Angular Momentum, Spin-½ Systems, and the Stern-Gerlach Experiment
Angular Momentum, Spin-½ Systems, and the Stern-Gerlach Experiment
Angular momentum occupies a central role in quantum mechanics, both as a conserved quantity arising from rotational symmetry and as the primary organizer of atomic structure. In classical mechanics, angular momentum is simply $\mathbf{L} = \mathbf{r} \times \mathbf{p}$, a smooth continuous vector that can take any value. In quantum mechanics, this familiar object is promoted to an operator $\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$, and the consequences of this promotion are profound: angular momentum becomes quantized, taking only discrete values, and its three components cannot simultaneously be measured with arbitrary precision.
The quantization of angular momentum is not imposed by hand but emerges inevitably from the algebraic structure of the commutation relations. The three components of $\hat{\mathbf{L}}$ satisfy $[\hat{L}_x, \hat{L}_y] = i\hbar\hat{L}_z$ (and cyclic permutations), and this algebra alone — without any reference to differential equations or wave functions — forces the eigenvalues of $\hat{L}^2$ to be $\hbar^2 l(l+1)$ where $l$ is a non-negative integer or half-integer, and the eigenvalues of $\hat{L}_z$ to be $m\hbar$ where $m \in \{-l, -l+1, \ldots, l-1, l\}$. The power of this algebraic approach cannot be overstated: it reveals that angular momentum quantization is a theorem of Lie algebra, not merely a peculiarity of differential equations.
Beyond orbital angular momentum, nature provides an intrinsic angular momentum called spin that has no classical analogue. The electron carries spin-$\frac{1}{2}$, meaning its spin quantum number is $s = \frac{1}{2}$ and its spin along any axis takes only two values: $+\hbar/2$ (spin-up, $|\uparrow\rangle$) or $-\hbar/2$ (spin-down, $|\downarrow\rangle$). This binary nature is described by the Pauli matrices and the two-dimensional Hilbert space $\mathbb{C}^2$. A general spin-½ state, called a spinor, is a superposition $|\psi\rangle = \alpha|\uparrow\rangle + \beta|\downarrow\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$.
The Stern-Gerlach experiment of 1922 provided the first direct evidence for space quantization and spin. Silver atoms (with one unpaired outer electron) were sent through an inhomogeneous magnetic field. Classical physics predicts a continuous smearing of the beam; instead, exactly two discrete spots appeared on the detector screen, corresponding to $m_s = \pm\frac{1}{2}$. This result was historically puzzling because the $l = 0$ ground state of silver should show no deflection at all — it was only later understood that the splitting arises from the intrinsic spin of the electron, not orbital angular momentum.
The geometry of spin states is elegantly captured by the Bloch sphere, a unit sphere in $\mathbb{R}^3$ where every pure state of a qubit (two-level system) corresponds to a unique point on the surface. The north pole represents $|\uparrow\rangle$, the south pole $|\downarrow\rangle$, and the general state $|\psi\rangle = \cos(\theta/2)|\uparrow\rangle + e^{i\phi}\sin(\theta/2)|\downarrow\rangle$ is parameterized by the polar angle $\theta \in [0,\pi]$ and the azimuthal angle $\phi \in [0,2\pi)$. The expectation values $\langle S_x \rangle$, $\langle S_y \rangle$, $\langle S_z \rangle$ give the Cartesian coordinates of the Bloch vector scaled by $\hbar/2$, providing a visual and geometric intuition for superposition, measurement, and unitary evolution.
When orbital and spin angular momenta are both present, they combine to give the total angular momentum $\hat{\mathbf{J}} = \hat{\mathbf{L}} + \hat{\mathbf{S}}$, which satisfies the same $\mathfrak{su}(2)$ algebra. The combined Hilbert space is a tensor product, and the eigenstates of $\hat{J}^2$ and $\hat{J}_z$ are related to the product states $|l, m_l\rangle \otimes |s, m_s\rangle$ through the Clebsch-Gordan coefficients $\langle j, m | l, m_l; s, m_s\rangle$. These coefficients encode the rules for how angular momenta add in quantum mechanics and appear throughout atomic, nuclear, and particle physics whenever composite systems are analyzed.
Spin precession under a magnetic field is described by the Larmor precession: a spin-½ particle in a field $\mathbf{B} = B_0\hat{z}$ evolves under the Hamiltonian $H = -\gamma \mathbf{B} \cdot \hat{\mathbf{S}} = -\gamma B_0 \hat{S}_z$, causing the Bloch vector to precess about the $z$-axis at the Larmor frequency $\omega_0 = \gamma B_0$. This precession underlies magnetic resonance imaging (MRI) and forms the basis of NMR spectroscopy, two of the most powerful tools in modern science and medicine.
