Spin ${1\over 2}$

Earlier, we showed that both integer and half integer angular momentum could satisfy the commutation relations for angular momentum operators but that there is no single valued functional representation for the half integer type.

Some particles, like electrons, neutrinos, and quarks have half integer internal angular momentum, also called spin. We will now develop a spinor representation for spin ${1\over 2}$. There are no coordinates $\theta$ and $\phi$ associated with internal angular momentum so the only thing we have is our spinor representation.

Electrons, for example, have total spin one half. There are no spin 3/2 electrons so there are only two possible spin states for an electron. The usual basis states are the eigenstates of $S_z$. We know from our study of angular momentum, that the eigenvalues of $S_z$ are $+{1\over 2}\hbar$ and $-{1\over 2}\hbar$. We will simply represent the $+{1\over 2}\hbar$ eigenstate as the upper component of a 2-component vector. The $-{1\over 2}\hbar$ eigenstate amplitude is in the lower component. So the pure eigenstates are.

$$\begin{eqnarray*} \chi_+=\pmatrix{1\cr 0} \\ \chi_-=\pmatrix{0\cr 1} \end{eqnarray*}$$

An arbitrary spin one half state can be represented by a spinor.

$$\bgroup\color{black} \chi=\pmatrix{a\cr b} \egroup$$

with the normalization condition that $\vert a\vert^2+\vert b\vert^2=1$.

It is easy to derive the matrix operators for spin.

$$\bgroup\color{black}S_x ={\hbar\over 2} \left(\matrix{0 &1\cr... ...z ={\hbar\over 2} \left(\matrix{1 &0\cr 0 &-1\cr}\right)\egroup$$

These satisfy the usual commutation relations from which we derived the properties of angular momentum operators. For example lets calculate the basic commutator.

$$\begin{eqnarray*}[S_x,S_y]&=&{\hbar^2\over 4} \left[\pmatrix{0 &1\cr 1 &0\cr}\pm... ...i\cr}=i\hbar {\hbar\over 2}\pmatrix{1 &0\cr 0 &-1\cr}=i\hbar S_z \end{eqnarray*}$$

The spin operators are an (axial) vector of matrices. To form the spin operator for an arbitrary direction $\hat{u}$, we simply dot the unit vector into the vector of matrices.

$$\bgroup\color{black} S_u=\hat{u}\cdot\vec{S} \egroup$$

The Pauli Spin Matrices, $\sigma_i$, are simply defined and have the following properties.

$$\begin{eqnarray*} S_i &\equiv & {\hbar\over 2} \sigma_i \\ \vec{S}&=&{\hbar\ove... ...sigma_j\right] & = & 2i\epsilon_{ijk}\sigma_k \\ \sigma_i^2&=&1 \end{eqnarray*}$$

They also anti-commute.

$$\begin{eqnarray*} \sigma_x\sigma_y=-\sigma_y\sigma_x \ \sigma_x\sigm... ...gma_y=-\sigma_y\sigma_z \\ \{ \sigma_i,\sigma_j\}=2\delta_{ij} \end{eqnarray*}$$

The $\sigma$ matrices are the Hermitian, Traceless matrices of dimension 2. Any 2 by 2 matrix can be written as a linear combination of the $\sigma$ matrices and the identity.

* Example: The expectation value of $S_x$.*
* Example: The eigenvectors of $S_x$.*
* Example: The eigenvectors of $S_y$.*

The (passive) rotation operators, for rotations of the coordinate axes can be computed from the formula $R_i(\theta_i)=e^{iS_i\theta_i/\hbar}$.

$$\bgroup\color{black}R_z(\theta)=\left(\matrix{e^{i\theta/2}&0... ...}\cr -\sin{\theta\over 2}&\cos{\theta\over 2}}\right)\egroup$$

Note that the operator for a rotation through $2\pi$ radians is minus the identity matrix for any of the axes (because ${\theta\over 2}$ appears everywhere). The surprising result is that the sign of the wave function of all fermions is changed if we rotate through 360 degrees.

