4.5 The Com­mu­ta­tor

As the pre­vi­ous sec­tion dis­cussed, the stan­dard de­vi­a­tion $\sigma$ is a mea­sure of the un­cer­tainty of a prop­erty of a quan­tum sys­tem. The larger the stan­dard de­vi­a­tion, the far­ther typ­i­cal mea­sure­ments stray from the ex­pected av­er­age value. Quan­tum me­chan­ics of­ten re­quires a min­i­mum amount of un­cer­tainty when more than one quan­tity is in­volved, like po­si­tion and lin­ear mo­men­tum in Heisen­berg's un­cer­tainty prin­ci­ple. In gen­eral, this amount of un­cer­tainty is re­lated to an im­por­tant math­e­mat­i­cal ob­ject called the com­mu­ta­tor, to be dis­cussed in this sec­tion.

4.5.1 Com­mut­ing op­er­a­tors

First, note that there is no fun­da­men­tal rea­son why sev­eral quan­ti­ties can­not have a def­i­nite value at the same time. For ex­am­ple, if the elec­tron of the hy­dro­gen atom is in a $\psi_{nlm}$ eigen­state, its to­tal en­ergy, square an­gu­lar mo­men­tum, and $z$-​com­po­nent of an­gu­lar mo­men­tum all have def­i­nite val­ues, with zero un­cer­tainty.

More gen­er­ally, two dif­fer­ent quan­ti­ties with op­er­a­tors $A$ and $B$ have def­i­nite val­ues if the wave func­tion is an eigen­func­tion of both $A$ and $B$. So, the ques­tion whether two quan­ti­ties can be def­i­nite at the same time is re­ally whether their op­er­a­tors $A$ and $B$ have com­mon eigen­func­tions. And it turns out that the an­swer has to do with whether these op­er­a­tors “com­mute”, in other words, on whether their or­der can be re­versed as in $AB$ $\vphantom0\raisebox{1.5pt}{$=$}$ $BA$.

In par­tic­u­lar, {D.18}:

Iff two Her­mit­ian op­er­a­tors com­mute, there is a com­plete set of eigen­func­tions that is com­mon to them both.

(For more than two op­er­a­tors, each op­er­a­tor has to com­mute with all oth­ers.)

For ex­am­ple, the op­er­a­tors $H_x$ and $H_y$ of the har­monic os­cil­la­tor of chap­ter 4.1.2 com­mute:

$$\begin{eqnarray*} H_x H_y \Psi & = & \left[ - \frac{\hbar^2}{2m} \frac{\parti... ...} cx^2 {\textstyle\frac{1}{2}} cy^2\Psi \\ & = & H_y H_x \Psi \end{eqnarray*}$$

This is true since it makes no dif­fer­ence whether you dif­fer­en­ti­ate $\Psi$ first with re­spect to $x$ and then with re­spect to $y$ or vice versa, and since the $\frac12cy^2$ can be pulled in front of the $x$-​dif­fer­en­ti­a­tions and the $\frac12cx^2$ can be pushed in­side the $y$-​dif­fer­en­ti­a­tions, and since mul­ti­pli­ca­tions can al­ways be done in any or­der.

The same way, $H_z$ com­mutes with $H_x$ and $H_y$, and that means that $H$ com­mutes with them all, since $H$ is just their sum. So, these four op­er­a­tors should have a com­mon set of eigen­func­tions, and they do: it is the set of eigen­func­tions $\psi_{n_xn_yn_z}$ de­rived in chap­ter 4.1.2.

Sim­i­larly, for the hy­dro­gen atom, the to­tal en­ergy Hamil­ton­ian $H$, the square an­gu­lar mo­men­tum op­er­a­tor $\L ^2$ and the $z$-​com­po­nent of an­gu­lar mo­men­tum $\L _z$ all com­mute, and they have the com­mon set of eigen­func­tions $\psi_{nlm}$.

