Spin | Physics - Wyatt's Notes

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:

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

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

$\langle S_z \rangle(t) = \frac{\hbar}{2}(|\alpha|^2 - |\beta|^2) = \mathrm{const}.$

The spin precesses at the Larmor frequency $\omega_L = \gamma B_0$.

For an electron, $\gamma = -e/(m_e)$ (negative charge), giving $\omega_L = eB_0/m_e$.

The Larmor frequency. For a typical laboratory field $B_0 = 1$ T:

$\omega_L = \frac{(1.602 \times 10^{-19})(1)}{9.109 \times 10^{-31}} = 1.76 \times 10^{11}\;\mathrm{rad}/s$

Corresponding to a frequency $\nu_L = \omega_L/(2\pi) = 28$ GHz (microwave range). This is the basis Of Electron Spin Resonance (ESR) and Nuclear Magnetic Resonance (NMR) spectroscopy, where transitions Between spin states are driven by oscillating magnetic fields at the Larmor frequency.

Example 7.1. An electron starts in the state $|\psi(0)\rangle = |\uparrow\rangle$. A magnetic field $\mathbf{B} = B_0\hat{x}$ is applied. Find $|\psi(t)\rangle$.

A beam of silver atoms passes through an inhomogeneous magnetic field and splits into two beams, Confirming the quantisation of angular momentum (spin-1/2 for the outer electron).

Detailed analysis. The force on a magnetic moment in an inhomogeneous field is:

$\mathbf{F} = \nabla(\boldsymbol{\mu}\cdot\mathbf{B})$

For a field $\mathbf{B} = B(z)\hat{z}$ with $\partial B_z/\partial z \neq 0$The $z$-component of force Is $F_z = \mu_z\,\partial B_z/\partial z$. Since $\mu_z = \gamma m_s\hbar$ and $m_s = \pm 1/2$:

$F_z = \pm\frac{\gamma\hbar}{2}\frac{\partial B_z}{\partial z}$

The beam splits into two, corresponding to $m_s = +1/2$ (deflected up) and $m_s = -1/2$ (deflected down).

Sequential Stern-Gerlach measurements. Consider three apparatuses in sequence:

1. First SG-Z: selects $|\uparrow\rangle$.
2. Second SG-X: splits into $|+\rangle_x$ and $|-\rangle_x$ with equal probability $1/2$.
3. Third SG-Z (on the $|-\rangle_x$ beam): again splits into $|\uparrow\rangle$ and $|\downarrow\rangle$ with equal probability $1/2$.

This demonstrates that the intermediate $S_x$ measurement erases the information about the Original $S_z$ state. The probabilities reflect the non-commutativity $[\hat{S}_x, \hat{S}_z] = i\hbar\hat{S}_y$.

Example 7.2. A spin-1/2 particle passes through SG-Z (selecting $|\uparrow\rangle$), then through SG-Z at angle $\theta$ from the $z$-axis. Find the probability of measuring $+1$ in the second Apparatus.

flowchart TD
    A[7_Spin] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Spin is the universe’s internal compass that has no classical counterpart. An electron is not literally spinning like a top, yet it carries angular momentum and a magnetic moment as if it were. The Pauli matrices encode the algebra of this internal angular momentum in a neat two-by-two package. When you put a spin in a magnetic field, it precesses like a wobbling top, but the energy comes in only two values: aligned or anti-aligned. The Stern-Gerlach experiment is the smoking gun: a beam of atoms splits into exactly two beams, not a continuous spread, proving that angular momentum is quantized. Measuring spin along one axis completely randomizes the result along a perpendicular axis, which is the quantum version of the uncertainty principle at work. Adding two spins together gives either a triplet (symmetric) or a singlet (antisymmetric), and the singlet state is so entangled that measuring one instantly tells you about the other, no matter how far apart they are.

Given two angular momenta $\hat{\mathbf{J}}_1$ and $\hat{\mathbf{J}}_2$ with quantum numbers $j_1, m_1$ and $j_2, m_2$Define the total $\hat{\mathbf{J}} = \hat{\mathbf{J}}_1 + \hat{\mathbf{J}}_2$.

Compatible observables: $\hat{J}^2$, $\hat{J}_z$, $\hat{J}_1^2$, $\hat{J}_2^2$ all commute. We label Simultaneous eigenstates as $|j_1, j_2; j, m\rangle$.

