Orbital angular momentum operators

Every problem we have solved so far has been one-dimensional: a particle moving along a line. But real atoms are three-dimensional — the electron orbits the nucleus. And the moment anything “goes around”, classical mechanics hands us a leading character: angular momentum.

First, why does it matter classically? A planet around the Sun, an electron around a nucleus — these are central-force problems: the force always points at the centre. In that situation $\vec L=\vec r\times\vec p$ is conserved — its direction locks the orbital plane in place, and its magnitude sets how “fat” or “thin” the orbit is. Kepler’s second law (equal areas in equal times) is nothing but angular momentum conservation in disguise.

Meanwhile, experiment had long been hinting that angular momentum in atoms is not continuous. Put a glowing atom in a magnetic field and its spectral lines split into a finite number of components (the Zeeman effect), as if angular momentum could only point in a few discrete directions in space. The Bohr model simply postulated $L=n\hbar$ — the specific values later turned out to be wrong, but the intuition of “quantised angular momentum” was pointing exactly the right way.

This chapter answers the questions: what does angular momentum actually look like in quantum mechanics, which values can it take, and why does it drag in something with no classical counterpart whatsoever — spin.

From classical quantity to operator

The postulates of chapter 3 give us the recipe for building operators: take the classical expression and replace $x$ and $p$ by operators. For angular momentum:

$$\hat{\vec L}=\hat{\vec r}\times\hat{\vec p} \quad\Longrightarrow\quad \begin{aligned} \hat L_x&=\hat y\hat p_z-\hat z\hat p_y\\ \hat L_y&=\hat z\hat p_x-\hat x\hat p_z\\ \hat L_z&=\hat x\hat p_y-\hat y\hat p_x \end{aligned}$$

The three formulas are really one formula: cycle $x\to y\to z\to x$ and each line turns into the next.

The key calculation: do the components commute?

The lesson of section 3.6 was that whether two observables can be simultaneously sharp is decided entirely by their commutator. So the first thing to do with three new operators is to compute their commutators.

Computing [L̂ₓ, L̂ᵧ]basic~6 minexpand

Goal: $[\hat L_x,\hat L_y]=[\hat y\hat p_z-\hat z\hat p_y,\;\hat z\hat p_x-\hat x\hat p_z]$.

Step 1: expand into four terms. The commutator is linear in both slots, so

$$[\hat L_x,\hat L_y] =[\hat y\hat p_z,\hat z\hat p_x] -[\hat y\hat p_z,\hat x\hat p_z] -[\hat z\hat p_y,\hat z\hat p_x] +[\hat z\hat p_y,\hat x\hat p_z]\tag{5.1.1}$$

Step 2: throw away the trivial terms first. The operators appearing in the second term are $\hat y,\hat p_z,\hat x$ — every pair among them commutes (coordinates and momenta along different directions leave each other alone), so the whole term vanishes. The third term contains $\hat z,\hat p_y,\hat p_x$ and dies for the same reason. Only the first and fourth terms survive.

Step 3: the first term. Grinding through the product rule for $[\hat A\hat B,\hat C\hat D]$ would be tedious. Better to notice that the only pair with any “spark” here is $\hat p_z$ and $\hat z$ ($[\hat z,\hat p_z]=\ii\hbar$); every other operator is a bystander. Pull the bystanders out front:

$$[\hat y\hat p_z,\hat z\hat p_x]=\hat y\,[\hat p_z,\hat z]\,\hat p_x =\hat y(-\ii\hbar)\hat p_x=-\ii\hbar\,\hat y\hat p_x\tag{5.1.2}$$

(Using $[\hat p_z,\hat z]=-[\hat z,\hat p_z]=-\ii\hbar$.)

Step 4: the fourth term. Here the sparking pair is $\hat z$ and $\hat p_z$:

$$[\hat z\hat p_y,\hat x\hat p_z]=\hat p_y\,[\hat z,\hat p_z]\,\hat x =\ii\hbar\,\hat x\hat p_y\tag{5.1.3}$$

Step 5: put it together.

$$[\hat L_x,\hat L_y]=-\ii\hbar\,\hat y\hat p_x+\ii\hbar\,\hat x\hat p_y =\ii\hbar(\hat x\hat p_y-\hat y\hat p_x)=\ii\hbar\hat L_z\tag{5.1.4}$$

Cycling $x\to y\to z\to x$ immediately yields the other two. All three together:

$$[\hat L_x,\hat L_y]=\ii\hbar\hat L_z,\qquad [\hat L_y,\hat L_z]=\ii\hbar\hat L_x,\qquad [\hat L_z,\hat L_x]=\ii\hbar\hat L_y\tag{5.1.5}$$

The result is not zero. That one small line of algebra has enormous consequences.

Is anything simultaneously sharp?