Note that such eigen­func­tions are not nec­es­sar­ily the only game in town. As a counter-ex­am­ple, for the hy­dro­gen atom $H$, $\L ^2$, and the $x$-​com­po­nent of an­gu­lar mo­men­tum $\L _x$ also all com­mute, and they too have a com­mon set of eigen­func­tions. But that will not be the $\psi_{nlm}$, since $\L _x$ and $\L _z$ do not com­mute. (It will how­ever be the $\psi_{nlm}$ af­ter you ro­tate them all 90 de­grees around the $y$-​axis.) It would cer­tainly be sim­pler math­e­mat­i­cally if each op­er­a­tor had just one unique set of eigen­func­tions, but na­ture does not co­op­er­ate.

Key Points

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Op­er­a­tors com­mute if you can change their or­der, as in $AB$ $\vphantom0\raisebox{1.5pt}{$=$}$ $BA$.

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For com­mut­ing op­er­a­tors, a com­mon set of eigen­func­tions ex­ists.

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For those eigen­func­tions, the phys­i­cal quan­ti­ties cor­re­spond­ing to the com­mut­ing op­er­a­tors all have def­i­nite val­ues at the same time.

4.5.1 Re­view Ques­tions

1.

The pointer state

$$\mbox{2p$_x$} = \frac 1{\sqrt 2}\left(-\psi_{211}+\psi_{21-1}\right).$$

is one of the eigen­states that $H$, $\L ^2$, and $\L _x$ have in com­mon. Check that it is not an eigen­state that $H$, $\L ^2$, and $\L _z$ have in com­mon.

So­lu­tion com­mutea-a

4.5.2 Non­com­mut­ing op­er­a­tors and their com­mu­ta­tor

Two quan­ti­ties with op­er­a­tors that do not com­mute can­not in gen­eral have def­i­nite val­ues at the same time. If one has a def­i­nite value, the other is in gen­eral un­cer­tain.

The qual­i­fi­ca­tion in gen­eral is needed be­cause there may be ex­cep­tions. The an­gu­lar mo­men­tum op­er­a­tors do not com­mute, but it is still pos­si­ble for the an­gu­lar mo­men­tum to be zero in all three di­rec­tions. But as soon as the an­gu­lar mo­men­tum in any di­rec­tion is nonzero, only one com­po­nent of an­gu­lar mo­men­tum can have a def­i­nite value.

A mea­sure for the amount to which two op­er­a­tors $A$ and $B$ do not com­mute is the dif­fer­ence be­tween $AB$ and $BA$; this dif­fer­ence is called their “com­mu­ta­tor” $[A,B]$:

$$\fbox{$\displaystyle [A,B] \equiv A B - B A $}$$ (4.45)

A nonzero com­mu­ta­tor $[A,B]$ de­mands a min­i­mum amount of un­cer­tainty in the cor­re­spond­ing quan­ti­ties $a$ and $b$. It can be shown, {D.19}, that the un­cer­tain­ties, or stan­dard de­vi­a­tions, $\sigma_a$ in $a$ and $\sigma_b$ in $b$ are at least so large that:

$$\fbox{$\displaystyle \sigma_a \sigma_b \mathrel{\raisebox{... ...$}}{\textstyle\frac{1}{2}} \vert\langle[A,B]\rangle\vert $} %$$ (4.46)

This equa­tion is called the “gen­er­al­ized un­cer­tainty re­la­tion­ship”.

Key Points

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The com­mu­ta­tor of two op­er­a­tors $A$ and $B$ equals $AB-BA$ and is writ­ten as $[A,B]$.

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The prod­uct of the un­cer­tain­ties in two quan­ti­ties is at least one half the mag­ni­tude of the ex­pec­ta­tion value of their com­mu­ta­tor.

