Ladder Operators

Ladder operators convert the angular momentum algebra into a practical way of constructing multiplets. They raise or lower the $m$ label while leaving $j$ fixed.

Define

$$J_+=J_x+iJ_y, \qquad J_-=J_x-iJ_y.$$

Commutation Relations

The ladder operators obey

$$[J_z,J_\pm]=\pm\hbar J_\pm,$$

and

$$[J^2,J_\pm]=0.$$

The first relation says $J_\pm$ changes the $J_z$ eigenvalue. The second says it does not change the $J^2$ eigenvalue.

Raising and Lowering

If

$$J_z\lvert j,m\rangle = \hbar m\lvert j,m\rangle,$$

then

$$J_z(J_\pm\lvert j,m\rangle) = \hbar(m\pm1)(J_\pm\lvert j,m\rangle).$$

If the result is nonzero, $J_+$ raises $m$ by one and $J_-$ lowers $m$ by one. The zero vector at an endpoint is not an additional physical state.

Normalization Factors

With normalized states and the standard Condon–Shortley convention,

$$J_\pm\lvert j,m\rangle = \hbar \sqrt{j(j+1)-m(m\pm1)} \lvert j,m\pm1\rangle.$$

The coefficient vanishes at the top or bottom of the multiplet:

$$J_+\lvert j,j\rangle=0, \qquad J_-\lvert j,-j\rangle=0.$$

Where the Coefficient Comes From

Use

$$J_-J_+ = J^2-J_z^2-\hbar J_z.$$

Then

$$\lVert J_+\lvert j,m\rangle\rVert^2 = \langle j,m|J_-J_+|j,m\rangle.$$

Substituting the eigenvalues gives

$$\lVert J_+\lvert j,m\rangle\rVert^2 = \hbar^2\left[j(j+1)-m(m+1)\right].$$

The square root gives the raising coefficient. The lowering coefficient follows similarly.

Allowed m Values

Within a unitary angular-momentum multiplet of fixed finite $J^2$, repeated raising and lowering must stop because norms cannot become negative. This algebraic termination gives a finite chain

$$m=-j,-j+1,\ldots,j.$$

The number of states is $2j+1$. Orbital angular momentum has integer $\ell$; spin allows half-integer $j$ as well.

Building a spin-one multiplet

Starting from the normalized highest state, positive ladder coefficients give

$$J_-|1,1\rangle=\sqrt2\hbar|1,0\rangle, \qquad J_-|1,0\rangle=\sqrt2\hbar|1,-1\rangle.$$

Thus $J_-^2|1,1\rangle=2\hbar^2|1,-1\rangle$, and a third application gives zero. Normalize after each nonzero application if you want a unit state vector.

The operators satisfy $J_+^\dagger=J_-$, so neither is a self-adjoint observable by itself. They are also not unitary: they change norms and annihilate an endpoint. Their coefficients are matrix elements, not transition probabilities without a specified interaction. Physical transverse rotations instead use the Hermitian combinations $J_x=(J_++J_-)/2$ and $J_y=(J_+-J_-)/(2i)$.

Exercises

1. Compute $J_+\lvert 1,0\rangle$.

Solution

Use

$$J_+\lvert j,m\rangle = \hbar\sqrt{j(j+1)-m(m+1)}\lvert j,m+1\rangle.$$

For $j=1,m=0$, the coefficient is $\hbar\sqrt{2}$, so

$$J_+\lvert 1,0\rangle = \hbar\sqrt2\,\lvert1,1\rangle.$$

2. Why does $J_+\lvert j,j\rangle=0$?

Solution

The coefficient is

$$\hbar\sqrt{j(j+1)-j(j+1)}=0.$$

There is no allowed state with $m=j+1$ in the fixed-$j$ multiplet.

References

• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020, chapter 3.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.