In position representation, the orbital angular momentum operator is $\hat{\mathbf{L}} = \hat{\mathbf{r}} \times \hat{\mathbf{p}}$ with components:
$$\hat{L}_x = \hat{y}\hat{p}_z - \hat{z}\hat{p}_y, \quad \hat{L}_y = \hat{z}\hat{p}_x - \hat{x}\hat{p}_z, \quad \hat{L}_z = \hat{x}\hat{p}_y - \hat{y}\hat{p}_x$$
These satisfy the fundamental commutation relations $[\hat{L}_i, \hat{L}_j] = i\hbar\,\epsilon_{ijk}\hat{L}_k$ where $\epsilon_{ijk}$ is the Levi-Civita symbol. The Casimir operator $\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$ commutes with all three components: $[\hat{L}^2, \hat{L}_i] = 0$.
The raising and lowering operators $\hat{L}_\pm = \hat{L}_x \pm i\hat{L}_y$ satisfy $[\hat{L}_z, \hat{L}_\pm] = \pm\hbar\hat{L}_\pm$ and act on simultaneous eigenstates $|l, m\rangle$ of $\hat{L}^2$ and $\hat{L}_z$ as:
$$\hat{L}_\pm|l,m\rangle = \hbar\sqrt{l(l+1) - m(m\pm 1)}\,|l, m\pm 1\rangle$$
The quantum numbers satisfy $l \in \{0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots\}$ and $m \in \{-l, -l+1, \ldots, l-1, l\}$, giving $2l+1$ values of $m$ for each $l$. For orbital angular momentum, $l$ must be a non-negative integer.
The spin-½ operators are $\hat{S}_i = \frac{\hbar}{2}\sigma_i$ where the Pauli matrices are:
$$\sigma_x = \begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}, \quad \sigma_y = \begin{pmatrix}0 & -i\\i & 0\end{pmatrix}, \quad \sigma_z = \begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$$
They satisfy $\sigma_i\sigma_j = \delta_{ij}I + i\epsilon_{ijk}\sigma_k$, so in particular $\{\sigma_i, \sigma_j\} = 2\delta_{ij}I$ and $[\sigma_i, \sigma_j] = 2i\epsilon_{ijk}\sigma_k$. The eigenvalues of each $\sigma_i$ are $\pm 1$; the eigenstates of $\sigma_z$ are $|\uparrow\rangle = \binom{1}{0}$ and $|\downarrow\rangle = \binom{0}{1}$.
Every pure state of a two-level system can be written as:
$$|\psi\rangle = \cos\frac{\theta}{2}|\uparrow\rangle + e^{i\phi}\sin\frac{\theta}{2}|\downarrow\rangle, \quad \theta \in [0,\pi],\; \phi \in [0,2\pi)$$
The expectation values of the spin operators give the Bloch vector $(n_x, n_y, n_z)$:
$$\langle S_x \rangle = \frac{\hbar}{2}\sin\theta\cos\phi, \quad \langle S_y \rangle = \frac{\hbar}{2}\sin\theta\sin\phi, \quad \langle S_z \rangle = \frac{\hbar}{2}\cos\theta$$
The probability of measuring spin-up is $P(\uparrow) = \cos^2(\theta/2)$. Mixed states correspond to points inside the sphere with $|\mathbf{n}| < 1$.
When two angular momenta $j_1$ and $j_2$ are combined, the tensor product space decomposes as:
$$\mathcal{H}_{j_1} \otimes \mathcal{H}_{j_2} = \bigoplus_{j=|j_1-j_2|}^{j_1+j_2} \mathcal{H}_j$$
The coupled basis states are related to the uncoupled basis by $|j,m\rangle = \sum_{m_1+m_2=m} \langle j_1,m_1;j_2,m_2|j,m\rangle\,|j_1,m_1\rangle|j_2,m_2\rangle$. For two spin-½ particles: $\frac{1}{2}\otimes\frac{1}{2} = 1 \oplus 0$, giving the triplet states $|1,m\rangle$ and singlet $|0,0\rangle = \frac{1}{\sqrt{2}}(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)$.
Find the eigenstates of $\hat{S}_x$ for a spin-½ particle and express them in terms of $|\uparrow\rangle$ and $|\downarrow\rangle$.
We need to diagonalize $\hat{S}_x = \frac{\hbar}{2}\sigma_x = \frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}$.
Characteristic equation: $\det(\sigma_x - \lambda I) = -\lambda^2 - (-1) = \lambda^2 - 1 = 0$, so $\lambda = \pm 1$, giving eigenvalues $\pm\hbar/2$.
