A.9 Wave func­tion sym­me­tries

Sym­me­tries are very im­por­tant in physics. For ex­am­ple, sym­me­tries in wave func­tions are of­ten quite help­ful to un­der­stand the physics qual­i­ta­tively.

As an ex­am­ple, the hy­dro­gen mol­e­c­u­lar ion is mir­ror sym­met­ric around its mid­plane. This mid­plane is the plane halfway in be­tween the two nu­clei, or­thog­o­nal to the line con­nect­ing them. To roughly un­der­stand what the mir­ror sym­me­try around this plane means, think of the mid­plane as an in­fi­nitely thin mir­ror. Take this mir­ror to be two-sided, so that you can look in it from ei­ther side. That al­lows you to see the mir­ror im­age of each side of the mol­e­cule. Sim­ply put, the mir­ror sym­me­try of the ion means that the mir­ror im­age looks ex­actly the same as the orig­i­nal ion.

(If you would place the en­tire mol­e­cule at one side of the mir­ror, its en­tire mir­ror im­age would be at the other side of it. But ex­cept for this ad­di­tional shift in lo­ca­tion, every­thing would re­main the same as in the case as­sumed here.)

Un­der the same terms, hu­man be­ings are roughly mir­ror sym­met­ric around the plane sep­a­rat­ing their left and right halves. But that sym­me­try is far from per­fect. For ex­am­ple, if you part your hair at one side, your mir­ror im­age parts it at the other side. And your heart changes sides too.

To de­scribe mir­ror sym­me­try more pre­cisely, take the line through the nu­clei to be the $z$-​axis. And take $z$ to be zero at the mir­ror. Then all that the mir­ror does math­e­mat­i­cally is re­place $z$ by $\vphantom0\raisebox{1.5pt}{$-$}$$z$. For ex­am­ple, the mir­ror im­age of the nu­cleus at pos­i­tive $z$ is lo­cated at the cor­re­spond­ing neg­a­tive $z$ value. And vice-versa.

The ef­fect of mir­ror­ing on any mol­e­c­u­lar wave func­tion $\Psi$ can be rep­re­sented by a “mir­ror op­er­a­tor” ${\cal M}$. Ac­cord­ing to the above, all this op­er­a­tor does is re­place $z$ by $\vphantom0\raisebox{1.5pt}{$-$}$$z$:

\begin{displaymath}
{\cal M}\Psi(x,y,z)=\Psi(x,y,-z)
\end{displaymath}

By de­f­i­n­i­tion a wave func­tion is mir­ror sym­met­ric if the mir­ror op­er­a­tor has no ef­fect on it. Math­e­mat­i­cally, if the mir­ror op­er­a­tor does not do any­thing, then ${\cal M}\Psi$ must be the same as $\Psi$. So mir­ror sym­me­try re­quires

\begin{displaymath}
{\cal M}\Psi(x,y,z)\equiv\Psi(x,y,-z)=\Psi(x,y,z)
\end{displaymath}

The fi­nal equal­ity above shows that a mir­ror-sym­met­ric wave func­tion is the same at pos­i­tive val­ues of $z$ as at the cor­re­spond­ing neg­a­tive val­ues. Math­e­mati­cians might sim­ply say that the wave func­tion is sym­met­ric around the $xy$-​plane, (i.e. the mir­ror). The ground state $\psi_{\rm {gs}}$ of the mol­e­c­u­lar ion is mir­ror sym­met­ric in this sense. The big ques­tion to be ad­dressed in this ad­den­dum is, why?

The fun­da­men­tal rea­son why the ion is mir­ror sym­met­ric is a math­e­mat­i­cal one. The mir­ror op­er­a­tor ${\cal M}$ com­mutes with the Hamil­ton­ian $H$. Re­call from chap­ter 4.5.1 what this means:

\begin{displaymath}
{\cal M}H = H {\cal M}
\end{displaymath}

In words, it does not make a dif­fer­ence in which or­der you ap­ply the two op­er­a­tors.

That can be seen from the physics. The Hamil­ton­ian con­sists of po­ten­tial en­ergy $V$ and ki­netic en­ergy $T$. Now it does not make a dif­fer­ence whether you mul­ti­ply a wave func­tion value by the po­ten­tial be­fore or af­ter you flip the value over to the op­po­site $z$-​po­si­tion. The po­ten­tial is the same at op­po­site $z$ val­ues, be­cause the nu­clei at the two sides of the mir­ror are the same. As far as the ki­netic en­ergy is con­cerned, if it in­volved a first-or­der $z$-​de­riv­a­tive, there would be a change of sign when you flip over the sign of $z$. But the ki­netic en­ergy has only a sec­ond or­der $z$-​de­riv­a­tive. A sec­ond or­der de­riv­a­tive does not change. So all to­gether it makes no dif­fer­ence whether you first mir­ror and then ap­ply the Hamil­ton­ian or vice-versa. The two op­er­a­tors com­mute.

