Coupling angular momenta and Clebsch–Gordan coefficients

The question posed at the end of the previous section: put two angular momenta together — what is the total?

First, why the question is inescapable. The electron in hydrogen carries both orbital $\hat{\vec L}$ and spin $\hat{\vec S}$; helium has the spins of two electrons; in the deuteron, the proton and neutron each carry spin. The moment there is more than one share of angular momentum, “how much altogether” becomes compulsory — how spectral lines split, whether two electrons can squeeze into the same orbital, where hydrogen’s 21 cm radio line comes from: all of it hangs on the answer.

Classical mechanics answers with primary-school arithmetic: $\vec J=\vec J_1+\vec J_2$, vector addition, with the length continuously adjustable between $|J_1-J_2|$ and $J_1+J_2$ depending on the angle. In quantum mechanics that road is blocked from the first step: no single “vector” even has three simultaneously sharp components (section 5.1), so “add them component by component” is a non-starter. We need a different question.

The right question: two bases

Work the problem to death on the smallest non-trivial example: two spin 1/2 particles (say, the electron’s spin and the proton’s spin in hydrogen). Each spin has two states, so the combined space is four-dimensional, and the most natural basis is “each particle reports its own state”:

$$\ket{\uparrow\uparrow},\quad\ket{\uparrow\downarrow},\quad \ket{\downarrow\uparrow},\quad\ket{\downarrow\downarrow}\tag{5.6.1}$$

(first arrow = particle 1, second = particle 2). This product basis consists of the joint eigenstates of $(\hat S_{1z},\hat S_{2z})$ and answers “which way does each one point”.

Now define the total spin operator:

$$\hat{\vec S}\equiv\hat{\vec S}_1+\hat{\vec S}_2\tag{5.6.2}$$

(operators acting on different particles commute, $[\hat S_{1i},\hat S_{2j}]=0$). It is easy to check that $\hat{\vec S}$ obeys the standard angular momentum commutation relations — so the whole machinery of section 5.2 applies automatically: $(\hat S^2,\hat S_z)$ have joint eigenstates $\ket{S,M}$ with eigenvalues $S(S+1)\hbar^2$ and $M\hbar$. This total angular momentum basis answers “how much altogether, and which way does the total point”.

Each basis answers its own question, but in general you cannot have both: $\hat S^2$ contains terms like $\hat S_{1x}\hat S_{2x}$ that flip individual spins, and it does not commute with $\hat S_{1z}$. “Definite total length” and “definite individual orientations” are an incompatible pair of facts — this is exactly where quantum addition parts ways with classical addition. What remains is a pure change of basis: expand $\ket{S,M}$ in the product basis. The expansion coefficients are called Clebsch–Gordan (CG) coefficients — an intimidating name for nothing more than a table of basis-change coefficients.

Building them by hand: singlet and triplet

Two spin 1/2 particles: constructing every |S, M⟩ from the product basisbasic~12 minexpand

Step 1: sort by M first. $\hat S_z=\hat S_{1z}+\hat S_{2z}$ is diagonal in the product basis — the two z components simply add (the one part that does behave like classical addition):

Product state	$M$
$\ket{\uparrow\uparrow}$	$+1$
$\ket{\uparrow\downarrow},\ \ket{\downarrow\uparrow}$	$0$
$\ket{\downarrow\downarrow}$	$-1$

The largest $M$ is $1$, so the largest $S$ is $1$; and $M=0$ holds two states, meaning one dimension does not belong to the $S=1$ family — a preview that an $S=0$ is also hiding here.

Step 2: pin down the top state. The $M=+1$ state is unique, so it can only be the top of the $S=1$ family:

$$\ket{1,1}=\ket{\uparrow\uparrow}\tag{5.6.3}$$

(One can verify directly that $\hat S^2\ket{\uparrow\uparrow}=2\hbar^2\ket{\uparrow\uparrow}$, i.e. $S(S+1)=2$.)

