Sub­sec­tions


9.1 The Vari­a­tional Method

Solv­ing the equa­tions of quan­tum me­chan­ics is typ­i­cally dif­fi­cult, so ap­prox­i­ma­tions must usu­ally be made. One very ef­fec­tive tool for find­ing ap­prox­i­mate so­lu­tions is the vari­a­tional prin­ci­ple. This sec­tion gives some of the ba­sic ideas, in­clud­ing ways to ap­ply it best.


9.1.1 Ba­sic vari­a­tional state­ment

Find­ing the state of a phys­i­cal sys­tem in quan­tum me­chan­ics means find­ing the wave func­tion $\Psi$ that de­scribes it. For ex­am­ple, at suf­fi­ciently low tem­per­a­tures, phys­i­cal sys­tems will be de­scribed by the ground state wave func­tion. The prob­lem is that if there are more than a cou­ple of par­ti­cles in the sys­tem, the wave func­tion is a very high-di­men­sional func­tion. It is far too com­plex to be crunched out us­ing brute force on any cur­rent com­puter.

How­ever, the ex­pec­ta­tion value of en­ergy is just a sim­ple sin­gle num­ber for any given wave func­tion. It is de­fined as

\begin{displaymath}
\left\langle{E}\right\rangle = \left\langle\vphantom{H\Psi}...
...ce{.03em}\right.\!\left\vert\vphantom{\Psi}H\Psi\right\rangle
\end{displaymath}

where $H$ is the Hamil­ton­ian of the sys­tem. The key ob­ser­va­tion on which the vari­a­tional method is based is that the ground state is the state among all al­low­able wave func­tions that has the low­est ex­pec­ta­tion value of en­ergy:
\begin{displaymath}
\fbox{$\displaystyle
\left\langle{E}\right\rangle \mbox{ is minimal for the ground state wave function.}
$} %
\end{displaymath} (9.1)

That means that if you would find $\left\langle{E}\right\rangle $ for all pos­si­ble sys­tem wave func­tions, you would be able to pick out the ground state sim­ply as the state that has the low­est value.

Of course, find­ing the ex­pec­ta­tion value of the en­ergy for all pos­si­ble wave func­tions is still an im­pos­si­ble task. But you may be able to guess a generic type of wave func­tion that you would ex­pect to be able to ap­prox­i­mate the ground state well, un­der suit­able con­di­tions. Nor­mally, suit­able con­di­tions means that the ap­prox­i­ma­tion will be good only if var­i­ous pa­ra­me­ters ap­pear­ing in the ap­prox­i­mate wave func­tion are well cho­sen.

That then leaves you with the much smaller task of find­ing good val­ues for this lim­ited set of pa­ra­me­ters. Here the key idea is:

\begin{displaymath}
\fbox{$\displaystyle
\left\langle{E}\right\rangle \mbox{ is lowest for the best approximation to the ground state.}
$} %
\end{displaymath} (9.2)

The vari­a­tional pro­ce­dure of in­ter­est here can be de­scribed as fol­lows: ad­just the set of pa­ra­me­ters so that the low­est value of ex­pec­ta­tion en­ergy is ob­tained that can be achieved by the ap­prox­i­mate wave func­tion. Take that as your best ap­prox­i­ma­tion. The ex­act ground state will have less en­ergy than the ap­prox­i­mate one, but if the ap­prox­i­mate one is well cho­sen, not that much less.

The vari­a­tional method as de­scribed above has al­ready been used ear­lier in this book to find the ground state for the hy­dro­gen mol­e­c­u­lar ion, chap­ter 4.6, and the one for the hy­dro­gen mol­e­cule, chap­ter 5.2. It will also be used to find the ground state for the he­lium atom, {A.38.2}. The method works quite well even for the crude ap­prox­i­mate wave func­tions used in those ex­am­ples.

Of course, it is not ob­vi­ous why get­ting the best en­ergy would also pro­duce the best com­plete wave func­tion. Af­ter all, best is a some­what tricky term for a com­plex ob­ject like a wave func­tion. To take an ex­am­ple from an­other field, surely you would not ar­gue that the best sprinter in the world must also be the best per­son in the world. But for ap­prox­i­mate ground state wave func­tions, if the ap­prox­i­mate en­ergy is close to the ex­act en­ergy, then the en­tire ap­prox­i­mate wave func­tion will also be close to the ex­act one. (This does as­sume that the ground state is unique; there must not be other wave func­tions with the same, or al­most the same, en­ergy. See ad­den­dum {A.7} for de­tails.)

