Stern-Gerlach experiment


The Stern-Gerlach experiment is an experiment that almost accidentally discovered the Spin of the electron. The true purpose of the experiment was to measure whether particles have an intrinsic angular momentum or not.

Theory

The idea behind the Stern-Gerlach experiment starts from the Bohr model of the hydrogen atom, which suggests that, since electrons orbit the nucleus, and rotating electric charges produce a magnetic field, then hydrogen atoms should, at least in principle, be a naturally occurring magnetic dipole with magnetic dipole moment determined exactly by the motion of the electron.

To analyze the situation, consider a circular loop of radius RR and area A=πR2A=\pi R^{2} traversed by an electric current II. This will be our "toy hydrogen atom".

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This loop has a magnetic dipole moment perpendicular to the loop given by μ=IAn^\boldsymbol{\mu}=IA \hat{\mathbf{n}}, where n^\hat{\mathbf{n}} is the outgoing normal vector from the surface. For a single charge qq moving circularly at a velocity vv on that radius, we have a current

I=λv=q2πRvI=\lambda v=\frac{q}{2\pi R} v

where λ\lambda is the linear charge density due to the charge particle. Of course, this density is highly variable in time, following the position of the particle. However, if we take the rotation of the particle to be very fast compared to the characteristic time of our system of interest1, we can consider λ\lambda to be the mean value over an entire rotation, so λ=q/2πR\lambda=q/2\pi R. If we plug this in the dipole moment formula we get

μ=IA=q2πRvπR2=12qvR\mu=IA=\frac{q}{2\pi R}v\pi R^{2}=\frac{1}{2}qvR

If the rotating charge has mass MM, then it must have angular momentum L=MvRL=MvR. We can invert this to find the velocity v=L/MRv=L/MR and by plugging this into the formula above

μ=q2ML\mu=\frac{q}{2M}L

This is a general relationship between magnetic dipole moment and angular momentum of a rotating charge. As it stands, we haven't used anything specific to atomic physics at all. This is just run of the mill electromagnetism. To make the connection to the hydrogen atom, we set the mass MM to be that of the electron mem_{e} and the charge to the Elementary charge ee2, but more importantly, we invoke the Bohr model. According to the third postulate of that model, the electron's angular momentum is discrete, so LL can't be any real number. We're "stuck" with a set of choices determined by an integer as L=nL=n\hbar, where \hbar is the reduced Planck constant. Substituting this in μ\mu leads to

μ=e2mn\mu=\frac{e}{2m}n\hbar

This leads to the natural conclusion that all possible dipole moments of the hydrogen atom are in units of

μB=e2m\boxed{\mu_{B}=\frac{e\hbar}{2m}}

This quantity is known as the Bohr magneton, and it allows us to find the magnetic dipole moment of a hydrogen atom as its multiples.

So far so good. What Stern and Gerlach instead asked is this: we know that an orbiting charged particle creates a magnetic dipole, but does a particle alone create a magnetic dipole? In other words, they asked if the spinning of a particle also created a dipole through the same mechanism.

In principle, the request makes sense. Any spinning charged object of nonzero radius will cause a charge density to move, thus triggering Ampere's law and creating a magnetic dipole. What wasn't known was which dipole moments could be created. Bohr showed that angular momentum appeared to be quantized, so "any real dipole moment" was no longer a safe bet. Their experiment wanted to answer this question.

Experiment

The experiment consisted of a source of silver atoms, which are then aimed and fired at through collimators to align the atoms into a thin beam. These atoms are then made to pass through two magnets (one with an exposed north pole, the other a south pole) irregularly shaped such that they create an inhomogeneous magnetic field B\mathbf{B}. The inhomogeneity produces a net force F=(μB)\mathbf{F}=\nabla(\boldsymbol{\mu}\cdot \mathbf{B}) on the atoms passing through3. The atoms then hit a detector that reveals their where they hit.

In the old way of thinking, there's no reason to think that we would see anything other than a smooth distribution of measured hits across the entire surface of the detector (see 4 in the figure). In the simple case of a hydrogen atom in the ground state, anything between μB\mu_{B} and μB-\mu_{B} would be fine, as it depends on how fast the atom is spinning. However, Bohr noted that this isn't the case. Only μB\mu_{B} and μB-\mu_{B} are allowed, but nothing in between. It's either spinning at a set (angular) speed in one direction, or at the same speed in the other. No in-between.

In fact, the observed distribution (see 5 in the figure) was only peaked around two points, at the two extremes, with near zero measurements everywhere else. Some atoms were being pushed up, other were being pushed down, but all of them with the same force, as they all landed near the same point. This confirmed Bohr's vision: there can only be discrete angular momenta. This behavior matched what would would expect out of a spinning particle and more. Not only did the atoms possess some sort of intrinsic magnetic dipole (and thus angular momentum), it was also quantized, just as Bohr had predicted. This intrinsic angular momentum was, in reference to the classical counterpart, named Spin.

A correction

The formula we derived for μ\mu works well enough for orbital momentum, but displays issues for spin. In reality, for spin specifically, this relationship is not quite correct, but the proof is not within the scope or historical context of this article. Quantum mechanics tells us that there is a factor, called the (electron spin) g-factor gg, that rescales this equation, such that

μe=geSe2me\mu_{e}=g\frac{eS_{e}}{2m_{e}}

This is the actually correct form of the equation, shown for the electron, where we replaced nn\hbar with the more general spin SeS_{e}. It is found that this number is approximately g=2g=2 (a couple thousandths more than 2, actually).

A more general form involving the total angular momentum J\mathbf{J} (not the current density!) gives us the total magnetic dipole moment of the system:

μ=gμBJ\boldsymbol{\mu}=-g\mu_{B} \frac{\mathbf{J}}{\hbar}

Footnotes

  1. In the context of electrons, this is a fair approximation. Even in classical descriptions, electrons move at relativistic speeds around the nucleus.

  2. Technically it should be e-e, but the sign doesn't matter: we can just make the electron orbit the other way and the sign disappears anyway, so we might as well ignoring from the start.

  3. Note that the inhomogeneity is obligatory, as since we assume that μ=0\nabla \boldsymbol{\mu}=0 (μ\boldsymbol{\mu} is a constant of the atom), the only term that can produce a nonzero force is B\nabla \mathbf{B}. A uniform field as zero Gradient everywhere, so no force would be felt.