Angular Momentum

All right—second-semester quantum mechanics begins here. We already know how to turn classical observables into operators, solve the Schrödinger equation for several potentials, and describe the hydrogen atom in terms of energy, position, and momentum. The next classical quantity to quantize is angular momentum.

One warning before we begin: the electron in an atom is not a little planet following a definite circular orbit. Orbital angular momentum is nevertheless a well-defined quantum observable, obtained by promoting the classical expression $\mathbf L=\mathbf r\times\mathbf p$ to an operator. Intrinsic spin, which appears in the next chapter, is a different kind of angular momentum.

This English edition keeps every historical source and localized raster in its established order. Where a preserved raster contains a known algebraic or labeling error, the accessible TeX and the explanation below give the corrected equation explicitly.

From the classical vector to the quantum operator

Classically,

$$ \mathbf L=\mathbf r\times\mathbf p. $$

Writing $\mathbf r=(x,y,z)$ and $\mathbf p=(p_x,p_y,p_z)$ gives

$$ \mathbf L=(yp_z-zp_y)\,\hat{\mathbf x} +(zp_x-xp_z)\,\hat{\mathbf y} +(xp_y-yp_x)\,\hat{\mathbf z}. $$

The source raster above writes $xp_y$ where the first component requires $zp_y$. The corrected component formula is the displayed TeX immediately above it.

In position space, momentum is the differential operator

$$ \mathbf p=-i\hbar\boldsymbol\nabla, \qquad p_j=-i\hbar\frac{\partial}{\partial x_j}. $$

Therefore

$$ \begin{aligned} L_x&=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right),\\ L_y&=-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right),\\ L_z&=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right). \end{aligned} $$

I will usually omit hats on operators when the context is unambiguous.

Angular-momentum commutators

The canonical commutation relations are

$$ [x_i,p_j]=i\hbar\delta_{ij},\qquad [x_i,x_j]=[p_i,p_j]=0. $$

Let us calculate one representative commutator. With $L_x=yp_z-zp_y$ and $L_y=zp_x-xp_z$,

The product rule for commutators,

$$ [A,BC]=[A,B]C+B[A,C], $$

and its corresponding form for $[AB,C]$ let us discard terms made entirely from mutually commuting coordinates or mutually commuting momenta.

Using the canonical relations,

the surviving terms combine to give

and cyclic permutation gives the general result

$$ [L_i,L_j]=i\hbar\sum_k\varepsilon_{ijk}L_k. $$

The crucial point is that these commutators are not zero:

Different Cartesian components are incompatible observables. No nonzero state can be a simultaneous eigenstate of two components with sharp nonzero angular-momentum structure in the ordinary way. More precisely, the uncertainty relation is

$$ \Delta L_i\,\Delta L_j\geq \frac{\hbar}{2}\left|\langle L_k\rangle\right| $$

for cyclic $(i,j,k)$. This is subtler than saying that one measurement mechanically “changes” both other components; the operator statement is that the components do not commute and therefore are not generally simultaneously sharp.

The square of the total angular momentum is

$$ L^2=L_x^2+L_y^2+L_z^2. $$

Direct use of the angular-momentum algebra gives

$$ [L^2,L_i]=0\qquad(i=x,y,z). $$

Thus $L^2$ and one chosen component—conventionally $L_z$—have common eigenstates. Write such a state as $f$ for the moment:

$$ L^2f=\lambda f,\qquad L_zf=\mu f. $$

Ladder operators and the spectrum

Define the angular-momentum ladder operators

$$ L_\pm=L_x\pm iL_y. $$

They obey

$$ [L^2,L_\pm]=0, \qquad [L_z,L_\pm]=\pm\hbar L_\pm. $$

Suppose $f$ is a simultaneous eigenstate:

Because $L^2$ commutes with $L_\pm$,

$$ L^2(L_\pm f)=L_\pm L^2f=\lambda(L_\pm f). $$

For $L_z$,

$$ \begin{aligned} L_z(L_\pm f) &=(L_\pm L_z+[L_z,L_\pm])f\\ &=(\mu\pm\hbar)(L_\pm f). \end{aligned} $$

This is why the name “ladder operator” is appropriate: $L_+$ raises $\mu$ by $\hbar$, while $L_-$ lowers it by $\hbar$, without changing $\lambda$.