4.5.3 The Heisen­berg un­cer­tainty re­la­tion­ship

This sec­tion will work out the un­cer­tainty re­la­tion­ship (4.46) of the pre­vi­ous sub­sec­tion for the po­si­tion and lin­ear mo­men­tum in an ar­bi­trary di­rec­tion. The re­sult will be a pre­cise math­e­mat­i­cal state­ment of the Heisen­berg un­cer­tainty prin­ci­ple.

To be spe­cific, the ar­bi­trary di­rec­tion will be taken as the $x$-​axis, so the po­si­tion op­er­a­tor will be ${\widehat x}$, and the lin­ear mo­men­tum op­er­a­tor ${\widehat p}_x$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\partial$$\raisebox{.5pt}{$/$}$​${\rm i}\partial{x}$. These two op­er­a­tors do not com­mute, ${\widehat p}_x{\widehat x}\Psi$ is sim­ply not the same as ${\widehat x}{\widehat p}_x\Psi$: ${\widehat p}_x{\widehat x}\Psi$ means mul­ti­ply func­tion $\Psi$ by $x$ to get the prod­uct func­tion $x\Psi$ and then ap­ply ${\widehat p}_x$ on that prod­uct, while ${\widehat x}{\widehat p}_x\Psi$ means ap­ply ${\widehat p}_x$ on $\Psi$ and then mul­ti­ply the re­sult­ing func­tion by $x$. The dif­fer­ence is found from writ­ing it out:

$${\widehat p}_x {\widehat x}\Psi = \frac{\hbar}{{\rm i}} \f... ...ial x} = -{\rm i}\hbar \Psi + {\widehat x}{\widehat p}_x \Psi$$

the sec­ond equal­ity re­sult­ing from dif­fer­en­ti­at­ing out the prod­uct.

Com­par­ing start and end shows that the dif­fer­ence be­tween ${\widehat x}{\widehat p}_x$ and ${\widehat p}_x{\widehat x}$ is not zero, but ${\rm i}\hbar$. By de­f­i­n­i­tion, this dif­fer­ence is their com­mu­ta­tor:

$$\fbox{$\displaystyle [{\widehat x},{\widehat p}_x] = {\rm i}\hbar $} %$$ (4.47)

This im­por­tant re­sult is called the “canon­i­cal com­mu­ta­tion re­la­tion.” The com­mu­ta­tor of po­si­tion and lin­ear mo­men­tum in the same di­rec­tion is the nonzero con­stant ${\rm i}\hbar$.

Be­cause the com­mu­ta­tor is nonzero, there must be nonzero un­cer­tainty in­volved. In­deed, the gen­er­al­ized un­cer­tainty re­la­tion­ship of the pre­vi­ous sub­sec­tion be­comes in this case:

$$\sigma_{x} \sigma_{p_x} \mathrel{\raisebox{-1pt}{$\geqslant$}}{\textstyle\frac{1}{2}} \hbar$$ (4.48)

This is the un­cer­tainty re­la­tion­ship as first for­mu­lated by Heisen­berg.

It im­plies that when the un­cer­tainty in po­si­tion $\sigma_{x}$ is nar­rowed down to zero, the un­cer­tainty in mo­men­tum $\sigma_{p_x}$ must be­come in­fi­nite to keep their prod­uct nonzero, and vice versa. More gen­er­ally, you can nar­row down the po­si­tion of a par­ti­cle and you can nar­row down its mo­men­tum. But you can never re­duce the prod­uct of the un­cer­tain­ties $\sigma_{x}$ and $\sigma_{p_x}$ be­low $\frac12\hbar$, what­ever you do.