For $\lambda = +1$: $(\sigma_x - I)\mathbf{v} = 0$ gives $\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\binom{a}{b} = 0$, so $a = b$. Normalized: $|+x\rangle = \frac{1}{\sqrt{2}}\binom{1}{1} = \frac{1}{\sqrt{2}}(|\uparrow\rangle + |\downarrow\rangle)$.
For $\lambda = -1$: $|{-x}\rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle - |\downarrow\rangle)$.
These are the $|\pm\rangle$ states on the Bloch sphere equator at $\phi = 0$, i.e., $\theta = \pi/2$, $\phi = 0$ and $\phi = \pi$ respectively.
A spin-½ particle is in the state $|\psi\rangle = \frac{\sqrt{3}}{2}|\uparrow\rangle + \frac{1}{2}|\downarrow\rangle$. Find $\langle S_z \rangle$, $\langle S_x \rangle$, and the probability of measuring $+\hbar/2$ along $z$.
Identify: $\alpha = \frac{\sqrt{3}}{2}$, $\beta = \frac{1}{2}$. Check normalization: $|\alpha|^2 + |\beta|^2 = \frac{3}{4} + \frac{1}{4} = 1$. ✓
Probability spin-up: $P(\uparrow) = |\alpha|^2 = \frac{3}{4}$.
$\langle S_z \rangle$: $\langle S_z \rangle = \frac{\hbar}{2}(|\alpha|^2 - |\beta|^2) = \frac{\hbar}{2}\left(\frac{3}{4} - \frac{1}{4}\right) = \frac{\hbar}{4}$.
$\langle S_x \rangle$: $\langle S_x \rangle = \frac{\hbar}{2}\cdot 2\,\text{Re}(\alpha^*\beta) = \hbar\,\text{Re}\!\left(\frac{\sqrt{3}}{2}\cdot\frac{1}{2}\right) = \frac{\sqrt{3}\hbar}{4}$.
In Bloch sphere coordinates: $\theta = 2\arccos(\sqrt{3}/2) = \pi/3 = 60°$, $\phi = 0$, giving $\langle \mathbf{S}\rangle = \frac{\hbar}{2}(\sin 60°, 0, \cos 60°) = \frac{\hbar}{2}(\frac{\sqrt{3}}{2}, 0, \frac{1}{2})$.
In the Stern-Gerlach experiment, silver atoms ($l=0$, $s=\frac{1}{2}$) pass through a magnetic field gradient $\partial B_z/\partial z = 15$ T/m over length $L = 0.05$ m with speed $v = 500$ m/s. The magnetic moment is $\mu = g_s\mu_B$ with $g_s \approx 2$, $\mu_B = 9.274\times10^{-24}$ J/T, mass $m = 1.79\times10^{-25}$ kg. Find the deflection of the $m_s = +\frac{1}{2}$ beam.
Force on the atom: $F_z = \mu_z \frac{\partial B_z}{\partial z}$ where $\mu_z = -g_s\mu_B m_s = -2\mu_B(+\frac{1}{2}) = -\mu_B$.
$F_z = -\mu_B \cdot 15 = -(9.274\times10^{-24})(15) = -1.391\times10^{-22}$ N (downward).
Transit time: $t = L/v = 0.05/500 = 10^{-4}$ s.
Transverse acceleration: $a_z = F_z/m = 1.391\times10^{-22}/1.79\times10^{-25} = 777$ m/s².
Deflection: $\Delta z = \frac{1}{2}a_z t^2 = \frac{1}{2}(777)(10^{-4})^2 = 3.9\times10^{-3}$ m $\approx 3.9$ mm.
The two beams ($m_s = \pm\frac{1}{2}$) are deflected by $\pm 3.9$ mm, producing two distinct spots separated by $\approx 7.8$ mm.
Two electrons have orbital angular momentum $l_1 = 1$ and $l_2 = 1$. List all possible total orbital angular momentum quantum numbers $L$ and the dimensions of each subspace. Verify the total dimension.
Using the Clebsch-Gordan rule for adding $l_1 = 1$ and $l_2 = 1$:
$$L = |l_1 - l_2|, \ldots, l_1 + l_2 = 0, 1, 2$$
Dimensions: Each $L$ subspace has dimension $2L+1$:
$L = 0$: dimension $1$; $L = 1$: dimension $3$; $L = 2$: dimension $5$.
Total dimension: $1 + 3 + 5 = 9 = (2\cdot 1+1)^2 = 3^2$. ✓
This matches $(2l_1+1)(2l_2+1) = 3 \times 3 = 9$, confirming the decomposition $1\otimes 1 = 0\oplus 1\oplus 2$ is complete. The Clebsch-Gordan coefficients give the explicit change-of-basis matrix between the uncoupled $|m_1,m_2\rangle$ and coupled $|L,M\rangle$ bases.