Why the ladder must end

The ladder cannot extend indefinitely. Since

and each $L_i^2$ is positive semidefinite,

$$ \langle L_z^2\rangle\leq \langle L^2\rangle=\lambda. $$

Let the largest $L_z$ eigenvalue be $\ell\hbar$ and the smallest be $\bar\ell\hbar$:

Denote the endpoint states by $f_{\mathrm{top}}$ and $f_{\mathrm{bottom}}$.

Repeated lowering produces

$$ \ell\hbar,(\ell-1)\hbar,(\ell-2)\hbar,\ldots,\bar\ell\hbar. $$

At the endpoints,

$$ L_+f_{\mathrm{top}}=0, \qquad L_-f_{\mathrm{bottom}}=0. $$

Both endpoint states still satisfy the simultaneous eigenvalue equations:

The useful operator identities are

$$ L_-L_+=L^2-L_z^2-\hbar L_z, \qquad L_+L_-=L^2-L_z^2+\hbar L_z. $$

Apply the first identity to the top state. Since $L_+f_{\mathrm{top}}=0$,

$$ 0=(\lambda-\hbar^2\ell^2-\hbar^2\ell)f_{\mathrm{top}}, $$

so

$$ \lambda=\hbar^2\ell(\ell+1). $$

Applying the second identity to the bottom state gives

$$ \lambda=\hbar^2\bar\ell(\bar\ell-1). $$

Equating the two expressions,

gives two algebraic possibilities:

The choice $\bar\ell=\ell+1$ contradicts the ordering of the endpoints, so

$$ \bar\ell=-\ell. $$

Therefore

$$ L^2\lvert \ell,m\rangle =\hbar^2\ell(\ell+1)\lvert \ell,m\rangle, \qquad L_z\lvert \ell,m\rangle =\hbar m\lvert \ell,m\rangle, $$

with

$$ m=-\ell,-\ell+1,\ldots,\ell-1,\ell. $$

The number of one-step intervals from $-\ell$ to $+\ell$ is $2\ell$, so $2\ell$ must be a nonnegative integer:

Thus the abstract angular-momentum algebra permits

$$ \ell=0,\frac12,1,\frac32,\ldots. $$

For orbital angular momentum, however, single-valued spherical harmonics require $\ell=0,1,2,\ldots$. Half-integer values describe intrinsic spin and other total-angular-momentum representations, not the orbital wavefunctions derived below.

Coordinate representation and spherical harmonics

Return to the classical definition

and quantize it:

$$ \mathbf L=-i\hbar\,\mathbf r\times\boldsymbol\nabla. $$

In spherical coordinates,

$$ \boldsymbol\nabla =\hat{\mathbf r}\frac{\partial}{\partial r} +\hat{\boldsymbol\theta}\frac1r\frac{\partial}{\partial\theta} +\hat{\boldsymbol\phi}\frac1{r\sin\theta}\frac{\partial}{\partial\phi}. $$

Since $\mathbf r=r\hat{\mathbf r}$, the radial derivative drops out of the cross product:

$$ \mathbf L=-i\hbar\left( \hat{\boldsymbol\phi}\frac{\partial}{\partial\theta} -\hat{\boldsymbol\theta}\frac1{\sin\theta}\frac{\partial}{\partial\phi} \right). $$

Converting the spherical unit vectors back to Cartesian components yields

$$ \begin{aligned} L_x&=i\hbar\left( \sin\phi\frac{\partial}{\partial\theta} +\cot\theta\cos\phi\frac{\partial}{\partial\phi} \right),\\ L_y&=i\hbar\left( -\cos\phi\frac{\partial}{\partial\theta} +\cot\theta\sin\phi\frac{\partial}{\partial\phi} \right),\\ L_z&=-i\hbar\frac{\partial}{\partial\phi}. \end{aligned} $$