Step 3: climb down with the lowering operator. The total lowering operator is the sum of the two particles’ operators: $\hat S_-=\hat S_{1-}+\hat S_{2-}$. On the left, the section 5.2 formula gives $\hat S_-\ket{1,1}=\hbar\sqrt{1\cdot2-1\cdot0}\ket{1,0}=\hbar\sqrt2\ket{1,0}$; on the right, act particle by particle (for spin 1/2, $\hat S_-\ket{\uparrow}=\hbar\ket{\downarrow}$):

$$(\hat S_{1-}+\hat S_{2-})\ket{\uparrow\uparrow} =\hbar\ket{\downarrow\uparrow}+\hbar\ket{\uparrow\downarrow}\tag{5.6.4}$$

Setting the two sides equal:

$$\ket{1,0}=\frac{\ket{\uparrow\downarrow}+\ket{\downarrow\uparrow}}{\sqrt2}\tag{5.6.5}$$

Step 4: one more rung. Repeat for $\ket{1,0}$: the left side gives $\hbar\sqrt2\ket{1,-1}$; acting on the expression above, $\ket{\uparrow\downarrow}\to\hbar\ket{\downarrow\downarrow}$ (only particle 1 can be lowered), $\ket{\downarrow\uparrow}\to\hbar\ket{\downarrow\downarrow}$, totalling $\frac{2\hbar}{\sqrt2}\ket{\downarrow\downarrow}=\hbar\sqrt2\ket{\downarrow\downarrow}$:

$$\ket{1,-1}=\ket{\downarrow\downarrow}\tag{5.6.6}$$

Bottom reached. The three members of the $S=1$ family are all present — the triplet.

Step 5: recover the missing dimension. Three of the four dimensions are spent; the remaining one must lie in the $M=0$ plane, orthogonal to $\ket{1,0}$. There is only one orthogonal combination (up to a phase convention):

$$\ket{0,0}=\frac{\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}}{\sqrt2}\tag{5.6.7}$$

Check that it really has $S=0$: $\hat S_-\ket{0,0} =\frac{\hbar\ket{\downarrow\downarrow}-\hbar\ket{\downarrow\downarrow}}{\sqrt2}=0$, and likewise $\hat S_+$ gives zero — it can climb neither up nor down, so it can only be a lone $S=0$. This is the singlet.

Summary: the four product states regroup as "$3+1$":

$$\underbrace{\ket{1,1},\ \ket{1,0},\ \ket{1,-1}}_{\text{triplet }S=1} \qquad \underbrace{\ket{0,0}}_{\text{singlet }S=0}\tag{5.6.8}$$

Written $\tfrac12\otimes\tfrac12=1\oplus0$. Dimension check: $2\times2=3+1$. ✓

The expansion coefficients in that derivation — $1$, $\tfrac1{\sqrt2}$, $-\tfrac1{\sqrt2}$ — are the entire true identity of the CG coefficients $\braket{m_1\,m_2}{S\,M}$: “the amplitude of product state $\ket{m_1 m_2}$ inside total state $\ket{S,M}$”. Example of use: the system is in $\ket{1,0}$ and you measure particle 1’s $S_{1z}$; the probability of “up” is the squared modulus of the amplitude $\tfrac1{\sqrt2}$, i.e. $50\%$. The page-long CG tables in textbook appendices are nothing but steps 2 through 5 repeated for every $(j_1,j_2)$ — the method is identical each time: pin the top, climb with the lowering operator, find the new family in the orthogonal complement.

