as it is the only meaningful, physical parameter in our problem. We might also try trigonometric functions, as  \eqref{ab} is very similar to the following relation:

\begin{equation}\label{trig}\cos^{2}x + \sin^{2}x = 1.\end{equation}

The na\"ive guess of $\alpha = \cos\varphi$ works well, but $\beta = \sin\varphi$ fails, because $\sin 90^{\circ} = 1$. For $\varphi = 90^{\circ}$, we must have an equal probability of finding the silver atom in either state $|\uparrow \;\rangle$ or $|\downarrow \;\rangle$.

The correct answer is given by a simple guess

$$\alpha = \cos\left(\frac{\varphi}{2}\right),\quad \beta = \sin\left(\frac{\varphi}{2}\right).$$

Because \eqref{trig} is generically true, \eqref{ab} is automatically satisfied. Moreover, if $\varphi = 90$ degrees, we have equal values

$$\alpha = \beta = \frac{1}{\sqrt{2}},$$

which implies equal probabilities

$$|\alpha|^{2} = |\beta|^{2} = \frac{1}{2}.$$

For the case of $\varphi = 45$, a quick calculation shows that

$$|\alpha|^{2} \sim 0.85,\quad |\beta|^{2}\sim 0.15,$$

which agrees with our observations from section 3.5.

Summary

We adjusted the Stern-Gerlach experimental setup to consider a different set of initial conditions. Rather than just randomly distributed atoms, we employed a uniform magnetic field to create a coherent beam of atoms whose spin angular momentum was all aligned in the same direction. Different choices of angle for the magnetic field lead to different observations on the detector screen.

Next time, we'll explore another variation, chaining multiple Stern-Gerlach experiments together.

Exercises

Exercise 1: In your own words, explain why the random, “thermal” initial conditions cause the beam to split when passed through the Stern-Gerlach magnet.

Exercise 2 : If instead of a silver atom, we passed an atom with spin angular momentum $h/2\pi$ - so that three angles can be measured - how could you arrange the uniform magnetic field to get three equal splits? How could you arrange it to get only one split. Is it possible to get the beam to split in two? If so, explain how.

Notes

$^{1}$: The magnetic dipole moment of the silver atom is generated by electrons. The electrons in silver are all paired up per the Pauli Exclusion Principle, except for the outermost, 5s electron. The paired electrons effectively cancel each other’s contribution to spin angular momentum out: each spin “up” is paired with a spin “down”. This is how atomic orbitals are filed out in chemistry. The nucleus in this case is not particularly relevant, as the only two naturally occurring isotopes 107Ag and 109Ag have nuclei that contribute nothing to the spin angular momentum. Other, unstable nuclei can! 95Ag - which lives for a little longer than a second, has nine times the angular momentum of its stable colleagues! This is certainly something of a happy accident.

$^{2}$: Here’s a convoluted comment for the pendants among us: The detector, in this instance, is a “classical” object, with more than sufficient information density (read: lots and lots of atoms) to handle being placed at an arbitrarily fine angular resolution with respect to the beam line. Of course, one might take umbrage with this assumption and demand that anything and everything in the universe must be quantum and therefore have a finite resolution to its information carrying capacity. In particular, you might try to argue that only a finite number of angles are available to the detector as well. While this is surely true - it is made up of atoms after all - it’s more than fair to say that we can place the detector at any angle up to a reasonable angular resolution for experimental purposes. Two or three decimal precision on the angle is more than enough to get our point across here. There is a reasonable way to discuss these kinds of issues - and we will - but not yet. Knowledge is built one step at a time. For now we can just operate on the theoretical assumption that detectors are classical.

$^{3}$: Exposing silver atoms to a strong magnet field still takes some time to get those atoms to align. The stronger the field, the less time it takes and the less strongly those thermally active silver atoms will deviate from the precise alignment. We assume in our discussion that all the silver atoms have become aligned with the magnetic field before they exit the oven.

$^{4}$: The name “ket” descends from a terrible pun forgivable only because its author was brilliant. We shall return to study ket vectors soon enough.