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In problem 1.17 of [1] we are to show that non-commuting operators that both commute with the Hamiltonian, have, in general, degenerate energy eigenvalues. It suggests considering \( L_x, L_z \) and a central force Hamiltonian \( H = \Bp^2/2m + V(r) \) as examples.

Let’s just demonstrate these commutators act as expected in these cases.

With \( \BL = \Bx \cross \Bp \), we have

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:20}

\begin{aligned}

L_x &= y p_z – z p_y \\

L_y &= z p_x – x p_z \\

L_z &= x p_y – y p_x.

\end{aligned}

\end{equation}

The \( L_x, L_z \) commutator is

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:40}

\begin{aligned}

\antisymmetric{L_x}{L_z}

&=

\antisymmetric{y p_z – z p_y }{x p_y – y p_x} \\

&=

\antisymmetric{y p_z}{x p_y}

-\antisymmetric{y p_z}{y p_x}

-\antisymmetric{z p_y }{x p_y}

+\antisymmetric{z p_y }{y p_x} \\

&=

x p_z \antisymmetric{y}{p_y}

+ z p_x \antisymmetric{p_y }{y} \\

&=

i \Hbar \lr{ x p_z – z p_x } \\

&=

– i \Hbar L_y

\end{aligned}

\end{equation}

cyclicly permuting the indexes shows that no pairs of different \( \BL \) components commute. For \( L_y, L_x \) that is

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:60}

\begin{aligned}

\antisymmetric{L_y}{L_x}

&=

\antisymmetric{z p_x – x p_z }{y p_z – z p_y} \\

&=

\antisymmetric{z p_x}{y p_z}

-\antisymmetric{z p_x}{z p_y}

-\antisymmetric{x p_z }{y p_z}

+\antisymmetric{x p_z }{z p_y} \\

&=

y p_x \antisymmetric{z}{p_z}

+ x p_y \antisymmetric{p_z }{z} \\

&=

i \Hbar \lr{ y p_x – x p_y } \\

&=

– i \Hbar L_z,

\end{aligned}

\end{equation}

and for \( L_z, L_y \)

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:80}

\begin{aligned}

\antisymmetric{L_z}{L_y}

&=

\antisymmetric{x p_y – y p_x }{z p_x – x p_z} \\

&=

\antisymmetric{x p_y}{z p_x}

-\antisymmetric{x p_y}{x p_z}

-\antisymmetric{y p_x }{z p_x}

+\antisymmetric{y p_x }{x p_z} \\

&=

z p_y \antisymmetric{x}{p_x}

+ y p_z \antisymmetric{p_x }{x} \\

&=

i \Hbar \lr{ z p_y – y p_z } \\

&=

– i \Hbar L_x.

\end{aligned}

\end{equation}

If these angular momentum components are also shown to commute with themselves (which they do), the commutator relations above can be summarized as

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:100}

\antisymmetric{L_a}{L_b} = i \Hbar \epsilon_{a b c} L_c.

\end{equation}

In the example to consider, we’ll have to consider the commutators with \( \Bp^2 \) and \( V(r) \). Picking any one component of \( \BL \) is sufficent due to the symmetries of the problem. For example

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:120}

\begin{aligned}

\antisymmetric{L_x}{\Bp^2}

&=

\antisymmetric{y p_z – z p_y}{p_x^2 + p_y^2 + p_z^2} \\

&=

\antisymmetric{y p_z}{{p_x^2} + p_y^2 + {p_z^2}}

-\antisymmetric{z p_y}{{p_x^2} + {p_y^2} + p_z^2} \\

&=

p_z \antisymmetric{y}{p_y^2}

-p_y \antisymmetric{z}{p_z^2} \\

&=

p_z 2 i \Hbar p_y

2 i \Hbar p_y

-p_y 2 i \Hbar p_z \\

&=

0.

\end{aligned}

\end{equation}

How about the commutator of \( \BL \) with the potential? It is sufficient to consider one component again, for example

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:140}

\begin{aligned}

\antisymmetric{L_x}{V}

&=

\antisymmetric{y p_z – z p_y}{V} \\

&=

y \antisymmetric{p_z}{V} – z \antisymmetric{p_y}{V} \\

&=

-i \Hbar y \PD{z}{V(r)} + i \Hbar z \PD{y}{V(r)} \\

&=

-i \Hbar y \PD{r}{V}\PD{z}{r} + i \Hbar z \PD{r}{V}\PD{y}{r} \\

&=

-i \Hbar y \PD{r}{V} \frac{z}{r} + i \Hbar z \PD{r}{V}\frac{y}{r} \\

&=

0.

\end{aligned}

\end{equation}

We’ve shown that all the components of \( \BL \) commute with a central force Hamiltonian, and each different component of \( \BL \) do not commute.

The next step will be figuring out how to use this to show that there are energy degeneracies.

# References

[1] Jun John Sakurai and Jim J Napolitano. *Modern quantum mechanics*. Pearson Higher Ed, 2014.