Once the states $\ket{j,m}$ are known, every angular momentum operator is a matrix in them. $\hat{J}_{z}$ is diagonal, with $m\hbar$ down the diagonal. $\hat{J}_{+}$ and $\hat{J}_{-}$ sit just above and just below it, with square roots that come from the lengths of raised states. $\hat{J}_{x}$ and $\hat{J}_{y}$ are their halves and differences. For $j=\tfrac12$ the matrices are two by two and turn out to be the Pauli matrices of the spin chapter; for $j=1$ they are three by three.

A timetable lists which stop follows which. The matrix of $\hat{J}_{+}$ is the timetable of the ladder: each state is sent to the next one up, and the entry says with what weight.

In the basis $\ket{j,j},\ket{j,j-1},\dots,\ket{j,-j}$, is diagonal and have one nonzero diagonal line each, with the coefficients below.

• $j=\tfrac12$: ${\widehat{J}}_z=\frac{\mathrm{\hslash }}{2}(\begin{array}{cc}1& 0\\ 0& \end{array}$, , .

The squared length of the raised state is the average of $\hat{J}_{-}\hat{J}_{+}$, since the two are adjoints. Written through and , that average is known on . Choosing the phase real and positive gives the coefficient.

The adjoint of $\hat{J}_{+}$ is $\hat{J}_{-}$.

From the lesson on ladder operators.

Each term is known on $\ket{j,m}$.

The raised state is a multiple of $\ket{j,m+1}$; the usual convention makes the multiple real and positive.

Raising $\ket{j,m}$ gives $\hbar\sqrt{j(j+1)-m(m+1)}\,\ket{j,m+1}$.

Lowering uses the same formula with $m-1$ in place of $m+1$.

In the basis ordered $m=1$, 0, $-1$, the matrix of $\hat{J}_{+}$ for has its two entries just above the diagonal: takes each state to the one listed before it.

For $j=\tfrac12$ the matrices are two by two. $\hat{J}_{x}$ has in both off-diagonal places.

Great work! You now know how to:

• compute raising and lowering coefficients
• write the matrices of $\hat{\mathbf{J}}$ for $j=\tfrac12$ and