Clebsch-Gordan decomposition. The total angular momentum quantum numbers range over:

$j = |j_1 - j_2|, |j_1 - j_2| + 1, \ldots, j_1 + j_2$

In integer steps. For each $j$The magnetic quantum number $m$ ranges from $-j$ to $j$.

The transformation between the product basis and the total-$j$ basis is:

$|j_1, j_2; j, m\rangle = \sum_{m_1, m_2} C(j_1\,m_1\,j_2\,m_2|j\,m)\,|j_1, m_1\rangle|j_2, m_2\rangle$

Where $C(j_1\,m_1\,j_2\,m_2|j\,m)$ are the Clebsch-Gordan coefficients.

Two spin-1/2 particles. The composite system has $j_1 = j_2 = 1/2$. The possible total spins are:

• Triplet ($j = 1$): three states with $m = 1, 0, -1$ $|1,1\rangle = |\uparrow\uparrow\rangle$ $|1,0\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle)$ $|1,-1\rangle = |\downarrow\downarrow\rangle$

• Singlet ($j = 0$): one state with $m = 0$ $|0,0\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)$

The triplet states are symmetric under particle exchange; the singlet is antisymmetric.

Total spin operator. $\hat{S}^2 = \hat{S}_1^2 + \hat{S}_2^2 + 2\hat{\mathbf{S}}_1\cdot\hat{\mathbf{S}}_2$ So:

$\hat{\mathbf{S}}_1\cdot\hat{\mathbf{S}}_2 = \frac{1}{2}(\hat{S}^2 - \hat{S}_1^2 - \hat{S}_2^2)$

For the triplet: $\hat{\mathbf{S}}_1\cdot\hat{\mathbf{S}}_2 = \hbar^2/4$. For the singlet: $\hat{\mathbf{S}}_1\cdot\hat{\mathbf{S}}_2 = -3\hbar^2/4$.

Complete set of commuting observables (CSCO). For a two-spin system, the set $\\{\hat{S}^2, \hat{S}_z, \hat{S}_1^2, \hat{S}_2^2\\}$ forms a CSCO: their simultaneous eigenstates are Uniquely labelled by the quantum numbers $(s, m_s, s_1, s_2)$. An alternative CSCO is $\\{\hat{S}_{1z}, \hat{S}_{2z}, \hat{S}_1^2, \hat{S}_2^2\\}$Which uses the product basis. The Clebsch-Gordan coefficients are the transformation matrix between these two bases.

Clebsch-Gordan table for $j_1 = j_2 = 1/2$:

Example 7.3. Two electrons are in the singlet state. If electron 1 is measured to have $S_z = +\hbar/2$What is the state of electron 2 immediately after? What is the probability of Measuring $S_x = +\hbar/2$ for electron 2?

Mistake 1: Confusing probability with probability amplitude. In quantum mechanics, the probability of a measurement outcome is the square of the absolute value of the probability amplitude. Do not confuse the amplitude (a complex number) with the probability (a real number between 0 and 1).

Mistake 2: Assuming that spin is a classical angular momentum. Spin is an intrinsic quantum property with no classical analogue. It does not correspond to physical rotation of the particle. Do not think of spin as the particle spinning like a top.

Mistake 3: Forgetting that measuring spin along one axis randomizes the result along a perpendicular axis. If a particle is in a definite state of $S_z$, measuring $S_x$ will give a random result (with equal probability for $+\hbar/2$ and $-\hbar/2$). The intermediate measurement destroys the information about the original state. Do not assume that measuring one component of spin tells you about the other components.

Mistake 4: Confusing the singlet and triplet states. The singlet state $|0,0\rangle$ is antisymmetric under particle exchange, while the triplet states $|1,m\rangle$ are symmetric. Do not confuse the two; they have different symmetry properties and different total spin.

Mistake 5: Assuming that spin-1/2 particles have only two states. A single spin-1/2 particle has two states ($|\uparrow\rangle$ and $|\downarrow\rangle$), but a system of two spin-1/2 particles has four states (triplet and singlet). Do not assume that the number of states is always two; the answer varies based on on the number of particles.

• Angular Momentum and the Hydrogen Atom: Orbital angular momentum provides the foundation for understanding spin as an intrinsic angular momentum.

• Operators and Observables: Operators represent physical observables and their commutation relations determine spin algebra.

• Identical Particles and Exchange Symmetry: Spin statistics connect spin to particle exchange symmetry and quantum statistics.

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