The addition rule

Intuition

Which totals can two angular momenta $j_{1}$ and $j_{2}$ make? Every value from the difference $|j_{1}-j_{2}|$ to the sum $j_{1}+j_{2}$ , in whole steps, each exactly once. Counting states settles it. The total projection $M=m_{1}+m_{2}$ runs from $-(j_{1}+j_{2})$ to $j_{1}+j_{2}$ , and the number of product states at each $M$ grows by one per step down from the top until it levels off. Each multiplet $J$ puts one state at every $M$ from $-J$ to $J$ , so a new multiplet must begin wherever the count grows. The classical picture agrees: the length of the sum of two vectors lies between the difference and the sum of their lengths.

Two sticks of lengths 2 and 1, joined by a hinge, can put their far end anywhere from 1 to 3 from the start. Quantum angular momenta do the same, but only in whole steps: $J=1$ , 2 or 3.

The classical picture of adding $j_{1}=2$ and $j_{2}=1$ : the tip of can lie anywhere on the dashed circle, so the length of runs from to . Quantum mechanics keeps the range and allows only .

Two angular momenta $j_{1}$ and $j_{2}$ combine into the totals, in whole steps:

\begin{aligned}J&=|j_{1}-j_{2}|,\ \dots,\ j_{1}+j_{2}\\ \sum_{J}(2J+1)&=(2j_{1}+1)(2j_{2}+1)\end{aligned}

Properties

Each allowed $J$ appears exactly once, with its $2J+1$ states $\ket{J,M}$ , $M=-J,\dots,J$ .

The possible totals

J\in\bigl\{|j_{1}-j_{2}|,\ \dots,\ j_{1}+j_{2}\bigr\},\ \text{each once}

Take $j_{1}\ge j_{2}$ and count the product states at each value of $M$ . Each multiplet contributes exactly one state to every it covers, so the number of multiplets with is the number of states at minus the number at . That difference is 1 from the top down to , and 0 below it.

Proof steps

M=m_{1}+m_{2},\qquad -(j_{1}+j_{2})\le M\le j_{1}+j_{2}

The projections add.

\begin{aligned}&k(M)=j_{1}+j_{2}+1-|M|\\ &\text{for}\ j_{1}-j_{2}\le|M|\le j_{1}+j_{2}\end{aligned}

Count the pairs $(m_{1},m_{2})$ on the line $m_{1}+m_{2}=M$ ; for the count stays at .

\text{a multiplet }J\ \text{has one state at each }M,\ |M|\le J

$\hat{J}_{\pm}$ move through its $2J+1$ states one step at a time.

\begin{aligned}&\#\{J=M\}=k(M)-k(M+1)=1\\ &\text{for}\ j_{1}-j_{2}\le M\le j_{1}+j_{2}\end{aligned}

The multiplets with $J>M$ already fill $k(M+1)$ of the states at $M$ ; below $M=j_{1}-j_{2}$ the difference is 0.

J=j_{1}-j_{2},\ j_{1}-j_{2}+1,\ \dots,\ j_{1}+j_{2}

One multiplet for each of these values, and no others.

Applications

Practice

The Rule

Two angular momenta combine into every total from the difference to the sum of their quantum numbers, in whole steps.

J=|j_{1}-j_{2}|,\ \dots,\ j_{1}+j_{2}

Try it

Which totals can $j_{1}=2$ and $j_{2}=1$ make?