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Stern–Gerlach experiment

Experiment using particles with +1⁄2 or -1⁄2 spin

If the experiment is conducted using charged particles like electrons, there will be a Lorentz force that tends to bend the trajectory in a circle (see cyclotron motion). This force can be cancelled by an electric field of appropriate magnitude oriented transverse to the charged particle's path.

Spin values for fermions

Electrons are spin-1⁄2 particles. These have only two possible spin angular momentum values measured along any axis, $+{\frac {h}{2}}$ or $-{\frac {h}{2}}$, a purely quantum mechanical phenomenon. Because its value is always the same, it is regarded as an intrinsic property of electrons, and is sometimes known as "intrinsic angular momentum" (to distinguish it from orbital angular momentum, which can vary and depends on the presence of other particles).

We can now seek to depict this experiment with spin $+{\frac {1}{2}}$ particles mathematically, it is easiest to use Dirac's bra–ket notation. Reviewing the data from the detector which shows two discrete clumps of atoms in the vertical plane we shall label the z-plane. Given that the deflection of the atoms is proportional to it's angular momentum these two discrete deflections equidistant from the meridian give us 4 conclusions:

1. The angular momentum of silver atoms is quantized into 2 possibilities varying in direction but not magnitude -- The central result of the experiment.
2. The angular momentum of the silver particle did not seem to change as it passed through the apparatus.
3. They showed equal deflection from the median, the angular momentum vary in sign but not magnitude.
4. We can assume the silver atoms coming out of the oven had isotropic spin since about half subsequently had up spin and about half down spin any spin state can be expressed in terms of some combination of spin up and spin down.

We can imagine that along the atoms path it felt a continuous consistent influence of the magnetic field due to it's spin. This tells us that once the spin was established as either up or down in the z-axis it did not change, and an up state therefore could not transition to a down state while in the apparatus. We will represent this as Orthogonal States in a vector representation of the possible spins of the silver atom. For convenience we will label these states, previously called up and down, + and -. Atoms with a positive angular momentum will be described as in the ket state $|+\rangle$ and $|-\rangle$ will be negative.

The general state of spin in 3 dimensions could be depicted as a linear combination of these two "basis" vectors

$|\psi \rangle =c_{1}\left|+\right\rangle +c_{2}\left|-\right\rangle$

where $c_{1},c_{2}$ are complex and satisfy $|c_{1}|^{2}+|c_{2}|^{2}=1$

Since this a linear combination we are free to pick that values of the coefficients but the ration is determined by the state of the system. It is a principle of quantum mechanics that the square of the magnitudes can be interpreted as a probability thus we would like it to lie between 0 and 1 to make calculations simpler. The reason we use the square of the magnitude rather than the complex numbers themselves is probabilities are real and non-negative, but by using complex numbers we gain more freedom in the state-space.

In quantum mechanics observables are the eigenvalues of some operator, a concrete representation for spin the operators are given by the Pauli matrices, we can take the any of these matrices but $\sigma _{3}$ is by convention used in the Z-direction ${\begin{aligned}\sigma _{z}&={\begin{pmatrix}1&0\\0&-1\end{pmatrix}}\end{aligned}}$ to be the spin operator in the axis relevant to our experiment. It has two eigenvalues +1 and -1, as required by us since we have 2 orthogonal observables. The eigenvectors are $(1,0),(0,1)$. Since the dot product of these vectors is 0 and they are therefore orthogonal, meeting our requirements in this regard too. We can now also give a concrete representations of the states $|+\rangle ={\begin{pmatrix}1\\0\end{pmatrix}},|-\rangle ={\begin{pmatrix}0\\1\end{pmatrix}}$.

Continuing with creating concrete vectors for the states we can now calculate a representation for $|\psi \rangle =c_{1}{\binom {1}{0}}+c_{2}{\binom {0}{1}}={\binom {c_{1}}{c_{2}}}$

We can now calculate the probability of the spins of the silver atom

${\begin{aligned}|\langle +|\psi \rangle |^{2}=|(1,0){\binom {c_{1}}{c_{2}}}|^{2}=|c_{1}|^{2}\\|\langle -|\psi \rangle |^{2}=|(0,1){\binom {c_{1}}{c_{2}}}|^{2}=|c_{2}|^{2}\end{aligned}}$