Practice Problems
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1. $[\hat{L}_x,\hat{L}_y] = [\hat{y}\hat{p}_z-\hat{z}\hat{p}_y, \hat{z}\hat{p}_x-\hat{x}\hat{p}_z]$. Expanding using $[\hat{q}_i,\hat{p}_j]=i\hbar\delta_{ij}$: the cross terms give $\hat{y}\hat{p}_x(-i\hbar) - (-i\hbar)\hat{x}\hat{p}_y = i\hbar(\hat{x}\hat{p}_y - \hat{y}\hat{p}_x) = i\hbar\hat{L}_z$. ✓
2. $m \in \{-2,-1,0,1,2\}$. $\langle L^2\rangle = \hbar^2 l(l+1) = 6\hbar^2$. Max $\langle L_z\rangle = 2\hbar$.
3. $\hat{L}_+|l,l\rangle = \hbar\sqrt{l(l+1)-l(l+1)}|l,l+1\rangle = 0$. There is no state above the maximum $m = l$; the ladder terminates.
4. $\hat{L}_-|2,1\rangle = \hbar\sqrt{2(3)-1(0)}|2,0\rangle = \hbar\sqrt{6}|2,0\rangle$.
5. $P(\uparrow)=1/4$, $P(\downarrow)=3/4$. $\langle S_z\rangle = \frac{\hbar}{2}(1/4-3/4) = -\hbar/4$. $\langle S_x\rangle = \hbar\,\text{Re}(\alpha^*\beta) = \hbar\,\text{Re}(-i/2\cdot\sqrt{3}/2) = 0$ (since $-i\sqrt{3}/4$ is purely imaginary).
6. Direct matrix multiplication: $\sigma_x^2 = \begin{pmatrix}0&1\\1&0\end{pmatrix}^2 = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$. Similarly for $\sigma_y,\sigma_z$. $\sigma_x\sigma_y = \begin{pmatrix}i&0\\0&-i\end{pmatrix} = i\sigma_z$. ✓
7. $|{-y}\rangle = \frac{1}{\sqrt{2}}|\uparrow\rangle + e^{i(-\pi/2)}\frac{1}{\sqrt{2}}|\downarrow\rangle$. So $\cos(\theta/2)=1/\sqrt{2}$, giving $\theta = \pi/2$. And $e^{i\phi}\sin(\pi/4) = -i/\sqrt{2}$, so $e^{i\phi} = -i = e^{-i\pi/2}$, giving $\phi = 3\pi/2$ (or $-\pi/2$).
8. $j = l+s = 3/2$ (4 states) and $j = l-s = 1/2$ (2 states). Total: $4+2=6=(2\cdot1+1)(2\cdot\frac{1}{2}+1)=3\times2$. ✓
9. $\omega_0 = g_s\mu_B B_0/\hbar = 2\times(9.274\times10^{-24})\times1/(1.055\times10^{-34}) \approx 1.76\times10^{11}$ rad/s (the electron cyclotron/Larmor frequency, ~28 GHz).
10. $\hat{S}_{\text{tot}}^2 = (\hat{S}_1+\hat{S}_2)^2 = \hat{S}_1^2+\hat{S}_2^2+2\hat{S}_1\cdot\hat{S}_2$. Each spin-½ gives $\hat{S}_i^2 = \frac{3}{4}\hbar^2 I$. For the singlet, $\hat{S}_1\cdot\hat{S}_2|0,0\rangle = -\frac{3}{4}\hbar^2|0,0\rangle$. So $\hat{S}_{\text{tot}}^2|0,0\rangle = (\frac{3}{4}+\frac{3}{4}-\frac{3}{2})\hbar^2|0,0\rangle = 0$. ✓
11. $|\uparrow\downarrow\rangle = \frac{1}{\sqrt{2}}|1,0\rangle + \frac{1}{\sqrt{2}}|0,0\rangle$ (using CG coefficients: $\langle 1,0|\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}\rangle = 1/\sqrt{2}$ and $\langle 0,0|\frac{1}{2},\frac{1}{2};\frac{1}{2},-\frac{1}{2}\rangle = 1/\sqrt{2}$).
12. $\langle L^2\rangle = \langle L_x^2\rangle + \langle L_y^2\rangle + \langle L_z^2\rangle$. Since $\langle L_z^2\rangle = m^2\hbar^2 = 0$: $\langle L_x^2\rangle + \langle L_y^2\rangle = 2\hbar^2$. By symmetry (isotropy in $x$-$y$ plane for $m=0$), $\langle L_x^2\rangle = \langle L_y^2\rangle = \hbar^2$.