And there are other ben­e­fits to specif­i­cally get­ting the en­ergy as ac­cu­rate as pos­si­ble. One prob­lem is of­ten to fig­ure out whether a sys­tem is bound. For ex­am­ple, can you add an­other elec­tron to a hy­dro­gen atom and have that elec­tron at least weakly bound? The an­swer may not be ob­vi­ous. But if us­ing your ap­prox­i­mate so­lu­tion, you man­age to show that the ap­prox­i­mate en­ergy of the bound sys­tem is less than that of hav­ing the ad­di­tional elec­tron at in­fin­ity, then you have proved that the bound state ex­ist. De­spite the fact that your so­lu­tion has er­rors. The rea­son is that, by de­f­i­n­i­tion, the ground state must have lower en­ergy than your ap­prox­i­mate wave func­tion, so is even more tightly bound to­gether than your ap­prox­i­mate wave func­tion says.

An­other rea­son to specif­i­cally get­ting the en­ergy as ac­cu­rate as pos­si­ble is that en­ergy val­ues are di­rectly re­lated to how fast sys­tems evolve in time when not in the ground state, chap­ter 7.

For the above rea­sons, it is also great that the er­rors in en­ergy turn out to be un­ex­pect­edly small in a vari­a­tional pro­ce­dure, when com­pared to the er­rors in the guessed wave func­tion, {A.7}.

To get the sec­ond low­est en­ergy state, you could search for the low­est en­ergy among all wave func­tions or­thog­o­nal to the ground state. But since you would not know the ex­act ground state, you would need to use your ap­prox­i­mate one in­stead. That would in­volve some er­ror, and it is no longer sure that the true sec­ond-low­est en­ergy level is no higher than what you com­pute, but any­way. The supris­ing ac­cu­racy in en­ergy will still ap­ply.

If you want to get truly ac­cu­rate re­sults in a vari­a­tional method, in gen­eral you will need to in­crease the num­ber of pa­ra­me­ters. The mol­e­c­u­lar ex­am­ple so­lu­tions were based on the atom ground states, and you could con­sider adding some ex­cited states to the mix. In gen­eral, a pro­ce­dure us­ing ap­pro­pri­ate guessed func­tions is called a Rayleigh-Ritz method. Al­ter­na­tively, you could just chop space up into lit­tle pieces, or el­e­ments, and use a sim­ple poly­no­mial within each piece. That is called a fi­nite-el­e­ment method. In ei­ther case, you end up with a fi­nite, but rel­a­tively large num­ber of un­knowns; the pa­ra­me­ters and/or co­ef­fi­cients of the func­tions, or the co­ef­fi­cients of the poly­no­mi­als.


9.1.2 Dif­fer­en­tial form of the state­ment

You might by now won­der about the wis­dom of try­ing to find the min­i­mum en­ergy by search­ing through the count­less pos­si­ble com­bi­na­tions of a lot of pa­ra­me­ters. Brute-force search worked fine for the hy­dro­gen mol­e­cule ex­am­ples since they re­ally only de­pended non­triv­ially on the dis­tance be­tween the nu­clei. But if you add some more pa­ra­me­ters for bet­ter ac­cu­racy, you quickly get into trou­ble. Semi-an­a­lyt­i­cal ap­proaches like Hartree-Fock even leave whole func­tions un­spec­i­fied. In that case, sim­ply put, every sin­gle func­tion value is an un­known pa­ra­me­ter, and a func­tion has in­fi­nitely many of them. You would be search­ing in an in­fi­nite-di­men­sion­al space, and could search for­ever. Maybe you could try some clever ge­netic al­go­rithm.

