A.8 Pos­i­tive ground state wave func­tion

This ad­den­dum dis­cusses why in at least the sim­plest cases a ground state wave func­tion can be as­sumed to be real, pos­i­tive, and unique (i.e. non­de­gen­er­ate). It is as­sumed that the po­ten­tial is a real func­tion of po­si­tion. That is true for the hy­dro­gen mol­e­c­u­lar ion. It is also true for a sin­gle hy­dro­gen atom and most other sim­ple sys­tems, at least in the non­rel­a­tivis­tic ap­prox­i­ma­tions nor­mally used in this book.

It should first be noted that if po­ten­tials are al­lowed that are pos­i­tive in­fin­ity in a fi­nite re­gion, nonunique ground states that can­not be taken pos­i­tive may in fact pos­si­ble. Such a po­ten­tial can pro­vide an im­pen­e­tra­ble bound­ary, com­pletely sep­a­rat­ing one re­gion of space from an­other. In that case the ground state wave func­tions at the two sides of the bound­ary be­come de­cou­pled, al­low­ing for in­de­ter­mi­nacy in the com­bined ground state. Such ar­ti­fi­cial cases are not cov­ered here. But you can read­ily find ex­am­ples in lots of books on quan­tum me­chan­ics, es­pe­cially in one di­men­sion. Here it will be as­sumed that the po­ten­tials stay away from pos­i­tive in­fin­ity. For prac­ti­cal pur­poses, it may also be noted that if the po­ten­tial be­comes pos­i­tive in­fin­ity at just a few points, it is usu­ally not a prob­lem un­less the ap­proach to sin­gu­lar­ity is very steep.

There is how­ever a much more im­por­tant re­stric­tion to the con­clu­sions in this note: ground states may not be pos­i­tive if you go to many-par­ti­cle sys­tems. That is dis­cussed fur­ther in the fi­nal para­graphs of this ad­den­dum.

First con­sider why the ground state, and any other en­ergy eigen­state, can be as­sumed to be real with­out loss of gen­er­al­ity. Sup­pose that you had a com­plex eigen­func­tion $\psi$ for the eigen­value prob­lem $H\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi$. Write the eigen­func­tion as

$$\psi = \psi_{\rm {r}} + {\rm i}\psi_{\rm {i}}$$

where the real part $\psi_{\rm {r}}$ and the imag­i­nary part $\psi_{\rm {i}}$ are real func­tions. Plug­ging this into the eigen­value prob­lem, you see that the real and imag­i­nary parts each sep­a­rately must sat­isfy the eigen­value prob­lem. (For the com­plex num­ber $H\psi-E\psi$ to be zero, both its real and imag­i­nary parts have to be zero.) So each of $\psi_{\rm {r}}$ and $\psi_{\rm {i}}$ is just as good an eigen­func­tion as $\psi$ and each is real. Since the orig­i­nal com­plex $\psi$ is a lin­ear com­bi­na­tion of the two, you do not need it sep­a­rately. (If ei­ther $\psi_{\rm {r}}$ or $\psi_{\rm {i}}$ is zero, it is not an eigen­func­tion, but then it is not needed to de­scribe $\psi$ ei­ther. And if it is nonzero, it can be nor­mal­ized.)

Next con­sider why the ground state can be taken to be pos­i­tive, as­sum­ing, for now, that it is unique. What char­ac­ter­izes the ground state $\psi_{\rm {gs}}$ is that it has the low­est pos­si­ble ex­pec­ta­tion value of the en­ergy. The ex­pec­ta­tion en­ergy can be writ­ten for ar­bi­trary, but nor­mal­ized, wave func­tions $\psi$ as

$$\left\langle{E}\right\rangle = \frac{\hbar^2}{2m} \int_{\rm... ...w0\vec r} + \int_{\rm all} V \psi^2 { \rm d}^3{\skew0\vec r}$$

In the first term, the ki­netic en­ergy, in­te­gra­tions by part have been used to get rid of the sec­ond or­der de­riv­a­tives. Note that $(\nabla\psi)^2$ stands for the sum of the square par­tial de­riv­a­tives of $\psi$. Now by de­f­i­n­i­tion $\psi_{\rm {gs}}$ has the low­est pos­si­ble value of the above ex­pec­ta­tion en­ergy among all pos­si­ble nor­mal­ized func­tions $\psi$. But only terms square in $\psi$ ap­pear in the ex­pec­ta­tion en­ergy. So $\vert\psi_{\rm {gs}}\vert$ has the same ex­pec­ta­tion en­ergy. That means that $\vert\psi_{\rm {gs}}\vert$ is a ground state wave func­tion too. Un­der the given as­sump­tion that the ground state is unique, $\vert\psi_{\rm {gs}}\vert$ can be taken to be the ground state. That makes the ground state pos­i­tive. (Note that a con­stant of mag­ni­tude one does not make a dif­fer­ence in an eigen­func­tion. So the orig­i­nal $\psi_{\rm {gs}}$ might well have been equal to $-\vert\psi_{\rm {gs}}\vert$. But $\vert\psi_{\rm {gs}}\vert$ is equiv­a­lent to that.)

Fi­nally, it needs to be shown that the ground state is in­deed unique as as­sumed above. That is re­ally messy, so it has been banned to de­riva­tion {D.22}. It is based on the same idea that the ab­solute value of a ground state is a ground state too.

Re­gret­tably the ar­gu­ments above stop work­ing for more than two elec­trons. To re­ally un­der­stand the rea­son, you will first need to read chap­ter 5.6 on mul­ti­ple-par­ti­cle sys­tems. But in a nut­shell, the wave func­tion for sys­tems with mul­ti­ple elec­trons must sat­isfy ad­di­tional re­quire­ments, called the an­ti­sym­metriza­tion re­quire­ments. These re­quire­ments nor­mally turn $\vert\psi_{\rm {gs}}\vert$ into an un­ac­cept­able wave func­tion. Then ob­vi­ously the above ar­gu­ments fall apart. For­tu­nately, for just two elec­trons, there is a loop­hole in the re­quire­ment called spin. That al­lows the hy­dro­gen mol­e­cule, with two elec­trons, still to be cov­ered.

The same prob­lem oc­curs for atomic nu­clei that con­tain mul­ti­ple pro­tons and/or neu­trons. (For atomic nu­clei, the po­ten­tials also tend to be far more com­pli­cated than as­sumed here. But that is an­other mat­ter.) In gen­eral, par­ti­cles for which an­ti­sym­metriza­tion re­quire­ments ap­ply are called fermi­ons.

There is how­ever a dif­fer­ent class of par­ti­cles called ”bosons.” For those, the wave func­tion has to sat­isfy sym­metriza­tion re­quire­ments. Sym­metriza­tion re­quire­ments are still OK if you re­place $\psi$ by $\vert\psi\vert$. So the ideas above are help­ful for un­der­stand­ing the ground state of large num­bers of bosons. For ex­am­ple, they are help­ful in un­der­stand­ing the su­per­flu­id­ity of liq­uid he­lium near its ground state, [18, pp. 321-323]. Com­plete he­lium atoms are bosons.