The three components fight each other — so is there any quantity that lives at peace with them? There is: the squared magnitude of the angular momentum,

$$\hat L^2\equiv\hat L_x^2+\hat L_y^2+\hat L_z^2\tag{5.1.7}$$

Verifying [L², L̂_z] = 0basic~4 minexpand

Work term by term. First a small general-purpose tool (the product rule):

$$[\hat A^2,\hat B]=\hat A[\hat A,\hat B]+[\hat A,\hat B]\hat A\tag{5.1.8}$$

First term $[\hat L_x^2,\hat L_z]$: using $[\hat L_x,\hat L_z]=-\ii\hbar\hat L_y$,

$$[\hat L_x^2,\hat L_z]=\hat L_x(-\ii\hbar\hat L_y)+(-\ii\hbar\hat L_y)\hat L_x =-\ii\hbar(\hat L_x\hat L_y+\hat L_y\hat L_x)\tag{5.1.9}$$

Second term $[\hat L_y^2,\hat L_z]$: using $[\hat L_y,\hat L_z]=\ii\hbar\hat L_x$,

$$[\hat L_y^2,\hat L_z]=\ii\hbar(\hat L_y\hat L_x+\hat L_x\hat L_y)\tag{5.1.10}$$

Third term $[\hat L_z^2,\hat L_z]=0$ (any operator commutes with its own powers).

Add the three: the first two are equal and opposite and cancel exactly:

$$[\hat L^2,\hat L_z]=0\tag{5.1.11}$$

By symmetry (the three directions enter on an equal footing), $\hat L^2$ commutes with $\hat L_x$ and $\hat L_y$ as well.

So the largest combination quantum mechanics lets us fix simultaneously is: the length plus one component. The conventional choice is $(\hat L^2,\hat L_z)$ — there is nothing physically special about $z$; it is pure notational tradition (in a real experiment, “the $z$ direction” is simply wherever your magnetic field points). This is the algebraic origin of the cone picture: the slant of the cone ($|\vec L|$) and its height ($L_z$) can be fixed together; the azimuthal angle cannot.

Why the commutator can determine the spectrum

The natural next question: what are the eigenvalues of $\hat L^2$ and $\hat L_z$?

The chapter-2 approach would be to write down a differential equation and solve it. But here there is an astonishing shortcut: from the commutation relations alone, without solving a single differential equation, every eigenvalue can be pinned down.

We have already seen the seed of the idea: in the algebraic solution of the harmonic oscillator, the single commutation relation $[\hat a,\hat a^\dagger]=1$ forces the levels into an evenly spaced ladder of $\hbar\omega$ — the commutator sets the rung spacing, positivity sets the bottom, and the spectrum is locked in. The angular momentum algebra is more intricate (the right-hand side is an operator, not a constant), but the same strategy works: build “ladder operators” $\hat L_x\pm\ii\hat L_y$ that climb the $L_z$ eigenvalues one rung at a time, then use the finiteness of $\hat L^2$ to cap the ladder at both ends.

∎Key formulas

• Angular momentum operators

  $$\hat L_z=\hat x\hat p_y-\hat y\hat p_x\quad(x\to y\to z\to x\ \text{cyclic})$$

  Coordinates and momenta in each term point along different directions: no ordering ambiguity, automatically Hermitian

• Fundamental commutation relations

  $$[\hat L_x,\hat L_y]=\ii\hbar\hat L_z\ \text{and cyclic}$$

  The true definition of angular momentum; anything satisfying it qualifies

• Uncertainty relation

  $$\sigma_{L_x}\sigma_{L_y}\ge\frac{\hbar}{2}\lvert\langle\hat L_z\rangle\rvert$$

  The bound depends on the state; when ⟨L_z⟩ = 0 both may vanish together

• Compatible partners

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

  Length plus one component can be fixed together; conventionally (L², L_z)

?Self-check4 questions

1. 1.

   Why is there no operator-ordering ambiguity in the construction L̂_z = x̂p̂_y − ŷp̂_x?

   • In each term the coordinate and the momentum belong to different directions and therefore commute
   • Because angular momentum is a conserved quantity
   • Because x̂ and p̂_y are both Hermitian operators
   • The quantisation rules require alphabetical ordering

   Submit
2. 2.

   Which statements about σ_{L_x}σ_{L_y} ≥ (ħ/2)|⟨L̂_z⟩| are correct? (Select all that apply.)

   Select all that apply

   • It is the generalised uncertainty relation applied directly to [L̂_x, L̂_y] = iħL̂_z
   • Its lower bound depends on the quantum state
   • It shows that L_x and L_y can never be simultaneously sharp in any state
   • In a state with ⟨L̂_z⟩ = 0, L_x and L_y may take definite values simultaneously

   Submit
3. 3.

   If the three components refuse to commute with one another, why can L̂² and L̂_z still be used together as labels?

   • Because [L̂², L̂_z] = 0, so they share common eigenstates
   • Because L̂² is not an observable and plays no role in measurement
   • Because the z direction is physically special
   • Because L̂² is a constant

   Submit
4. 4.

   A state has L̂² eigenvalue 6ħ² (next section: this corresponds to l = 2). What is the magnitude |L| in units of ħ? (Give three decimal places.)

   ħ1% relative tolerance

   Submit

What comes next

We have placed our bet on the commutation relations: the claim is that

$$[\hat L_x,\hat L_y]=\ii\hbar\hat L_z\ \text{and its cyclic partners}\tag{5.1.13}$$

alone determine every eigenvalue of $\hat L^2$ and $\hat L_z$.

The next section makes good on that promise — climbing out the entire spectrum rung by rung with ladder operators. And the answer hides a surprise: the algebra allows half again as many eigenvalues as orbital motion actually uses. That extra half is the seat reserved for spin.