* Example: The eigenvectors of $S_u$.*

As for orbital angular momentum ( $\vec{L}$), there is also a magnetic moment associated with internal angular momentum ( $\vec{S}$).

$$\bgroup\color{black}\vec{\mu}_{spin}=-{eg\over 2mc} \vec{S}\egroup$$

This formula has an additional factor of $g$, the gyromagnetic ratio, compared to the formula for orbital angular momenta. For point-like particles, like the electron, $g$ has been computed in Quantum ElectroDynamics to be a bit over 2, $g=2+{\alpha\over\pi}+...$. For particles with structure, like the proton or neutron, $g$ is hard to compute, but has been measured. Because the factor of 2 from $g$ cancels the factor of 2 from $s={1\over 2}$, the magnetic moment due to the spin of an electron is almost exactly equal to the magnetic moment due to the orbital angular momentum in an $\ell=1$ state. Both are 1 Bohr Magneton, $\mu_B={e\hbar\over 2mc}$.

$$\bgroup\color{black}H=-\vec{\mu}\cdot\vec{B}={eg\hbar\over 4mc}\vec{\sigma}\cdot\vec{B}=\mu_B\vec{\sigma}\cdot\vec{B}\egroup$$

If we choose the $z$ axis to be in the direction of $B$, then this reduces to

$$\bgroup\color{black}H=\mu_BB\sigma_z.\egroup$$

* Example: The time development of an arbitrary electron state in a magnetic field.*
* Example: Nuclear Magnetic Resonance (NMR and MRI).*

A beam of spin one-half particles can also be separated by a Stern-Gerlach apparatus which uses a large gradient in the magnetic field to exert a force on particles proprtional to the component of spin along the field gradient. Thus, we can measure the component of spin along a direction we choose. A field gradient will separate a beam of spin one-half particles into two beams. The particles in each of those beams will be in a definite spin state, the eigenstate with the component of spin along the field gradient direction either up or down, depending on which beam the particle is in.

We may represent a Stern-Gerlach appartatus which blocks the lower beam by the symbol below.

$$\bgroup\color{black} \left\{\matrix{+\cr - \vert}\right\}_z\rightarrow \egroup$$

This apparatus is equivalent to the operator that projects out the $+{\hbar\over 2}$ eigenstate.

$$\bgroup\color{black} \vert+\rangle\;\langle+\vert = \pmatrix{1\cr 0}\pmatrix{1^* & 0^*}=\pmatrix{1 & 0\cr 0 &0} \egroup$$

We can perform several thought experiments. The appartus below starts with an unpolarized beam. In such a beam we don't know the state of any of the particles. For a really unpolarized beam, half of the particles will go into each of the separated beams. (Note that an unpolarized beam cannot be simply represented by a state vector.) In the apparatus below, we block the upper beam so that only half of the particles come out of the first part of the apparatus and all of those particles are in the definite state having spin down along the z axis. The second part of the apparatus blocks the lower separated beam. All of the particles are in the lower beam so nothing is left coming out of the apparatus.

$$\bgroup\color{black} \mathrm{Unpolarized Beam (N particles... ...rrow \left\{\matrix{+\cr -\vert}\right\}_z\rightarrow 0 \egroup$$

The result is unaffected if we insert an additional apparatus that separates in the x direction in the middle of the apparatus above. While the apparatus separates, neither beam is blocked (and we assume we cannot observe which particles go into which beam). This apparatus does not change the state of the beam!

$$\bgroup\color{black} \mathrm{(N particles)}\rightarrow \lef... ...rrow \left\{\matrix{+\cr - \vert}\right\}_z\rightarrow 0\egroup$$

Now if we block one of the beams separated according to the x direction, particles can get through the whole apparatus. The middle part of the apparatus projects the state onto the positive eigenstate of $S_x$. This state has equal amplitudes to have spin up and spin down along the z direction. So now, 1/8 of the particles come out of the apparatus. By blocking one beam, the number of particles coming out increased from 0 to $N/8$. This seems a bit strange but the simple explanation is that the upper and lower beams of the middle part of the apparatus were interfering to give zero particles. With one beam blocked, the inteference is gone.

$$\bgroup\color{black} (N)\rightarrow \left\{\matrix{+ \vert\c... ...ght\}_z\rightarrow {N\over 8}\pmatrix{1\cr 0}\rightarrow\egroup$$

Note that we can compute the number of particles coming out of the second (and third) part by squaring the amplitude to go from the input state to the output state

$$\bgroup\color{black} {N\over 2}\left\vert\pmatrix{{1\over\sqr... ...ver\sqrt{2}}}\pmatrix{0\cr 1}\right\vert^2 = {N\over 4} \egroup$$

or we can just use the projection operator $\pmatrix{{1\over\sqrt{2}}\cr {1\over\sqrt{2}}}\pmatrix{{1\o... ...\sqrt{2}}}= \pmatrix{{1\over 2} & {1\over 2}\cr {1\over 2} & {1\over 2}}$.

$$\bgroup\color{black} \pmatrix{{1\over 2} & {1\over 2}\cr {1\o... ...N\over 4}\pmatrix{{1\over\sqrt{2}}\cr {1\over\sqrt{2}}} \egroup$$