Angular Momentum, Spin and Hydrogen Atom

Spin-1/2 states and magnetic energy

A spin-1/2 state is a two-component column \chi=(a,b)^T normalised so that |a|^2+|b|^2=1, and the squared moduli give the probabilities of S_z=+\hbar/2 and S_z=-\hbar/2. To find the probability of another spin component, project the state onto the eigenvector of that component and take the squared modulus: P=|\langle\chi_{\text{target}}|\chi\rangle|^2. The eigenvectors of S_x are \frac{1}{\sqrt2}(1,1)^T for +\hbar/2 and \frac{1}{\sqrt2}(1,-1)^T for -\hbar/2. The projection is defined only for a unit vector, so an unnormalised state must be divided by its norm before projecting.

For magnetic moments \mu_i in a field H, the energy is E=-\sum_i\mu_i H, so a moment parallel to the field lowers the energy. A given total energy is met by a fixed set of spin configurations. If these configurations are equally likely, the probability that a particular particle is up is the fraction of them in which that particle is up. The mean moment of a configuration is the average of the individual moment values.

In a product state |\uparrow\rangle_1|\downarrow\rangle_2, the expectation value of \mathbf S_1\cdot\mathbf S_2 is found component by component. The transverse terms S_{1x}S_{2x} and S_{1y}S_{2y} flip both spins, so they have zero expectation value in a product of S_z eigenstates. Only S_{1z}S_{2z} survives, and for opposite projections it equals (\hbar/2)(-\hbar/2)=-\hbar^2/4. Multiplying by the coupling constant gives the energy.

Pauli spin matrix algebra

The Pauli matrices \sigma_x,\sigma_y,\sigma_z satisfy \sigma_i^2=I and \sigma_\alpha\sigma_\beta=\delta_{\alpha\beta}I+i\epsilon_{\alpha\beta\gamma}\sigma_\gamma. From this follow the commutator [\sigma_\alpha,\sigma_\beta]=2i\epsilon_{\alpha\beta\gamma}\sigma_\gamma and the anticommutator \{\sigma_\alpha,\sigma_\beta\}=2\delta_{\alpha\beta}I. Distinct Pauli matrices anticommute, and the commutator carries the factor 2i, which is often lost in careless checks.

For two spin-1/2 particles, (\boldsymbol\sigma_1+\boldsymbol\sigma_2)^2=6+2\,\boldsymbol\sigma_1\cdot\boldsymbol\sigma_2. Since (\boldsymbol\sigma_1+\boldsymbol\sigma_2)^2=4S(S+1) with S the total spin, the singlet with S=0 gives \boldsymbol\sigma_1\cdot\boldsymbol\sigma_2=-3. This is the standard way to evaluate the spin dot product from the total spin.

A spin-up state along \hat x is \frac{1}{\sqrt2}(1,1)^T, and a spin-up state along a direction at angle \theta in the xy-plane is \frac{1}{\sqrt2}(1,e^{i\theta})^T. The overlap gives the probability P=\cos^2(\theta/2), which equals 1 for \theta=0 and 0 for \theta=\pi.

Hydrogen radial wavefunction

The radial probability density is P(r)=r^2R^2(r) for a normalised radial function R(r). Its maximum follows from dP/dr=0. For a density of the form P\propto r^k e^{-r/a}, the derivative gives kr^{k-1}-(r^k/a)=0, so the most probable radius is r=ka, where a is the Bohr radius.

For a 2p state, R_{21}\propto r\,e^{-r/2a}, so R^2\propto r^2e^{-r/a} and P\propto r^4e^{-r/a}. The maximum therefore lies at r=4a, which is the peak of the n=2, l=1 radial distribution.

The ground state of hydrogen is the nondegenerate 1s state with l=0 and definite parity (-1)^l=+1. Its first-order linear Stark shift vanishes because it is nondegenerate, and the spin-orbit term \mathbf L\cdot\mathbf S vanishes because L=0. The electron-proton hyperfine interaction is nonzero and gives the 21 cm splitting of about 5.9\times10^{-6} eV.