Also ac­cord­ing to chap­ter 4.5.1, that has a con­se­quence. It im­plies that you can take en­ergy eigen­func­tions to be mir­ror eigen­func­tions too. And the ground state is an en­ergy eigen­func­tion. So it can be taken to be an eigen­func­tion of ${\cal M}$ too:

\begin{displaymath}
{\cal M}\psi_{\rm gs}(x,y,z) = \lambda \psi_{\rm gs}(x,y,z)
\end{displaymath}

Here $\lambda$ is a con­stant called the eigen­value. But what would this eigen­value be?

To an­swer that, ap­ply ${\cal M}$ twice. That mul­ti­plies the wave func­tion by the square eigen­value. But if you ap­ply ${\cal M}$ twice, you al­ways get the orig­i­nal wave func­tion back, be­cause $-(-z)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$. So the square eigen­value $\lambda^2$ must be 1, in or­der that the wave func­tion does not change when mul­ti­plied by it. That means that the eigen­value $\lambda$ it­self can be ei­ther 1 or $\vphantom0\raisebox{1.5pt}{$-$}$1. So for the ground state wave func­tion $\psi_{\rm {gs}}$, ei­ther

\begin{displaymath}
{\cal M}\psi_{\rm gs}(x,y,z)\equiv\psi_{\rm gs}(x,y,-z)= 1\times\psi_{\rm gs}(x,y,z)
\end{displaymath}

or

\begin{displaymath}
{\cal M}\psi_{\rm gs}(x,y,z)\equiv\psi_{\rm gs}(x,y,-z)=-1\times\psi_{\rm gs}(x,y,z)
\end{displaymath}

If the first pos­si­bil­ity ap­plies, the wave func­tion does not change un­der the mir­ror­ing. So by de­f­i­n­i­tion it is mir­ror sym­met­ric. If the sec­ond pos­si­bil­ity ap­plies, the wave func­tion changes sign un­der the mir­ror­ing. Such a wave func­tion is called “mir­ror antisym­met­ric.” But the sec­ond pos­si­bil­ity has wave func­tion val­ues of op­po­site sign at op­po­site val­ues of $z$. That is not pos­si­ble, be­cause the pre­vi­ous ad­den­dum showed that the ground state wave func­tion is every­where pos­i­tive. So it must be pos­si­bil­ity one. That means that the ground state must in­deed be mir­ror sym­met­ric as claimed.

It may be noted that the state of sec­ond low­est en­ergy will be an­ti­sym­met­ric. You can see the same thing hap­pen­ing for the eigen­func­tions of the par­ti­cle in a pipe. The ground state fig­ure 3.8, (or 3.11 in three di­men­sions), is sym­met­ric around the cen­ter cross-sec­tion of the pipe. The first ex­cited state, at the top of fig­ures 3.9, (or 3.12), is an­ti­sym­met­ric. (Note that the grey tones show the square wave func­tion. If the wave func­tion is an­ti­sym­met­ric, the square wave func­tion is sym­met­ric. But it will be zero at the sym­me­try plane.)

Next con­sider the ro­ta­tional sym­me­try of the hy­dro­gen mol­e­c­u­lar ion around the axis through the nu­clei. The ground state of the mol­e­c­u­lar ion does not change if you ro­tate the ion around the $z$-​axis through the nu­clei. That makes it ro­ta­tion­ally sym­met­ric. The big ques­tion is again, why?

In this case, let ${\cal R}_\varphi$ be the op­er­a­tor that ro­tates a wave func­tion $\Psi$ over an an­gle $\varphi$ around the $z$-​axis. This op­er­a­tor too com­mutes with the Hamil­ton­ian. Af­ter all, the only phys­i­cally mean­ing­ful di­rec­tion is the $z$-​axis through the nu­clei. The an­gu­lar ori­en­ta­tion of the $xy$ axes sys­tem nor­mal to it is a com­pletely ar­bi­trary choice. So it should not make a dif­fer­ence at what an­gle around the $z$ axis you ap­ply the Hamil­ton­ian.

