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

$$[\L _x,\L _y]= [{\widehat y}{\widehat p}_z - {\widehat z}{\w... ...p}_y, {\widehat z}{\widehat p}_x - {\widehat x}{\widehat p}_z]$$

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:

$$x\to y, \quad y\to z, \quad z\to x$$

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

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

For the com­mu­ta­tors with square an­gu­lar mo­men­tum, work out

$$[\L _x,\L _x^2+\L _y^2+\L _z^2]$$

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

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

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

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

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 $\L _y$:

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

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:

$$[x\L _y-y\L _x,\L ^2]= [x,\L ^2]\L _y - [y,\L ^2]L_x$$

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

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

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

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

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 $[{\skew0\vec r},\L ^2]$.