D.20 Derivation of the commutator rules

This note explains where the formulae of chapter 4.5.4 come from.

The general assertions are readily checked by simply writing out both sides of the equation and comparing. And some are just rewrites of earlier ones.

Position and potential energy operators commute since they are just ordinary numerical multiplications, and these commute.

The linear momentum operators commute because the order in which differentiation is done is irrelevant. Similarly, commutators between angular momentum in one direction and position in another direction commute since the other directions are not affected by the differentiation.

The commutator between the position $X$ and linear momentum $p_x$ in the $x$-​direction was worked out in the previous subsection to figure out Heisenberg's uncertainty principle. Of course, three-di­men­sion­al space has no preferred direction, so the result applies the same in any direction, including the $y$ and $z$ directions.

The angular momentum commutators are simplest obtained by just grinding out

\begin{displaymath}[\L _x,\L _y]=
[{\widehat y}{\widehat p}_z - {\widehat z}{\...
...}_y, {\widehat z}{\widehat p}_x - {\widehat x}{\widehat p}_z]
\end{displaymath}

using the linear combination and product manipulation rules and the commutators for linear angular momentum. To generalize the result you get, you cannot just arbitrarily swap $x$, $y$, and $z$, since, as every mechanic knows, a right-handed screw is not the same as a left-handed one, and some axes swaps would turn one into the other. But you can swap axes according to the $xyzxyzx\ldots$” “cyclic permutation scheme, as in:

\begin{displaymath}
x\to y, \quad y\to z, \quad z\to x
\end{displaymath}

which produces the other two commutators if you do it twice:

\begin{displaymath}[\L _x,\L _y]= {\rm i}\hbar\L _z
\quad\longrightarrow\quad
...
...\quad\longrightarrow\quad
[\L _z,\L _x] = {\rm i}\hbar\L _y
\end{displaymath}

For the commutators with square angular momentum, work out

\begin{displaymath}[\L _x,\L _x^2+\L _y^2+\L _z^2]
\end{displaymath}

using the manipulation rules and the commutators between angular momentum components.

A commutator like $[{\widehat x},\L _x]$ $\vphantom0\raisebox{1.5pt}{$=$}$ $[{\widehat x},{\widehat y}{\widehat p}_z-{\widehat z}{\widehat p}_y]$ is zero because everything commutes in it. However, in a commutator like $[{\widehat x},\L _y]$ $\vphantom0\raisebox{1.5pt}{$=$}$ $[{\widehat x},{\widehat z}{\widehat p}_x-{\widehat x}{\widehat p}_z]$, ${\widehat x}$ does not commute with ${\widehat p}_x$, so multiplying out and taking the ${\widehat z}$ out of $[{\widehat x},{\widehat z}{\widehat p}_x]$ at its own side, you get ${\widehat z}[{\widehat x},{\widehat p}_x]$, and the commutator left is the canonical one, which has value ${\rm i}\hbar$. Plug these results and similar into $[{\widehat x}^2+{\widehat y}^2+{\widehat z}^2,L_x]$ and you get zero.

For a commutator like $[{\widehat x},\L ^2]$ $\vphantom0\raisebox{1.5pt}{$=$}$ $[{\widehat x},\L _x^2+\L _y^2+\L _z^2]$, the $L_x^2$ term produces zero because $\L _x$ commutes with ${\widehat x}$, and in the remaining term, taking the various factors out at their own sides of the commutator produces

\begin{eqnarray*}[{\widehat x},\L ^2]
& = & \L _y[{\widehat x},\L _y] + [{\wid...
...{\rm i}\hbar \L _z {\widehat y}- {\rm i}\hbar {\widehat y}\L _z
\end{eqnarray*}

the final equality because of the commutators already worked out. Now by the nature of the commutator, you can swap the order of the terms in $\L _y{\widehat z}$ as long as you add the commutator $[\L _y,{\widehat z}]$ to make up for it, and that commutator was already found to be ${\rm i}\hbar{\widehat x}$, The same way the order of $\L _z{\widehat y}$ can be swapped to give

\begin{displaymath}[{\widehat x},\L ^2]= -2\hbar^2 {\widehat x}-2{\rm i}\hbar({\widehat y}\L _z-{\widehat z}\L _y)
\end{displaymath}

and the parenthetical expression can be recognized as the $x$-​component of ${\skew 2\widehat{\skew{-1}\vec r}}$ $\times$ ${\skew 4\widehat{\vec L}}$, giving one of the expressions claimed.

Instead you can work out the parenthetical expression further by substituting in the definitions for $\L _z$ and $\L _y$:

\begin{displaymath}[{\widehat x},\L ^2]= -2\hbar^2 {\widehat x}-2{\rm i}\hbar
...
...({\widehat x}{\widehat p}_x-{\widehat x}{\widehat p}_x)\bigg)
\end{displaymath}

where the third term added within the big parentheses is self-evidently zero. This can be reordered to the $x$-​component of the second claimed expression. And as always, the other components are of course no different.

The commutators between linear and angular momentum go almost identically, except for additional swaps in the order between position and momentum operators using the canonical commutator.

To derive the first commutator in (4.73), consider the $z$-​component as the example:

\begin{displaymath}[x\L _y-y\L _x,\L ^2]= [x,\L ^2]\L _y - [y,\L ^2]L_x
\end{displaymath}

because $L^2$ commutes with ${\skew 4\widehat{\vec L}}$, and using (4.68)

\begin{displaymath}[x\L _y-y\L _x,\L ^2]= -2\hbar^2x\L _y -2{\rm i}\hbar(y\L _z\...
... _y^2)
+2\hbar^2y\L _x +2{\rm i}\hbar(z\L _x^2-x\L _z\L _x)
\end{displaymath}

Now use the commutator $[\L _y,\L _z]$ to get rid of $\L _z\L _y$ and $[\L _z,\L _x]$ to get rid of $\L _z\L _x$ and clean up to get

\begin{displaymath}[x\L _y-y\L _x,\L ^2]= 2{\rm i}\hbar
\left(- y\L _y\L _z + z\L _y^2 + z\L _x^2- x\L _x\L _z\right)
\end{displaymath}

Now ${\skew0\vec r}\cdot{\skew 4\widehat{\vec L}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\skew0\vec r}\cdot({\skew0\vec r}\times{\skew 4\widehat{\skew{-.5}\vec p}})$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 so $x\L _x+y\L _y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-z\L _z$, which gives the claimed expression. To verify the second equation of (4.73), use (4.68), the first of (4.73), and the definition of $[{\skew0\vec r},\L ^2]$.