Angular momentum eigenvalues and spherical harmonics

The previous section placed a bet: the commutation relations $[\hat L_x,\hat L_y]=\ii\hbar\hat L_z$ (and their cyclic partners) alone can determine every eigenvalue. This section pays it off. The whole argument uses just three ingredients: the commutation relations, the Hermiticity of the operators, and the plain truth that “a squared length cannot be negative”.

Building the ladder

The lesson of the harmonic oscillator: to “hop rungs” among the eigenvalues, you need an operator $\hat A$ whose commutator with $\hat L_z$ is proportional to $\hat A$ itself — then acting with $\hat A$ on an eigenstate produces another eigenstate, with the eigenvalue shifted by one rung. Try this combination:

$$\hat L_\pm\equiv\hat L_x\pm\ii\hat L_y\tag{5.2.1}$$

They are not Hermitian ($\hat L_+^\dagger=\hat L_-$) and represent no observable — they are tools, not objects of study. Compute the two key commutators (both expand directly from the fundamental relations):

$$[\hat L_z,\hat L_\pm]=[\hat L_z,\hat L_x]\pm\ii[\hat L_z,\hat L_y] =\ii\hbar\hat L_y\pm\ii(-\ii\hbar\hat L_x)=\pm\hbar\hat L_\pm\tag{5.2.2}$$ $$[\hat L^2,\hat L_\pm]=0\quad(\hat L^2\ \text{commutes with every component})\tag{5.2.3}$$

The first line says: $\hat L_+$ raises the $L_z$ eigenvalue by $\hbar$ and $\hat L_-$ lowers it by $\hbar$ — that is the ladder. The second line says: while you climb, the eigenvalue of $\hat L^2$ does not budge — the ladder stands on a cone of fixed length.

Climbing out the full spectrum

The spectrum by pure algebra: from commutators to l(l+1)ħ²advanced~12 minexpand

Set up notation. Let $\ket{\lambda,m}$ be a joint eigenstate:

$$\hat L^2\ket{\lambda,m}=\lambda\hbar^2\ket{\lambda,m},\qquad \hat L_z\ket{\lambda,m}=m\hbar\ket{\lambda,m}\tag{5.2.4}$$

(Factoring out $\hbar$ makes $\lambda$ and $m$ pure numbers. At this point we know nothing about either.)

Step 1: check that the ladder really climbs. Ask what $\hat L_+\ket{\lambda,m}$ is an eigenstate of:

$$\hat L_z\big(\hat L_+\ket{\lambda,m}\big) =\big(\hat L_+\hat L_z+[\hat L_z,\hat L_+]\big)\ket{\lambda,m} =\big(m\hbar+\hbar\big)\hat L_+\ket{\lambda,m}\tag{5.2.5}$$

So $\hat L_+\ket{\lambda,m}$ is either the zero vector or an eigenstate with $m+1$. Likewise $\hat L_-$ gives $m-1$. And $[\hat L^2,\hat L_\pm]=0$ guarantees $\lambda$ stays put.

Step 2: the ladder must have a top. The crucial physical input: a component cannot exceed the total length. Precisely,

$$\lambda\hbar^2-m^2\hbar^2 =\bra{\lambda,m}\hat L^2-\hat L_z^2\ket{\lambda,m} =\bra{\lambda,m}\hat L_x^2+\hat L_y^2\ket{\lambda,m}\ge0\tag{5.2.6}$$

The last step holds because the expectation of $\hat L_x^2$ is $\|\hat L_x\ket{\lambda,m}\|^2$, a squared length, hence non-negative. So $m^2\le\lambda$: $m$ is trapped in a finite interval and the ladder cannot climb forever. There must be a top rung $m_{\max}$ with

$$\hat L_+\ket{\lambda,m_{\max}}=0\tag{5.2.7}$$

(Not “acts and survives” but outright annihilated — that is the only way a ladder can terminate.)

