At the end of last time, we derived the eigenvalue spectrum of the simultaneously diagonalized operators \( \hat{J}^2 \) and \( \hat{J}_z \), finding the results
\[ \begin{aligned} \hat{J}{}^2 \ket{j,m} = \hbar^2 j(j+1) \ket{j,m} \\ \hat{J}_z \ket{j,m} = \hbar m \ket{j,m}, \end{aligned} \]
where \( j \) is integer or half-integer, and \( -j \leq m \leq j \).
The raising and lowering operators \( \hat{J}_{\pm} \) were instrumental in this derivation, but although we know that they act to give us \( \hat{J}_z \) eigenstates, we still need to fix the normalization. The best way to do this is to consider matrix elements, assuming the basic states \( \ket{j,m} \) to be properly normalized:
\[ \begin{aligned} \bra{j', m'} \hat{J}{}^2 \ket{j,m} = \hbar^2 j(j+1) \delta_{j,j'} \delta_{m,m'}, \\ \bra{j', m'} \hat{J}_z \ket{j,m} = \hbar m \delta_{j,j'} \delta_{m,m'}. \end{aligned} \]
We can now recycle our earlier rewriting of the product \( \hat{J}+^\dagger \hat{J}+ \), starting first with the diagonal matrix elements:
\[ \begin{aligned} \bra{j,m} \hat{J}_+^\dagger \hat{J}_+ \ket{j,m} = \bra{j,m} (\hat{J}{}^2 - \hat{J}_z{}^2 - \hbar \hat{J}_z) \ket{j,m} \\ = \hbar^2 [j(j+1) - m^2 - m]. \end{aligned} \]
But we can also rewrite this product by letting \( \hat{J}_+ \) act on the state directly:
\[ \begin{aligned} \hat{J}_+ \ket{j,m} = c_{jm}^+ \ket{j,m+1} \end{aligned} \]
and so
\[ \begin{aligned} \bra{j,m} \hat{J}_+^\dagger \hat{J}_+ \ket{j,m} = |c_{jm}^+|^2 \sprod{j,m+1}{j,m+1} = |c_{jm}^+|^2. \end{aligned} \]
This fixes \( c_{jm}^+ \), up to an arbitrary phase; by convention we take it to be real and positive. Thus, repeating the same derivation for \( \hat{J}_- \), we arrive at the formulas
\[ \begin{aligned} \hat{J}_+ \ket{j,m} = \hbar \sqrt{(j-m)(j+m+1)} \ket{j,m+1} \\ \hat{J}_- \ket{j,m} = \hbar \sqrt{(j+m)(j-m+1)} \ket{j,m-1}. \end{aligned} \]
It can be easy to mix these similar-looking formulas up, but an easy way to keep them straight is to remember that the \( \hat{J}+ \) formula should vanish when \( m=j \), and the \( \hat{J}- \) one must vanish when \( m=-j \). Thus, we have for the matrix elements
\[ \begin{aligned} \bra{j',m'} \hat{J}_{\pm} \ket{j,m} = \sqrt{(j \mp m) (j \pm m + 1)} \hbar \delta_{j,j'} \delta_{m',m\pm 1}. \end{aligned} \]
Let's see a few examples of what these various operators we've defined look like for systems with small, fixed \( j \).
This gives us \( m = 0 \) as the only possibility, so this is completely trivial: the operators \( \hat{J}^2 \) and \( \hat{J}_z \) give zero when acting on any state. Physically, this makes sense - a \( j=0 \) object has no angular momentum at all.
This is a familiar example in some new notation. With \( j=1/2 \) we find two possible states \( \ket{\frac{1}{2}, \pm \frac{1}{2}} \). We can write out the matrix expressions for the various operators we've defined:
\[ \begin{aligned} \hat{J}_x = \frac{\hbar}{2} \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)\ \ \hat{J}_y = \frac{\hbar}{2} \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right)\ \ \hat{J}_z = \frac{\hbar}{2} \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \\ \hat{J}_+ = \hbar \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)\ \ \hat{J}_- = \hbar \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right)\ \ \hat{J}{}^2 = \frac{3\hbar^2}{4} \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right). \end{aligned} \]
This is the same algebra we saw for spin-1/2 systems; since our derivation was totally general, our results apply to both spin angular momentum and orbital angular momentum (which we'll come to soon.)
For the next highest \( j \), we find three states: \( m=-1,0,1 \). To write things in matrix form we can order them, as is conventional, from highest to lowest:
\[ \begin{aligned} \ket{1,+1} = \left(\begin{array}{c}1\\ 0\\ 0\end{array} \right) \ \ \ket{1,0} = \left(\begin{array}{c}0\\ 1\\ 0\end{array} \right) \ \ \ket{1,-1} = \left(\begin{array}{c}0\\ 0\\ 1\end{array} \right) \ \ \end{aligned} \]
Then the \( \hat{J}_z \) operator and the raising and lowering operators are
\[ \begin{aligned} \hat{J}{}^2 = 2\hbar^2 \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \\ \hat{J}_z = \hbar \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right) \\ \hat{J}_+ = \sqrt{2} \hbar \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) \ \ \hat{J}_- = \sqrt{2} \hbar \left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right) \end{aligned} \]
Of course, since \( \hat{J}^2 \) doesn't depend on \( m \), it's easy to see that when we write things out in this way, the matrix \( \hat{J}^2 \) will always be proportional to the identity. This is an artifact of our current focus on systems with fixed \( j \); when we come to consider addition of angular momentum, this operator won't always be so trivial.
