Part 4 · Spin and angular momentum

4.2 Angular momentum

Source: the commutators, the states $|j,m⟩$, the ladder operators, the $j=1$ matrices, the cone and the $j=1$ exercise from the 2024/25 boards #14–#16; this year’s course has not reached it yet

$$\hat{\bar{J}}=({\hat{J}}_x,{\hat{J}}_y,{\hat{J}}_z),{\hat{J}}^2={\hat{J}}_x^2+{\hat{J}}_y^2+{\hat{J}}_z^2.$$ Two physical kinds:

• Spin $\hat{\bar{S}}$, the intrinsic angular momentum. It depends on no coordinates: its states are $|\pm ⟩$, not functions of $x,y,z$ (§4.1 Spin).
• Orbital $\hat{\bar{L}}=\hat{\bar{r}}\wedge \hat{\bar{p}}$. It depends on motion, and its eigenfunctions are functions of position (§4.3 Orbital angular momentum).

The total is $\hat{\bar{J}}=\hat{\bar{L}}+\hat{\bar{S}}$ (§4.4 Addition of angular momenta). ==Everything here uses only the commutation relations==, so it holds for $\hat{\bar{L}}$, for $\hat{\bar{S}}$ and for their sum alike. Only the allowed $j$ differ: orbital $\ell$ is always an integer, spin can be half-integer.

Commutation relations

==No two components of an angular momentum are compatible.== The squared length, though, commutes with every component: $$[{\hat{J}}^2,{\hat{J}}_i]=0.$$ So the length and one component can be sharp together; by convention that component is $J_z$.

The states $|j,m⟩$

$j$ is the angular momentum quantum number, fixing the length $|\bar{J}|=\hslash \sqrt{j(j+1)}$; $m$ (or $m_j$) the magnetic quantum number, the projection on $z$ in units of $\hslash$. Since $m$ goes from $-j$ to $j$ in whole steps, $2j$ is an integer (beyond the boards). Spin ½ is $j=\frac{1}{2}$: ${\hslash }^2\cdot \frac{1}{2}\cdot \frac{3}{2}=\frac{3}{4}{\hslash }^2$, the value of $\hat S^2$ and the cone.

${\hat{J}}_{+}$ raises $m$ by one, ${\hat{J}}_{-}$ lowers it, and $j$ stays the same. At the ends the coefficient vanishes, ${\hat{J}}_{+}|j,j⟩=0$: that is why the list of $m$ stops at $\pm j$. For $j=m=\frac{1}{2}$: $C^{+}=0$, and $C^{-}=\hslash \sqrt{\frac{3}{4}+\frac{1}{4}}=\hslash$, so ${\hat{S}}_{-}|+⟩=\hslash |-⟩$. ${\hat{J}}_{-}$ is the adjoint of ${\hat{J}}_{+}$, since ${\hat{J}}_x$ and ${\hat{J}}_y$ are Hermitian. (In the matrix section the board writes $+m(m\pm 1)$ under the root; the values it uses are those of the minus sign.)

Matrices for $j=1$

$j=1$ has three states, columns $|1,1⟩=(1,0,0{)}^T$, $|1,0⟩$, $|1,-1⟩$. An operator is the table of its matrix elements $⟨1,m^{\prime }|\hat{A}|1,m⟩$, row $m^{\prime }$, column $m$, in the order $1,0,-1$.

For $j=1$, both steps up have $\hslash \sqrt{2}$: $${\hat{J}}_z=\hslash \left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& -1\end{array}\right),{\hat{J}}_x=\frac{\hslash }{\sqrt{2}}\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right),{\hat{J}}_y=\frac{\hslash }{\sqrt{2}}\left(\begin{array}{ccc}0& -i& 0\\ i& 0& -i\\ 0& i& 0\end{array}\right),$$ and ${\hat{J}}^2=2{\hslash }^2\hat{I}$. ==${\hat{J}}_x$ and ${\hat{J}}_y$ are not diagonal: $|1,m⟩$ are not their eigenstates.== That is the matrix form of “only one component can be sharp”. ${\hat{J}}_x/\hslash$ is the $\hat{A}$ of Problem 3.8, whose eigenvalues came out $-1,0,1$ (beyond the boards).

The board sets up the check ${\hat{J}}_x{\hat{J}}_y-{\hat{J}}_y{\hat{J}}_x\stackrel{?}{=}i\hslash {\hat{J}}_z$ and stops. Finished (beyond the boards): with ${\hat{J}}_x=\frac{\hslash }{\sqrt{2}}A$, ${\hat{J}}_y=\frac{\hslash }{\sqrt{2}}B$, $AB-BA=\mathrm{diag}(2i,0,-2i)$, so the commutator is $i{\hslash }^2\mathrm{diag}(1,0,-1)=i\hslash {\hat{J}}_z$.

The cone

The angle is $\cos \alpha =m/\sqrt{j(j+1)}$ (beyond the boards). For $j=1$ (board 15), $|\bar{J}|=\sqrt{2}\hslash$, and the top cone is at 45°.

How wide the cone is (beyond the boards). In $|j,m⟩$, $⟨J_x⟩=⟨J_y⟩=0$ and $⟨J_x^2⟩=⟨J_y^2⟩=\frac{1}{2}⟨{\hat{J}}^2-{\hat{J}}_z^2⟩$, so $$\Delta J_x=\hslash \sqrt{\frac{j(j+1)-m^2}{2}},$$ never zero, since $m^2\le j^2<j(j+1)$.

==A quantum angular momentum cannot point exactly along any axis==; the leftover ${\hslash }^2[j(j+1)-m^2]$ is the uncertainty in $J_x$ and $J_y$, the bound $\Delta J_x\Delta J_y\ge \frac{\hslash }{2}|⟨J_z⟩|$. As $j$ grows the top cone closes, $5.7°$ at $j=100$: the classical limit, where a vector can point along an axis.

For orbital angular momentum the same states are written $|\ell ,m_{\ell }⟩$. With the energy they give the labels $n,\ell ,m$ of an atomic orbital (§4.3 Orbital angular momentum).

Exercises