It should be noted that the un­cer­tainty re­la­tion­ship is of­ten writ­ten as $\Delta{p}_x\Delta{x}$ $\raisebox{-.5pt}{$\geqslant$}$ $\frac12\hbar$ or even as $\Delta{p_x}\Delta{x}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $\hbar$ where $\Delta{p}$ and $\Delta{x}$ are taken to be vaguely de­scribed un­cer­tain­ties in mo­men­tum and po­si­tion, rather than rig­or­ously de­fined stan­dard de­vi­a­tions. And peo­ple write a cor­re­spond­ing un­cer­tainty re­la­tion­ship for time, $\Delta{E}\Delta{t}$ $\raisebox{-.5pt}{$\geqslant$}$ $\frac12\hbar$, be­cause rel­a­tiv­ity sug­gests that time should be treated just like space. But note that un­like the lin­ear mo­men­tum op­er­a­tor, the Hamil­ton­ian is not at all uni­ver­sal. So, you might guess that the de­f­i­n­i­tion of the un­cer­tainty $\Delta{t}$ in time would not be uni­ver­sal ei­ther, and you would be right, chap­ter 7.2.2.

Key Points

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The canon­i­cal com­mu­ta­tor $[{\widehat x},{\widehat p}_x]$ equals ${\rm i}\hbar$.

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If ei­ther the un­cer­tainty in po­si­tion in a given di­rec­tion or the un­cer­tainty in lin­ear mo­men­tum in that di­rec­tion is nar­rowed down to zero, the other un­cer­tainty blows up.

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The prod­uct of the two un­cer­tain­ties is at least the con­stant $\frac12\hbar$.

4.5.3 Re­view Ques­tions

1.

This sounds se­ri­ous! If I am dri­ving my car, the po­lice re­quires me to know my speed (lin­ear mo­men­tum). Also, I would like to know where I am. But nei­ther is pos­si­ble ac­cord­ing to quan­tum me­chan­ics.

So­lu­tion com­mutec-a

4.5.4 Com­mu­ta­tor ref­er­ence

It is a fact of life in quan­tum me­chan­ics that com­mu­ta­tors pop up all over the place. Not just in un­cer­tainty re­la­tions, but also in the time evo­lu­tion of ex­pec­ta­tion val­ues, in an­gu­lar mo­men­tum, and in quan­tum field the­ory, the ad­vanced the­ory of quan­tum me­chan­ics used in solids and rel­a­tivis­tic ap­pli­ca­tions. This sec­tion can make your life eas­ier deal­ing with them. Browse through it to see what is there. Then come back when you need it.

Re­call the de­f­i­n­i­tion of the com­mu­ta­tor $[A,B]$ of any two op­er­a­tors $A$ and $B$:

$$[A,B]= AB - BA %$$ (4.49)

By this very de­f­i­n­i­tion , the com­mu­ta­tor is zero for any two op­er­a­tors $A_1$ and $A_2$ that com­mute, (whose or­der can be in­ter­changed):

$$[A_1,A_2]= 0 \quad \mbox{if $A_1$ and $A_2$ commute; } A_1 A_2 = A_2 A_1. %$$ (4.50)

If op­er­a­tors all com­mute, all their prod­ucts com­mute too:

$$[A_1 A_2 \ldots A_k, A_{k+1}\ldots A_n]= 0 \quad \mbox{if } ... ... A_2, \ldots, A_k, A_{k+1},\ldots, A_n \mbox{ all commute.} %$$ (4.51)

Every­thing com­mutes with it­self, of course:

$$[A,A]= 0, %$$ (4.52)

and every­thing com­mutes with a nu­mer­i­cal con­stant; if $A$ is an op­er­a­tor and $a$ is some num­ber, then:

$$[A,a]= [a,A] = 0. %$$ (4.53)

The com­mu­ta­tor is an­ti­sym­met­ric; or in sim­pler words, if you in­ter­change the sides; it will change the sign, {D.20}:

$$[B,A]= - [A,B]. %$$ (4.54)

For the rest how­ever, lin­ear com­bi­na­tions mul­ti­ply out just like you would ex­pect:

$$[aA+bB,cC+dD]= ac[A,C] + ad[A,D] + bc[B,C] + bd[B,D], %$$ (4.55)

(in which it is as­sumed that $A$, $B$, $C$, and $D$ are op­er­a­tors, and $a$, $b$, $c$, and $d$ nu­mer­i­cal con­stants.)