Step 3: solve for λ from the top-rung condition. We need an identity, obtained by direct expansion:

$$\hat L_-\hat L_+=(\hat L_x-\ii\hat L_y)(\hat L_x+\ii\hat L_y) =\hat L_x^2+\hat L_y^2+\ii[\hat L_x,\hat L_y] =\hat L^2-\hat L_z^2-\hbar\hat L_z\tag{5.2.8}$$

Apply it to the top-rung state: the left side is zero, and the right side gives

$$0=\big(\lambda-m_{\max}^2-m_{\max}\big)\hbar^2 \quad\Longrightarrow\quad \lambda=m_{\max}(m_{\max}+1)\tag{5.2.9}$$

Step 4: treat the bottom the same way. With $\hat L_-\ket{\lambda,m_{\min}}=0$, the identity $\hat L_+\hat L_-=\hat L^2-\hat L_z^2+\hbar\hat L_z$ gives

$$\lambda=m_{\min}(m_{\min}-1)\tag{5.2.10}$$

Setting the two expressions equal: $m_{\max}(m_{\max}+1)=m_{\min}(m_{\min}-1)$. Read as a quadratic in $m_{\min}$, the two roots are $m_{\min}=-m_{\max}$ and $m_{\min}=m_{\max}+1$. The latter is absurd (a bottom above the top), so

$$m_{\min}=-m_{\max}\tag{5.2.11}$$

The ladder is symmetric about zero.

Step 5: integers or half-integers? Climbing from bottom to top in steps of $+1$ takes some whole number of steps $N$:

$$m_{\max}-m_{\min}=2m_{\max}=N \quad\Longrightarrow\quad m_{\max}=\frac{N}{2}=0,\ \tfrac12,\ 1,\ \tfrac32,\ 2,\ \dots\tag{5.2.12}$$

Conclusion. Writing $l\equiv m_{\max}$ (the conventional symbol):

$$\hat L^2\ket{l,m}=l(l+1)\hbar^2\ket{l,m},\qquad \hat L_z\ket{l,m}=m\hbar\ket{l,m}\tag{5.2.13}$$

where $l$ is a non-negative integer or half-integer, and $m$ runs from $-l$ to $+l$ one rung at a time — $2l+1$ values in all. Not one differential equation solved, and the whole spectrum is in hand.

While we are at it, fix the ladder-operator coefficients (we will need them for building matrices and for Clebsch–Gordan coefficients). From the identity above, $\|\hat L_\pm\ket{l,m}\|^2=\big[l(l+1)-m(m\pm1)\big]\hbar^2$; with the positive-real phase convention:

$$\hat L_\pm\ket{l,m}=\hbar\sqrt{l(l+1)-m(m\pm1)}\;\ket{l,m\pm1}\tag{5.2.14}$$

Substitute $m=\pm l$ and the square root vanishes on the spot — the formula seals the top and bottom of the ladder all by itself.

Back to the sphere: orbital angular momentum drops half the values

The algebra allows half-integer $l$. But orbital angular momentum — the kind genuinely built from $\vec r\times\vec p$ — realises only the integer half. The reason lives in the wavefunction. In spherical coordinates $(r,\theta,\phi)$ (with $\theta$ the angle from the z axis and $\phi$ the azimuthal angle around it), $\hat L_z$ takes a remarkably clean form:

$$\hat L_z=-\ii\hbar\frac{\partial}{\partial\phi}\tag{5.2.15}$$

Its eigenfunctions are $\ee^{\ii m\phi}$. Now the decisive blow: $\phi$ and $\phi+2\pi$ are the same point in space, and a wavefunction must have a single value at a single point:

$$\ee^{\ii m(\phi+2\pi)}=\ee^{\ii m\phi} \quad\Longrightarrow\quad \ee^{2\pi\ii m}=1 \quad\Longrightarrow\quad m\ \text{must be an integer}\tag{5.2.16}$$

Were $m=\tfrac12$, the wavefunction would change sign after one full turn — two values at the same point, which no function on space is allowed to do. So orbital angular momentum has only $l=0,1,2,\dots$. The half-integer seats stand empty: the algebra plainly permits them, but a position wavefunction cannot accommodate them. Hold on to this loose end — next section it finds its owner.

