# Angular Momentum and Harmonic Oscillators

There are many analogies that can be drawn between spin angular momentum and orbital angular momentum. This is because they obey identical commutation relations:

$[L_x, L_y] = i\hbar L_z$     &     $[S_x, S_y] = i\hbar S_z$.

One can circularly permute the indices to obtain the other commutation relations. However, there is one crucial difference between the orbital and spin angular momenta: components of the orbital angular momentum cannot take half-integer values, whereas this is permitted for spin angular momentum.

The forbidden half-integer quantization stems from the fact that orbital angular momentum can be expressed in terms of the position and momentum operators:

$\textbf{L} = \textbf{r} \times \textbf{p}$.

While in most textbooks the integer quantization of the orbital angular momentum is shown by appealing to the Schrodinger equation, Schwinger demonstrated that by mapping the angular momentum problem to that of two uncoupled harmonic oscillators (pdf!), integer quantization easily follows.

I’m just going to show this for the $z$-component of the angular momentum since the $x$– and $y$-components can easily be obtained by permuting indices. $L_z$ can be written as:

$L_z = xp_y - yp_x$

As Schwinger often did effectively, he made a canonical transformation to a different basis and wrote:

$x_1 = \frac{1}{\sqrt{2}} [x+(a^2/\hbar)p_y]$

$x_2 = \frac{1}{\sqrt{2}} [x-(a^2/\hbar)p_y]$

$p_1 = \frac{1}{\sqrt{2}} [p_x-(\hbar/a^2)y]$

$p_2 = \frac{1}{\sqrt{2}} [p_x+(\hbar/a^2)y]$,

where $a$ is just some variable with units of length. Now, since the transformation is canonical, these new operators satisfy the same commutation relations, i.e. $[x_1,p_1]=i\hbar, [x_1,p_2]=0$, and so forth.

If we now write $L_z$ in terms of the new operators, we find something rather amusing:

$L_z = (\frac{a^2}{2\hbar}p_1^2 + \frac{\hbar}{2a^2}x_1^2) - ( \frac{a^2}{2\hbar}p_2^2 + \frac{\hbar}{2a^2}x_2^2)$.

With the substitution $\hbar/a^2 \rightarrow m$, $L_z$ can be written as so:

$L_z = (\frac{1}{2m}p_1^2 + \frac{m}{2}x_1^2) - ( \frac{1}{2m}p_2^2 + \frac{m}{2}x_2^2)$.

Each of the two terms in brackets can be identified as Hamiltonians for harmonic oscillators with angular frequency, $\omega$, equal to one. The eigenvalues of the harmonic oscillator problem can therefore be used to obtain the eigenvalues of the $z$-component of the orbital angular momentum:

$L_z|\psi\rangle = (H_1 - H_2)|\psi\rangle = \hbar(n_1 - n_2)|\psi\rangle = m\hbar|\psi\rangle$,

where $H_i$ denotes the Hamiltonian operator of the $i^{th}$ oscillator. Since the $n_i$ can only take integer values in the harmonic oscillator problem, integer quantization of Cartesian components of the angular momentum also naturally follows.

How do we interpret all of this? Let’s imagine that we have $n_1$ spins pointing up and $n_2$ spins pointing down. Now consider the angular momentum raising and lowering operators. The angular momentum raising operator in this example, $L_+ = \hbar a_1^\dagger a_2$, corresponds to flipping a spin of angular momentum, $\hbar/2$ from down to up. The $a_1^\dagger (a_2)$ corresponds to the creation (annihilation) operator for oscillator 1 (2). The change in angular momentum is therefore $+\hbar/2 -(-\hbar/2) = \hbar$. It is this constraint, that we cannot “destroy” these spins, but only flip them, that results in the integer quantization of orbital angular momentum.

I find this solution to the forbidden half-integer problem much more illuminating than with the use of the Schrodinger equation and spherical harmonics. The analogy between the uncoupled oscillators and angular momentum is very rich and actually extends much further than this simple example. It has even been used in papers on supersymmetry (which, needless to say, extends far beyond my realm of knowledge).