The angular part of the hydrogen Hamiltonian already contains $L^2$:

After separation of variables, the azimuthal and polar factors are

and their product is the spherical harmonic,

$$ Y_\ell^m(\theta,\phi)=\Theta_\ell^m(\theta)\Phi_m(\phi), $$

which satisfies

$$ \begin{aligned} L^2Y_\ell^m&=\hbar^2\ell(\ell+1)Y_\ell^m,\\ L_zY_\ell^m&=\hbar mY_\ell^m. \end{aligned} $$

The explicit differential operator is

$$ L^2=-\hbar^2\left[ \frac1{\sin\theta}\frac{\partial}{\partial\theta} \left(\sin\theta\frac{\partial}{\partial\theta}\right) +\frac1{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} \right]. $$

Worked problem 4.18: normalize the ladder action

Suppose

$$ L_\pm\lvert \ell,m\rangle=A_\pm\lvert \ell,m\pm1\rangle. $$

First show that the ladder operators are Hermitian conjugates:

Because $L_x^\dagger=L_x$ and $L_y^\dagger=L_y$,

$$ L_\pm^\dagger=(L_x\pm iL_y)^\dagger=L_x\mp iL_y=L_\mp. $$

Use

and take the norm:

The result is

$$ \boxed{ L_\pm\lvert \ell,m\rangle =\hbar\sqrt{\ell(\ell+1)-m(m\pm1)} \,\lvert \ell,m\pm1\rangle } $$

up to a conventional phase choice for the states.

Worked problem 4.19: commutators with position and momentum

Start from $L_z=xp_y-yp_x$ and the canonical commutation relations. Applying $[A,BC]=[A,B]C+B[A,C]$ gives

$$ [L_z,x]=i\hbar y, $$

$$ [L_z,y]=-i\hbar x, $$

$$ [L_z,z]=0, $$

and

$$ [L_z,p_x]=i\hbar p_y. $$

The historical exercise text asks for $[L_z,L_x]=i\hbar L_z$. That right-hand side is a source typo. The angular-momentum algebra requires

$$ \boxed{[L_z,L_x]=i\hbar L_y}. $$

Indeed, with $L_x=yp_z-zp_y$,

$$ \begin{aligned} [L_z,L_x] &=[L_z,y]p_z+y[L_z,p_z]-[L_z,z]p_y-z[L_z,p_y]\\ &=-i\hbar xp_z+i\hbar zp_x\\ &=i\hbar(zp_x-xp_z)=i\hbar L_y. \end{aligned} $$

Rotations about the $z$ axis leave both $r^2=x^2+y^2+z^2$ and $p^2=p_x^2+p_y^2+p_z^2$ invariant. Direct calculation confirms

$$ [L_z,r^2]=0, \qquad [L_z,p^2]=0. $$

Worked problem 4.24: the three-dimensional rigid rotor

Two particles of mass $m$ are attached to the ends of a massless rod of length $a$, and the center of the rod is fixed. Each particle is a distance $a/2$ from the center, so the moment of inertia is

$$ I=2m\left(\frac a2\right)^2=\frac{ma^2}{2}. $$

The rotational Hamiltonian is

$$ H=\frac{L^2}{2I}=\frac{L^2}{ma^2}. $$

The same result follows from the classical kinetic energy of the two masses:

For each mass, $r=a/2$ and the velocity is perpendicular to the rod, so the total angular momentum magnitude is $L=amv$.

Substituting the orbital-angular-momentum eigenvalue gives the energy spectrum

$$ \boxed{E_\ell=\frac{\hbar^2}{ma^2}\ell(\ell+1), \qquad \ell=0,1,2,\ldots} $$

Each level has degeneracy $2\ell+1$, corresponding to $m=-\ell,-\ell+1,\ldots,\ell$.

That completes the basic orbital-angular-momentum story: the noncommuting components form a closed algebra, $L^2$ and one component can be diagonalized together, the ladder terminates, and the coordinate eigenfunctions are spherical harmonics. Next comes the genuinely new half-integer angular momentum: spin.