The general rule and dimension counting

For arbitrary $j_1,j_2$ the method is exactly as above, and the conclusion just as clean. The allowed values of the total angular momentum $J$ are

$$J=|j_1-j_2|,\ |j_1-j_2|+1,\ \dots,\ j_1+j_2\qquad(\text{each }J\text{ appearing once})\tag{5.6.11}$$

— from “antiparallel” to “parallel” in steps of 1. The classical “length continuously adjustable between difference and sum” becomes “between difference and sum, adjustable in discrete integer steps”. Each $J$ family has $2J+1$ members, and the total dimension must balance:

$$(2j_1+1)(2j_2+1)=\sum_{J=|j_1-j_2|}^{j_1+j_2}(2J+1)\tag{5.6.12}$$

A quick self-check tool. One example we will need immediately: an electron in a hydrogen p orbital, $l=1$ coupled to spin $s=\tfrac12$:

$$1\otimes\tfrac12=\tfrac32\oplus\tfrac12,\qquad 3\times2=4+2\ \checkmark\tag{5.6.13}$$

The six states of a p electron regroup into four with $j=\tfrac32$ and two with $j=\tfrac12$. Remember this "$4+2$" — next section it becomes two real lines in a spectrum.

∎Key formulas

• Total angular momentum and two bases

  $$\hat{\vec S}=\hat{\vec S}_1+\hat{\vec S}_2,\qquad \braket{m_1 m_2}{S\,M}=\text{CG coefficients}$$

  Product basis answers "which way each"; total basis answers "how much altogether" — the two are incompatible

• Triplet

  $$\ket{1,\pm1}=\ket{\uparrow\uparrow},\ket{\downarrow\downarrow},\quad \ket{1,0}=\tfrac{1}{\sqrt2}(\ket{\uparrow\downarrow}+\ket{\downarrow\uparrow})$$

  Exchange symmetric; S = 1

• Singlet

  $$\ket{0,0}=\tfrac{1}{\sqrt2}(\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow})$$

  Exchange antisymmetric, rotationally invariant, entangled; opposite outcomes along every direction

• Coupling rule

  $$J=|j_1-j_2|,\dots,j_1+j_2;\qquad (2j_1{+}1)(2j_2{+}1)=\textstyle\sum_J(2J{+}1)$$

  Dimension-count self-check; e.g. 1⊗½=3/2⊕½, 6=4+2

?Self-check4 questions

1. 1.

   Why can "definite total length S" and "definite S_{1z} of particle 1" generally not be had together?

   • Because Ŝ² does not commute with Ŝ_{1z} — Ŝ² contains terms like Ŝ₁ₓŜ₂ₓ that flip individual spins
   • Because measurement always disturbs the system
   • Because the two particles are too close together
   • They actually can be had together; it is just inconvenient to compute

   Submit
2. 2.

   Which statements about the singlet |0,0⟩ = (|↑↓⟩ − |↓↑⟩)/√2 are correct? (Select all that apply.)

   Select all that apply

   • Measured along any direction, the two particles' spin components are strictly opposite
   • It changes sign under exchange of the two particles
   • It can be written as a product of two single-particle states
   • It is an entangled state

   Submit
3. 3.

   The system is in |↑↓⟩. Measuring total spin S², the possible outcomes and probabilities are:

   • 2ħ² (probability 1/2) or 0 (probability 1/2)
   • Certainly 0, because up and down cancel
   • Certainly 2ħ²
   • ħ² (probability 1)

   Submit
4. 4.

   Couple j₁ = 2 with j₂ = 1. How many allowed values of J are there? (Count using |j₁−j₂| ≤ J ≤ j₁+j₂.)

   values0% relative tolerance

   Submit

What comes next

We can now add $\hat{\vec L}$ and $\hat{\vec S}$ into a total angular momentum $\hat{\vec J}$, and we know a p electron’s six states regroup into a $j=\tfrac32$ family and a $j=\tfrac12$ family.

But so far this is only a mathematical regrouping — the two families still have exactly the same energy. The next section divides the estate between them: as the electron moves around the nucleus, its own magnetic moment feels an internally generated magnetic field, and the coupling term $\hat{\vec L}\cdot\hat{\vec S}$ pushes the energies of different-$j$ states apart. The two D lines in a sodium lamp’s yellow glow, 0.6 nm apart, are exactly this inheritance being split.