Clebsch–Gordan coefficients

Intuition

The numbers that write each coupled state in the product basis are the Clebsch–Gordan coefficients. They are found the way the triplet was: start at the top state, the one product with the largest $M$ ; lower it with the total lowering operator to fill the multiplet of the largest $J$ ; at the next value of $M$ the state orthogonal to what has been built is the top of the next multiplet; and so on down. With the standard choice of phases they are real numbers. They vanish unless $M=m_{1}+m_{2}$ and the triangle rule holds, and their squares are probabilities: the chance of finding each pair $(m_{1},m_{2})$ in a state of definite $J$ .

Converting a recipe from cups to grams uses a fixed table: each amount in the new units is a set mixture of the old. Clebsch–Gordan coefficients are the conversion table between the uncoupled and the coupled bases, worked out once and used forever.

The six product states of $j_{1}=1$ and $j_{2}=\tfrac12$ at . The coupled state lies on the line and is made of its two products, found with probabilities and , the squares of its Clebsch–Gordan coefficients.

The coupled states in terms of the uncoupled ones:

\begin{aligned}\ket{J,M}&=\sum_{m_{1},m_{2}}C^{JM}_{m_{1}m_{2}}\,\ket{j_{1},m_{1}}\ket{j_{2},m_{2}}\\ C^{JM}_{m_{1}m_{2}}&=\braket{j_{1}m_{1};j_{2}m_{2}|J\,M}\end{aligned}

Properties

$C^{JM}_{m_{1}m_{2}}=0$ unless and .

Coupling 1 and \tfrac12

\ket{\tfrac32,\tfrac12}=\sqrt{\tfrac23}\,\ket{1,0}\ket{\uparrow}+\sqrt{\tfrac13}\,\ket{1,1}\ket{\downarrow}

Start from the top product, which is $\ket{\tfrac32,\tfrac32}$ . Lower it in two ways: with the total lowering operator acting on the coupled state, and with the sum of the parts’ lowering operators acting on the product. Equating the two gives the next state of the multiplet; the combination orthogonal to it is the top of .

Proof steps

\ket{\tfrac32,\tfrac32}=\ket{1,1}\ket{\uparrow}

The only product with $M=\tfrac32$ .

\hat{J}_{-}\ket{\tfrac32,\tfrac32}=\hbar\sqrt3\,\ket{\tfrac32,\tfrac12}

$\sqrt{J(J+1)-M(M-1)}=\sqrt{\tfrac{15}{4}-\tfrac34}=\sqrt3$ .

\begin{aligned}(\hat{J}_{1-}+\hat{J}_{2-})\ket{1,1}\ket{\uparrow}&=\hbar\sqrt2\,\ket{1,0}\ket{\uparrow}\\ &\quad+\hbar\,\ket{1,1}\ket{\downarrow}\end{aligned}

Each part lowers its own factor.

\ket{\tfrac32,\tfrac12}=\sqrt{\tfrac23}\,\ket{1,0}\ket{\uparrow}+\sqrt{\tfrac13}\,\ket{1,1}\ket{\downarrow}

Equate the two and divide by $\hbar\sqrt3$ .

\ket{\tfrac12,\tfrac12}=-\sqrt{\tfrac13}\,\ket{1,0}\ket{\uparrow}+\sqrt{\tfrac23}\,\ket{1,1}\ket{\downarrow}

Orthogonal to the last state, with the coefficient of $\ket{1,1}\ket{\downarrow}$ positive by convention.

Applications

Practice

Squares Are Probabilities

Each Clebsch–Gordan coefficient is the amplitude for one pair $(m_{1},m_{2})$ in a coupled state, so its square is the probability of finding that pair.

\ket{\tfrac32,\tfrac12}=\sqrt{\tfrac23}\,\ket{1,0}\ket{\uparrow}+\sqrt{\tfrac13}\,\ket{1,1}\ket{\downarrow}

Try it

In the state $\ket{\tfrac32,\tfrac12}$ of $1\otimes\tfrac12$ , what is the probability of finding the spin up? Give three decimal places.