Vector model of angular momentum

In the vector model, the magnitude of the total angular momentum is |\mathbf J|=\sqrt{j(j+1)}\,\hbar, where j is the total angular momentum quantum number. A Stern-Gerlach apparatus separates a beam according to the 2j+1 allowed values of m_j, so the number of beams equals 2j+1. For example, \sqrt{12}\,\hbar corresponds to j=3 because 3\times4=12, which gives 7 beams.

For an orbital angular momentum with quantum number l, the z-component is L_z=m_l\hbar and the angle with the z-axis satisfies \cos\theta=m_l/\sqrt{l(l+1)}. The smallest angle comes from the largest allowed m_l. For a p electron, |\mathbf L|=\sqrt2\,\hbar and m_l=+1 give \cos\theta=1/\sqrt2.

A wavefunction proportional to z\,e^{-\alpha r^2} can be written as r\cos\theta\,e^{-\alpha r^2}, so its angular part is \cos\theta\propto Y_1^0. This corresponds to l=1 and m=0, giving L^2=2\hbar^2 and L_z=0.

LS coupling term symbols

In LS coupling an atomic term is written ^{2S+1}L_J, where L and S are the total orbital and spin quantum numbers. The allowed values of J run in integer steps from |L-S| to L+S, so a symbol is possible only if its J lies in this range. For example, ^4S has S=3/2 and L=0, so only J=3/2 is allowed.

Hund's rules determine the ground term of a configuration. The first rule maximises the total spin S, the second maximises L, and the third sets J=|L-S| when the subshell is no more than half filled and J=L+S when it is more than half filled. For two equivalent p electrons, as in the carbon ground state, the first rule gives S=1 and the second gives L=1, so the term is ^3P. Since the p shell is less than half filled, J=|L-S|=0.

Zeeman effect splitting

In a weak magnetic field the anomalous Zeeman effect splits each level into 2J+1 sublevels with energy shifts g_J m_J\mu_B B. The Landé factor is g_J=1+\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}. For a level with L=2, S=1/2, J=3/2 this gives g_J=4/5, and for L=1, S=1/2, J=1/2 it gives g_J=2/3.

The selection rule for a transition is \Delta m_J=0,\pm1. To count the lines, list the allowed pairs (m_u,m_l) between the upper and lower sublevels and check whether the energy shifts (g_u m_u-g_l m_l)\mu_B B coincide. In the case of a J=3/2 upper level and a J=1/2 lower level, the allowed pairs give six distinct lines.

Adiabatic precession in changing magnetic field

When a magnetic field changes slowly, the total angular momentum \mathbf J of an atom precesses about the instantaneous field direction. If the field changes slowly compared with the precession frequency, the projection m_j on the field direction is conserved. The angle between \mathbf J and the field then follows from \cos\theta=m_j/\sqrt{j(j+1)}.

The total quantum number from orbital and spin parts satisfies |l-s|\le j\le l+s. The largest j is l+s, and its largest m_j equals j, so the angle is the smallest possible. The precession picture of the vector model gives this relation directly.

Angular momentum commutation relations

Angular momentum components satisfy [J_x,J_y]=i\hbar J_z and cyclic permutations. The total square \mathbf J^2 commutes with each component J_i, so [\mathbf J^2,J_z]=0. The square of the orbital momentum \mathbf L^2 commutes with \mathbf L and with \mathbf S, so it commutes with \mathbf J=\mathbf L+\mathbf S and with \mathbf J^2.

A component of one angular momentum does not generally commute with the total square. Since \mathbf J^2=\mathbf L^2+\mathbf S^2+2\mathbf L\cdot\mathbf S, the operator \mathbf J^2 fails to commute with L_z. The component L_z and the total J_z=L_z+S_z commute, because L_z commutes with L_z and S_z.