There­fore the ground state must be an eigen­func­tion of the ro­ta­tion op­er­a­tor just like it is one of the mir­ror op­er­a­tor:

\begin{displaymath}
{\cal R}_\varphi \psi_{\rm gs} = \lambda \psi_{\rm gs}
\end{displaymath}

But now what is that eigen­value $\lambda$? First note that the mag­ni­tude of all eigen­val­ues of ${\cal R}_\varphi$ must be 1. Oth­er­wise the mag­ni­tude of the wave func­tion would change cor­re­spond­ingly dur­ing the ro­ta­tion. How­ever, the mag­ni­tude of a wave func­tion does not change if you sim­ply ro­tate it. And if the eigen­value is a com­plex num­ber of mag­ni­tude 1, then it can al­ways be writ­ten as $e^{{\rm i}\alpha}$ where $\alpha$ is some real num­ber. So the ro­tated ground state is some mul­ti­ple $e^{{\rm i}\alpha}$ of the orig­i­nal ground state. But the val­ues of the ro­tated ground state are real and pos­i­tive just like that of the orig­i­nal ground state. That can only be true if the mul­ti­ply­ing fac­tor $e^{{\rm i}\alpha}$ is real and pos­i­tive too. And if you check the Euler for­mula (2.5), you see that $e^{{\rm i}\alpha}$ is only real and pos­i­tive if it is 1. Since mul­ti­ply­ing by 1 does not change the wave func­tion, the ground state does not change when ro­tated. That then makes it ro­ta­tion­ally sym­met­ric around the $z$-​axis through the nu­clei as claimed.

You might of course won­der about the ro­ta­tional changes of ex­cited en­ergy states. For those a cou­ple of ad­di­tional ob­ser­va­tions ap­ply. First, the num­ber $\alpha$ must be pro­por­tional to the ro­ta­tion an­gle $\varphi$, since ro­tat­ing $\Psi$ twice is equiv­a­lent to ro­tat­ing it once over twice the an­gle. That means that, more pre­cisely, the eigen­val­ues are of the form $e^{{{\rm i}}m\varphi}$, where $m$ is a real con­stant in­de­pen­dent of $\varphi$. Sec­ond, ro­tat­ing the ion over a $2\pi$ full turn puts each point back to where it came from. That should re­pro­duce the orig­i­nal wave func­tion. So an eigen­value $e^{{{\rm i}}m2\pi}$ for a full turn must be 1. Ac­cord­ing to the Euler for­mula, that re­quires $m$ to be an in­te­ger, one of ..., $\vphantom0\raisebox{1.5pt}{$-$}$2, $\vphantom0\raisebox{1.5pt}{$-$}$1, 0, 1, 2, .... For the ground state, $m$ will have to be zero; that is the only way to get $e^{{{\rm i}}m\varphi}$ equal to 1 for all an­gles $\varphi$. But for ex­cited states, $m$ can be a nonzero in­te­ger. In that case, these states do not have ro­ta­tional sym­me­try.

Re­call­ing the dis­cus­sion of an­gu­lar mo­men­tum in chap­ter 4.2.2, you can see that $m$ is re­ally the mag­netic quan­tum num­ber of the state. Ap­par­ently, there is a con­nec­tion be­tween ro­ta­tions around the $z$-​axis and the an­gu­lar mo­men­tum in the $z$-​di­rec­tion. That will be ex­plored in more de­tail in chap­ter 7.3.

For the neu­tral hy­dro­gen mol­e­cule dis­cussed in chap­ter 5.2, there is still an­other sym­me­try of rel­e­vance. The neu­tral mol­e­cule has two elec­trons, in­stead of just one. This al­lows an­other op­er­a­tion: you can swap the two elec­trons. That is called “par­ti­cle ex­change.” Math­e­mat­i­cally, what the par­ti­cle ex­change op­er­a­tor ${\cal P}$ does with the wave func­tion is swap the po­si­tion co­or­di­nates of elec­tron 1 with those of elec­tron 2. Ob­vi­ously, phys­i­cally this does not do any­thing at all; the two elec­trons are ex­actly the same. It does not make a dif­fer­ence which of the two is where. So par­ti­cle ex­change com­mutes again with the Hamil­ton­ian.

The math­e­mat­ics of the par­ti­cle ex­change is sim­i­lar to that of the mir­ror­ing dis­cussed above. In par­tic­u­lar, if you ex­change the par­ti­cles twice, they are back to where they were orig­i­nally. From that, just like for the mir­ror­ing, it can be seen that swap­ping the par­ti­cle po­si­tions does noth­ing to the ground state. So the ground state is sym­met­ric un­der par­ti­cle ex­change.

It should be noted that the ground state of sys­tems in­volv­ing three or more elec­trons is not sym­met­ric un­der ex­chang­ing the po­si­tions of the elec­trons. Wave func­tions for mul­ti­ple elec­trons must sat­isfy spe­cial par­ti­cle-ex­change re­quire­ments, chap­ter 5.6. In fact they must be antisym­met­ric un­der an ex­panded de­f­i­n­i­tion of the ex­change op­er­a­tor. This is also true for sys­tems in­volv­ing three or more pro­tons or neu­trons. How­ever, for some par­ti­cle types, like three or more he­lium atoms, the sym­me­try un­der par­ti­cle ex­change con­tin­ues to ap­ply. This is very help­ful for un­der­stand­ing the prop­er­ties of su­per­fluid he­lium, [18, p. 321].