Spin

We begin by considering angular momentum in quantum mechanics. Angular momentum is defined classically as the cross product of position and momentum, that is \begin{equation} \ll = \rr \times \pp \label{classical} \end{equation} In quantum mechanics, momentum is replaced by a differential operator, that is \begin{equation} \pp \longmapsto -i\hbar\grad \end{equation} Inserting this into (\ref{classical}) we obtain the components of the angular momentum operator as \begin{align} L_x &= -i\hbar (y\partial_z - z\partial_y) \\ L_y &= -i\hbar (z\partial_x - x\partial_z) \\ L_z &= -i\hbar (x\partial_y - y\partial_x) \end{align}

These operators have the following characteristic properties. First of all, the commutator of any two is proportional to the third, for instance \begin{equation} [L_x,L_y] = L_x L_y - L_y L_x = i\hbar L_z \label{algebra} \end{equation} and cyclic permutations. The second property is more subtle. The total angular momentum operator is \begin{equation} L^2 = |\ll|^2 = \ll\cdot\ll \end{equation} When one separates variables in the Schrödinger equation in a central potential, the differential equation corresponding to the spherical angle $\theta$ reduces to an eigenvalue equation for the operator $L^2$. Imposing the boundary condition that the solutions should be well-behaved at the poles, the eigenvalues take the form $l(l+1)\hbar^2$, where $l$ is a non-negative integer.

The famous Stern-Gerlach experiment split a beam of atoms into separate beams for each of the $2l+1$ allowed angular momentum states. If $l$ is an integer, there should therefore be an odd number of beams. But the experiment produced 2 beams of silver atoms, thus showing that the angular momentum of the valence electron corresponds to $l=\frac12$. How can we describe this?

The commutation relations (\ref{algebra}) (and cyclic permutations) can also be represented using matrices. A particularly nice choice is given in terms of the Pauli matrices \begin{equation} \SIGMA_x = \begin{pmatrix}0& 1\cr \noalign{\smallskip} 1& 0\cr\end{pmatrix} \qquad \SIGMA_y = \begin{pmatrix}0& -i\cr \noalign{\smallskip} i&   0\cr\end{pmatrix} \qquad \SIGMA_z = \begin{pmatrix}1&   0\cr \noalign{\smallskip} 0& -1\cr\end{pmatrix} \end{equation} that were introduced in § 7.3. If we now introduce \begin{equation} \LL_a = {\hbar\over2} \, \SIGMA_a \label{spinop} \end{equation} where $a=x,y,z$, then \begin{equation} [\LL_x,\LL_y] = \LL_x \LL_y - \LL_y \LL_x = i\hbar \LL_z \end{equation} If we interpret the $\LL_m$ as the components of some sort of angular momentum, and compute the corresponding total angular momentum \begin{equation} \LL^2 = \LL_x^2 + \LL_y^2 + \LL_z^2 \end{equation} we discover that \begin{equation} \LL^2 = \frac34 \hbar^2 \II \end{equation} since each of the Pauli matrices squares to the identity matrix. But this corresponds to $l=\frac12$ (since $\frac34=\frac12(\frac12+1)$).

This new concept of angular momentum is however distinct from orbital angular momentum, and is therefore called spin angular momentum, or simply spin. 1)

Spinors represent states of spin-$\frac12$ particles such as the electron. Spinors are important because they are needed to give a spin-$\frac12$ representation of the angular momentum algebra (\ref{algebra}).

Here's a crash course in quantum mechanics. Physical states are represented by elements in a Hilbert space, which in this case consists of 2-component spinors. Any multiple of a given state $v$ represents the same physical state, so we usually consider only normalized states, that is, we demand that $v^\dagger v=1$. Operators such as $\LL_z$ act on the state $v$ by ordinary matrix multiplication, which takes $v$ to some other state, given in this case by $\LL_z v$. The expectation value of the operator $\LL_z$ in the state $v$ is given by $v^\dagger\LL_z v$, 2) and gives the (average) result of (a large number of) physical measurements, in this case of the $z$-component of spin.

There are special states which satisfy the eigenvalue equation \begin{equation} \LL_z v_\lambda = \lambda v_\lambda \end{equation} for some $\lambda$. The eigenvalues $\lambda$ are the only possible results of physical measurements; it is only in an eigenstate $v_\lambda$ that the measured value actually equals the expectation value. By requiring physical quantities (“observables”) to correspond to Hermitian matrices, we ensure that the result of any measurement is real, since the eigenvalues of Hermitian matrices are real—at least, over $\CC$. (See § 13 for a discussion of what happens over $\HH$ and $\OO$.)

What are the eigenstates of $\LL_z$? We have \begin{equation} \LL_z v_\pm = \pm{\hbar\over2} \, v_\pm \end{equation} where \begin{equation} v_+ = \begin{pmatrix}1\cr 0\cr\end{pmatrix} \qquad v_- = \begin{pmatrix}0\cr 1\cr\end{pmatrix} \end{equation} The expectation value of either $\LL_x$ or $\LL_y$ is $0$ in these states: A “spin-up” electron is equally likely to be “spin-left” as “spin-right”. Furthermore, $v_\pm$ are not eigenstates of either $\LL_x$ or $\LL_y$. Only operators which commute with each other correspond to observables which can be measured simultaneously. This means that it is not possible to simultaneously measure more than one component of angular momentum!

As discussed in § 7.3, the Pauli matrices are also closely related to rotations in three dimensions—as they should be if they are to describe angular momentum. Using (\ref{spinop}), we see that \begin{equation} \LL_z = i\hbar\,{dR_z\over d\alpha} \Bigg|_{\alpha=0} \end{equation} and similarly for $\LL_y$ and $\LL_x$. The spin operators are thus infinitesimal rotations; they live in the Lie algebra corresponding to the Lie group of three-dimensional rotations.

1) There is a well-known $3\times3$ representation of (\ref{algebra}), involving matrices of the form \[ \hbar\begin{pmatrix}0& -i& 0\cr i&  0& 0\cr 0&  0& 0\cr\end{pmatrix} \] It is easily checked that the total angular momentum operator in this case — the sum of the squares of the 3 matrices — is $2\hbar^2$ times the identity, corresponding to $l=1$; this is a spin-1 representation of (\ref{algebra}), corresponding to ordinary vectors in $\RR^3$.
2) Those familiar with the Dirac “bra/ket” notation can make the identifications \begin{align*} | v \rangle & \longleftrightarrow v \\ \langle v | & \longleftrightarrow v^\dagger \end{align*}

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