To deal with com­mu­ta­tors that in­volve prod­ucts of op­er­a­tors, the rule to re­mem­ber is: “the first fac­tor comes out at the front of the com­mu­ta­tor, the sec­ond at the back”. More pre­cisely:

$$\rule[-10pt]{7pt}{0pt} \raisebox{-7pt}{ \begin{picture}(0... ...}} \end{picture} } [\ldots,AB] = A[\ldots,B] + [\ldots,A]B. %$$ (4.56)

So, if $A$ or $B$ com­mutes with the other side of the op­er­a­tor, it can sim­ply be taken out at at its side; (the sec­ond com­mu­ta­tor will be zero.) For ex­am­ple,

$$[A_1B, A_2]= A_1 [B, A_2], \qquad [B A_1, A_2] = [B, A_2] A_1$$

if $A_1$ and $A_2$ com­mute.

Now from the gen­eral to the spe­cific. Be­cause chang­ing sides in a com­mu­ta­tor merely changes its sign, from here on only one of the two pos­si­bil­i­ties will be shown. First the po­si­tion op­er­a­tors all mu­tu­ally com­mute:

$$[{\widehat x},{\widehat y}]=[{\widehat y},{\widehat z}]=[{\widehat z},{\widehat x}]=0 %$$ (4.57)

as do po­si­tion-de­pen­dent op­er­a­tors such as a po­ten­tial en­ergy $V(x,y,z)$:

$$[{\widehat x},V(x,y,z)]=[{\widehat y},V(x,y,z)]=[{\widehat z},V(x,y,z)]=0 %$$ (4.58)

This il­lus­trates that if a set of op­er­a­tors all com­mute, then all com­bi­na­tions of those op­er­a­tors com­mute too.

The lin­ear mo­men­tum op­er­a­tors all mu­tu­ally com­mute:

$$[{\widehat p}_x,{\widehat p}_y]= [{\widehat p}_y,{\widehat p}_z] = [{\widehat p}_z,{\widehat p}_x] = 0 %$$ (4.59)

How­ever, po­si­tion op­er­a­tors and lin­ear mo­men­tum op­er­a­tors in the same di­rec­tion do not com­mute; in­stead:

$$[{\widehat x},{\widehat p}_x]= [{\widehat y},{\widehat p}_y] = [{\widehat z},{\widehat p}_z] = {\rm i}\hbar %$$ (4.60)

As seen in the pre­vi­ous sub­sec­tion, this lack of com­mu­ta­tion causes the Heisen­berg un­cer­tainty prin­ci­ple. Po­si­tion and lin­ear mo­men­tum op­er­a­tors in dif­fer­ent di­rec­tions do com­mute:

$$[{\widehat x},{\widehat p}_y]= [{\widehat x},{\widehat p}_z]... ...ehat z},{\widehat p}_x] = [{\widehat z},{\widehat p}_y] = 0 %$$ (4.61)

A gen­er­al­iza­tion that is fre­quently very help­ful is:

$$[f,{\widehat p}_x]= {\rm i}\hbar \frac{\partial f}{\partial x... ...{\widehat p}_z] = {\rm i}\hbar \frac{\partial f}{\partial z} %$$ (4.62)

where $f$ is any func­tion of $x$, $y$, and $z$.

Un­like lin­ear mo­men­tum op­er­a­tors, an­gu­lar mo­men­tum op­er­a­tors do not mu­tu­ally com­mute. The com­mu­ta­tors are given by the so-called “ fun­da­men­tal com­mu­ta­tion re­la­tions:”

$$[\L _x,\L _y]= {\rm i}\hbar\L _z \quad [\L _y,\L _z] = {\rm i}\hbar\L _x \quad [\L _z,\L _x] = {\rm i}\hbar\L _y %$$ (4.63)

Note the $\ldots{x}yzxyz\ldots$ or­der of the in­dices that pro­duces pos­i­tive signs; a re­versed $\ldots{z}yxzy\ldots$ or­der adds a mi­nus sign. For ex­am­ple $[\L _z,\L _y]$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-{\rm i}\hbar\L _x$ be­cause $y$ fol­low­ing $z$ is in re­versed or­der.

