Spin | Physics - Wyatt's Notes

Attempt these from memory before reading.

• Write down the three Pauli matrices.
• What are their eigenvalues, and what is $[\sigma_x, \sigma_y]$?
• How many independent parameters specify a general spin-1/2 state?
• What does the Stern-Gerlach experiment actually demonstrate?
• After a $360^\circ$ rotation, does a spin-1/2 state return to itself?

Check your answers against the text. Prompt 5 is the one that is almost always answered wrong, and it is not a formality: the sign change is measurable.

The angular momentum page argued that the algebra, not the geometry, is the physics. Spin is where that claim stops being a slogan, because the geometric route is unavailable.

For orbital angular momentum there is an object to rotate: $\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$. For spin there is no such object — an electron is not a spinning ball, and the naive picture fails twice over. Treating the electron as a sphere of radius $\sim e^2/mc^2$ rotating fast enough to produce the observed magnetic moment puts a point on its surface well past the speed of light. And a point particle has no surface to rotate.

So spin is postulated as an independent degree of freedom, and the only structure it is required to have is the angular momentum algebra. From $[S_x, S_y] = i\hbar S_z$ alone, plus Hermiticity and tracelessness in two dimensions, the Pauli matrices follow — and the derivation in section 7.3 does exactly that.

Two things to notice in that derivation. First, it produces $2\times2$ matrices without anyone choosing a dimension: the smallest non-trivial representation of the algebra is two-dimensional, which is what “spin-1/2” means. Second, it produces three Hermitian, traceless, unitary matrices whose squares are the identity — a set with no classical counterpart, because no classical quantity squares to one while failing to commute with the others.

The Bloch sphere. A general spin-1/2 state is

\lvert\psi\rangle = \cos\frac{\theta}{2}\lvert\uparrow\rangle + e^{i\phi}\sin\frac{\theta}{2}\lvert\downarrow\rangle ,

which is two angles — a point on a sphere. The global phase has already been divided out, so the relative phase $\phi$ is the only phase left, and it is half the story of what a qubit is.

Spin is an intrinsic form of angular momentum with no classical analogue. For spin-$1/2$ particles (e.g., electrons):

\hat{S}_x = \frac{\hbar}{2}\sigma_x, \quad \hat{S}_y = \frac{\hbar}{2}\sigma_y, \quad \hat{S}_z = \frac{\hbar}{2}\sigma_z

Where $\sigma_x, \sigma_y, \sigma_z$ are the Pauli matrices:

\sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix}, \quad \sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix}

\sigma_i^2 = I, \quad \sigma_i \sigma_j = i\epsilon_{ijk}\sigma_k \quad (i \neq j)

[\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k, \quad {\sigma_i, \sigma_j} = 2\delta_{ij}I

Spin states: $|\uparrow\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (spin up, $m_s = +1/2$) and $|\downarrow\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ (spin down, $m_s = -1/2$).

The Pauli matrices are uniquely determined (up to unitary equivalence) by the angular momentum algebra For $j = 1/2$.

Requirements. We seek $2 \times 2$ matrices $\sigma_x, \sigma_y, \sigma_z$ such that:

1. $\sigma_i^2 = I$ (eigenvalues are $\pm 1$ Corresponding to $S_z = \pm\hbar/2$)
2. $\sigma_i^\dagger = \sigma_i$ (Hermitian)
3. $\mathrm{Tr}(\sigma_i) = 0$ (traceless, since eigenvalues sum to zero)
4. $[\sigma_x, \sigma_y] = 2i\sigma_z$ (and cyclic permutations)

Step 1: Fix $\sigma_z$. A traceless Hermitian matrix with eigenvalues $\pm 1$ is:

\sigma_z = \begin{pmatrix} 1 & 0 \ 0 & -1 \end{pmatrix}

(up to an overall unitary transformation, which corresponds to choosing the quantisation axis).

Step 2: Determine $\sigma_x$. A general traceless Hermitian matrix is:

\sigma_x = \begin{pmatrix} a & b \ b^* & -a \end{pmatrix}

Where $a \in \mathbb{R}$ and $b \in \mathbb{C}$. From $\sigma_x^2 = I$: $a^2 + |b|^2 = 1$ and $2ab = 0$. Since $|b| \neq 0$ (otherwise $\sigma_x$ is diagonal and commutes with $\sigma_z$ Violating $[\sigma_x, \sigma_z] \neq 0$), we have $a = 0$ and $|b| = 1$. Choosing $b = 1$ (by convention):

\sigma_x = \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}

Step 3: Determine $\sigma_y$. From $[\sigma_x, \sigma_y] = 2i\sigma_z$:

\sigma_x\sigma_y - \sigma_y\sigma_x = 2i\sigma_z

Writing $\sigma_y = \begin{pmatrix} c & d \\ d^* & -c \end{pmatrix}$ and imposing $\sigma_y^2 = I$, $\sigma_y^\dagger = \sigma_y$ And the commutation relation, we find $c = 0$, $|d| = 1$ And the commutator gives $d - d^* = 2i$ So $d = i$:

\sigma_y = \begin{pmatrix} 0 & -i \ i & 0 \end{pmatrix}

$\blacksquare$

A particle with magnetic moment $\boldsymbol{\mu} = \gamma\mathbf{S}$ (where $\gamma$ is the Gyromagnetic ratio) in a magnetic field $\mathbf{B} = B_0\hat{z}$ has Hamiltonian:

\hat{H} = -\boldsymbol{\mu}\cdot\mathbf{B} = -\gamma B_0\hat{S}_z = -\frac{\hbar\gamma B_0}{2}\sigma_z

The eigenstates are $|\uparrow\rangle$ and $|\downarrow\rangle$ with energies $E_\uparrow = -\hbar\gamma B_0/2$ And $E_\downarrow = +\hbar\gamma B_0/2$. The energy splitting is $\Delta E = \hbar\gamma B_0$.

Time evolution. For an arbitrary initial state:

|\psi(0)\rangle = \alpha|\uparrow\rangle + \beta|\downarrow\rangle

The state at time $t$ is:

|\psi(t)\rangle = \alpha e^{i\gamma B_0 t/2}|\uparrow\rangle + \beta e^{-i\gamma B_0 t/2}|\downarrow\rangle

Larmor precession. The expectation values precess around the $z$-axis:

\langle S_x \rangle(t) = \frac{\hbar}{2}(\alpha^\beta,e^{-i\gamma B_0 t} + \alpha\beta^,e^{i\gamma B_0 t})