Spherical harmonics: what the eigenstates look like

The joint eigenfunctions of $(\hat L^2,\hat L_z)$ depend only on the angles. Written $Y_l^m(\theta,\phi)$, they are the spherical harmonics — “harmonics on the sphere”, playing the role that $\sin$ and $\cos$ play on a line. The construction copies the algebra: the top-rung condition $\hat L_+Y_l^l=0$ is a first-order differential equation with solution $Y_l^l\propto\sin^l\theta\,\ee^{\ii l\phi}$, and $\hat L_-$ walks you down from there. The first few:

$$Y_0^0=\frac{1}{\sqrt{4\pi}},\qquad Y_1^0=\sqrt{\frac{3}{4\pi}}\cos\theta,\qquad Y_1^{\pm1}=\mp\sqrt{\frac{3}{8\pi}}\sin\theta\,\ee^{\pm\ii\phi}\tag{5.2.17}$$

∎Key formulas

• Eigenvalue spectrum

  $$\hat L^2\ket{l,m}=l(l+1)\hbar^2\ket{l,m},\quad \hat L_z\ket{l,m}=m\hbar\ket{l,m}$$

  A purely algebraic result; m = −l, …, +l gives 2l+1 values

• Ladder operators

  $$\hat L_\pm\ket{l,m}=\hbar\sqrt{l(l+1)-m(m\pm1)}\,\ket{l,m\pm1}$$

  The coefficient vanishes at m = ±l: the ladder caps itself at both ends

• Key identity

  $$\hat L_\mp\hat L_\pm=\hat L^2-\hat L_z^2\mp\hbar\hat L_z$$

  The engine of the spectrum derivation; also yields the ladder coefficients

• Orbital angular momentum: integers only

  $$\ee^{2\pi\ii m}=1\ \Rightarrow\ m\in\mathbb{Z}$$

  Single-valuedness of the wavefunction excludes half-integers; those seats are reserved for spin

• Spherical harmonics (l=1)

  $$Y_1^0=\sqrt{\tfrac{3}{4\pi}}\cos\theta,\quad Y_1^{\pm1}=\mp\sqrt{\tfrac{3}{8\pi}}\sin\theta\,\ee^{\pm\ii\phi}$$

  m=0 is a dumbbell along the axis, |m|=1 a doughnut at the equator

?Self-check4 questions

1. 1.

   In the ladder-operator derivation, what is the physical basis for "the ladder must have a top"?

   • ⟨L_x² + L_y²⟩ ≥ 0, so m² ≤ l(l+1) and m cannot grow without bound
   • Energy conservation forbids the angular momentum from growing without bound
   • ħ is the minimum action, and climbing cannot exceed it
   • After finitely many applications the operator L_+ itself becomes zero

   Submit
2. 2.

   Why can orbital angular momentum not have l = 1/2, even though the algebra allows it?

   • With m = 1/2 the wavefunction e^{imφ} changes sign after one full turn about z, but a spatial wavefunction must be single-valued at each point
   • Half-integer l would make the energy negative
   • l = 1/2 has never been observed, so it is excluded by hand
   • Half-integer l breaks the commutativity of L² and L_z

   Submit
3. 3.

   Acting with L̂_+ on |l=1, m=0⟩ gives ħ·c·|1,1⟩. What is the coefficient c?

   • √2
   • 1
   • 2
   • √6

   Submit
4. 4.

   A beam of particles is in an l = 3 state. At most how many distinct outcomes can a measurement of L_z give?

   outcomes0% relative tolerance

   Submit

What comes next

The algebra allows $l=\tfrac12,\tfrac32,\dots$, yet wavefunctions on the sphere cannot house them. Mathematics reserved the seats; space cannot fill them — so were those seats reserved for nothing?

In 1922, two German physicists fired a beam of silver atoms through an inhomogeneous magnetic field and found two separated traces on a glass plate — precisely the value of $2l+1$ at $l=\tfrac12$. The half-integer seat has an owner, and the angular momentum it carries has nothing to do with motion through space.