The an­gu­lar mo­men­tum com­po­nents do all com­mute with the square an­gu­lar mo­men­tum op­er­a­tor:

$$[\L _x,\L ^2]= [\L _y,\L ^2] = [\L _z,\L ^2] = 0 \quad\mbox{where } \L ^2 = \L _x^2 + \L _y^2 + \L _z^2 %$$ (4.64)

Just the op­po­site of the sit­u­a­tion for lin­ear mo­men­tum, po­si­tion and an­gu­lar mo­men­tum op­er­a­tors in the same di­rec­tion com­mute,

$$[{\widehat x}, \L _x]= [{\widehat y}, \L _y] = [{\widehat z}, \L _z] = 0 %$$ (4.65)

but those in dif­fer­ent di­rec­tions do not:

$$[{\widehat x},\L _y]=[\L _x,{\widehat y}]={\rm i}\hbar {\wide... ...hat z},\L _x]=[\L _z,{\widehat x}]={\rm i}\hbar {\widehat y} %$$ (4.66)

Square po­si­tion com­mutes with all com­po­nents of an­gu­lar mo­men­tum,

$$[{\widehat r}^2,\L _x]= [{\widehat r}^2,\L _y] = [{\widehat r}^2,\L _z] = [{\widehat r}^2,\L ^2] = 0 %$$ (4.67)

The com­mu­ta­tor be­tween po­si­tion and square an­gu­lar mo­men­tum is, us­ing vec­tor no­ta­tion for con­cise­ness,

$$[{\skew 2\widehat{\skew{-1}\vec r}},\L ^2]= - 2\hbar^2 {\skew... ...t{\skew{-1}\vec r}}\cdot{\skew 4\widehat{\skew{-.5}\vec p}}) %$$ (4.68)

The com­mu­ta­tors be­tween lin­ear and an­gu­lar mo­men­tum are very sim­i­lar to the ones be­tween po­si­tion and an­gu­lar mo­men­tum:

$$[{\widehat p}_x, \L _x]= [{\widehat p}_y, \L _y] = [{\widehat p}_z, \L _z] = 0 %$$ (4.69)

$$[{\widehat p}_x,\L _y]=[\L _x,{\widehat p}_y]={\rm i}\hbar {\... ..._z,\L _x]=[\L _z,{\widehat p}_x]={\rm i}\hbar {\widehat p}_y %$$ (4.70)

$$[{\widehat p}^2,\L _x]= [{\widehat p}^2,\L _y] = [{\widehat p}^2,\L _z] = [{\widehat p}^2,\L ^2] = 0 %$$ (4.71)

$$[{\skew 4\widehat{\skew{-.5}\vec p}},\L ^2]= - 2\hbar^2 {\ske... ...{\skew{-.5}\vec p}}\cdot{\skew 4\widehat{\skew{-.5}\vec p}}) %$$ (4.72)

The fol­low­ing com­mu­ta­tors are also use­ful:

$$[{\skew0\vec r}\times{\skew 4\widehat{\vec L}},\L ^2]= 2 {\rm... ...],\L ^2] = 2\hbar^2({\skew0\vec r}\L ^2+\L ^2{\skew0\vec r}) %$$ (4.73)

Com­mu­ta­tors in­volv­ing spin are dis­cussed in a later chap­ter, 5.5.3.

Key Points

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Rules for eval­u­at­ing com­mu­ta­tors were given.

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Re­turn to this sub­sec­tion if you need to fig­ure out some com­mu­ta­tor or the other.