You might start to see a pattern here in the form of the other operators as matrices, but it may not be quite as simple as you think; let's do one more example.
We've now added one more state, for four possibilities: \( m=-3/2, -1/2, +1/2, +3/2 \). Going in order from large to small \( m \) again, we have
\[ \begin{aligned} \hat{J}_z = \hbar \left( \begin{array}{cccc} \frac{3}{2} &&& \\ &\frac{1}{2}&& \\ &&-\frac{1}{2}& \\ &&&-\frac{3}{2} \end{array} \right) \\ \hat{J}_+ = \hbar \left( \begin{array}{cccc} 0& \sqrt{3} &0&0\\ 0&0&2&0\\ 0&0&0&\sqrt{3} \\ 0&0&0&0 \end{array} \right) \\ \hat{J}_- = \hbar \left( \begin{array}{cccc} 0&0&0&0\\ \sqrt{3}&0&0&0\\ 0&2&0&0 \\ 0&0&\sqrt{3}&0 \end{array} \right) \\ \end{aligned} \]
Although the raising and lowering matrices are now more interesting - we can't just pull an overall factor out front anymore - they still have the same off-diagonal structure, with \( \hat{J}_+ \) occupying the upper diagonal (due to our choice of \( m=+3/2 \) as the "top" vector state.)
So far, we've introduced the idea of a generic Hermitian angular momentum operator \( \hat{J}_i \) as the infinitesmal generator of rotations about axis \( i \). But there's another way we could have defined angular momentum: by taking the classical angular-momentum operator
\[ \begin{aligned} \vec{L} = \vec{r} \times \vec{p} \end{aligned} \]
and promoting \( \vec{r} \) and \( \vec{p} \) to operators, as we've done before. What do we find if we take this approach?
It's easy to see that \( \hat{\vec{L}} \) is indeed an angular momentum operator as we've defined it, in particular its components satisfy the angular momentum commutation relation
\[ \begin{aligned} [\hat{L}_i, \hat{L}_j] = i \hbar \epsilon_{ijk} \hat{L}_k. \end{aligned} \]
You can convince yourself that this is true by expanding out the definition in terms of \( \hat{\vec{x}} \) and \( \hat{\vec{p}} \). Furthermore, we can see that if we let this operator act on a position wavefunction, we do get an infinitesmal rotation as expected. For example, trying to generate an infinitesmal rotation \( \delta \phi \) about the \( z \)-axis,
\[ \begin{aligned} \bra{x,y,z} \left[ 1 - i \frac{\delta \phi}{\hbar} \hat{L}_z \right] \ket{\psi} = \left[ 1 - i \frac{\delta \phi}{\hbar} \hat{p}_y \hat{x} + i \frac{\delta \phi}{\hbar} \hat{p}_x \hat{y} \right] \ket{\psi} \\ = \bra{x,y,z}\left[1 - (\delta \phi) x \frac{\partial}{\partial y} + (\delta \phi) y \frac{\partial}{\partial x} \right] \ket{\psi} \\ = \psi(x,y,z) - x (\delta \phi) \frac{\partial \psi}{\partial y} + y (\delta \phi) \frac{\partial \psi}{\partial x} \\ = \psi(x+y(\delta \phi), y - x (\delta \phi), z). \end{aligned} \]
This is exactly what we expect for spatial rotation of a scalar function (here it's a backwards rotation, since it's acting on the position bra to the left.) In fact, this result is nicer if we switch to spherical coordinates,
\[ \begin{aligned} x = r \sin \theta \cos \phi \\ y = r \sin \theta \sin \phi \\ z = r \cos \theta. \end{aligned} \]
Then we have from above
\[ \begin{aligned} \bra{x,y,z} \left[ 1 - i \frac{\delta \phi}{\hbar} \hat{L}_z \right] \ket{\psi} = \psi(r, \theta, \phi - \delta \phi) \end{aligned} \]
which lets us identify
\[ \begin{aligned} \bra{x,y,z} \hat{L}_z \ket{\psi} = -i \hbar \frac{\partial}{\partial \phi} \sprod{x,y,z}{\psi}. \end{aligned} \]
It's important to emphasize that this version of angular momentum is independent of the internal properties of whatever particle we're considering; we've constructed it simply from the position and momentum operators. As a result, \( \hat{\vec{L}} \) is known as the orbital angular momentum. All of the results we've derived for general \( \hat{\vec{J}} \) still apply, with the one exception that the eigenvalues of \( \hat{\vec{L}}^2 \) can only take on integer values, and never half-integer; we'll prove this soon.
Now, going through the same exercise for the other directions is a little tedious, but with a little work it's easy to show that again in spherical coordinates, we can express the other two angular momentum operators as the derivatives
\[ \begin{aligned} \bra{x,y,z} \hat{L}_x \ket{\psi} = -i \hbar \left( -\sin \phi \frac{\partial}{\partial \theta} - \cot \theta \cos \phi \frac{\partial}{\partial \phi} \right) \sprod{x,y,z}{\psi}, \\ \bra{x,y,z} \hat{L}_y \ket{\psi} = -i \hbar \left( \cos \phi \frac{\partial}{\partial \theta} - \cot \theta \sin \phi \frac{\partial}{\partial \phi} \right) \sprod{x,y,z}{\psi}. \\ \end{aligned} \]
Next time: we continue with orbital angular momentum.