Usu­ally it is a much bet­ter idea to write some equa­tions for the min­i­mum en­ergy first. From cal­cu­lus, you know that if you want to find the min­i­mum of a func­tion, the so­phis­ti­cated way to do it is to note that the par­tial de­riv­a­tives of the func­tion must be zero at the min­i­mum. Less rig­or­ously, but a lot more in­tu­itive, at the min­i­mum of a func­tion the changes in the func­tion due to small changes in the vari­ables that it de­pends on must be zero. In the sim­plest pos­si­ble ex­am­ple of a func­tion $f(x)$ of one vari­able $x$, a rig­or­ous math­e­mati­cian would say that at a min­i­mum, the de­riv­a­tive $f'(x)$ must be zero. In­stead a typ­i­cal physi­cist would say that the change $\delta{f}$, (or ${\rm d}{f}$,) in $f$ due to a small change $\delta{x}$ in $x$ must be zero. It is the same thing, since $\delta{f}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f'\delta{x}$, so that if $f'$ is zero, then so is $\delta{f}$. But math­e­mati­cians do not like the word small, since it has no rig­or­ous mean­ing. On the other hand, in physics you may not like to talk about de­riv­a­tives, for if you say de­riv­a­tive, you must say with re­spect to what vari­able; you must say what $x$ is as well as what $f$ is, and there is of­ten more than one pos­si­ble choice for $x$, with none pre­ferred un­der all cir­cum­stances. (And in prac­tice, the word small does have an un­am­bigu­ous mean­ing: it means that you must ig­nore every­thing that is of square mag­ni­tude or more in terms of the small quan­ti­ties.)

In physics terms, the fact that the ex­pec­ta­tion en­ergy must be min­i­mal in the ground state means that you must have:

\begin{displaymath}
\fbox{$\displaystyle
\delta \left\langle{E}\right\rangle =...
...mbox{ for all acceptable small changes in wave function}
$} %
\end{displaymath} (9.3)

The changes must be ac­cept­able; you can­not al­low that the changed wave func­tion is no longer nor­mal­ized. Also, if there are bound­ary con­di­tions, the changed wave func­tion should still sat­isfy them. (There may be ex­cep­tions per­mit­ted to the lat­ter un­der some con­di­tions, but these will be ig­nored here.) So, in gen­eral you have con­strained min­i­miza­tion; you can­not make your changes com­pletely ar­bi­trary.


9.1.3 Us­ing La­grangian mul­ti­pli­ers

As an ex­am­ple of how you can ap­ply the vari­a­tional for­mu­la­tion of the pre­vi­ous sub­sec­tion an­a­lyt­i­cally, and how it can also de­scribe eigen­states of higher en­ergy, this sub­sec­tion will work out a very ba­sic ex­am­ple. The idea is to fig­ure out what you get if you truly zero the changes in the ex­pec­ta­tion value of en­ergy $\langle{E}\rangle$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\langle\psi\vert H\vert\psi\rangle$ over all ac­cept­able wave func­tions $\psi$. (In­stead of just over all pos­si­ble ver­sions of a nu­mer­i­cal ap­prox­i­ma­tion, say.) It will il­lus­trate how you can deal with the con­straints.

The dif­fer­en­tial state­ment is:

\begin{displaymath}
\delta \langle\psi\vert H\vert\psi\rangle = 0
\mbox{ for all acceptable changes $\delta\psi$\ in $\psi$}
\end{displaymath}

But ac­cept­able is not a math­e­mat­i­cal con­cept. What does it mean? Well, if it is as­sumed that there are no bound­ary con­di­tions, (like the har­monic os­cil­la­tor, but un­like the par­ti­cle in a pipe,) then ac­cept­able just means that the wave func­tion must re­main nor­mal­ized un­der the change. So the change in $\langle\psi\vert\psi\rangle$ must be zero, and you can write more specif­i­cally:

\begin{displaymath}
\delta \langle\psi\vert H\vert\psi\rangle = 0
\mbox{ whenever } \delta\langle\psi\vert\psi\rangle = 0.
\end{displaymath}

But how do you crunch a state­ment like that down math­e­mat­i­cally? Well, there is a very im­por­tant math­e­mat­i­cal trick to sim­plify this. In­stead of rig­or­ously try­ing to en­force that the changed wave func­tion is still nor­mal­ized, just al­low any change in wave func­tion. But add penalty points to the change in ex­pec­ta­tion en­ergy if the change in wave func­tion goes out of al­lowed bounds:

\begin{displaymath}
\delta \langle\psi\vert H\vert\psi\rangle
- \epsilon \delta\langle\psi\vert\psi\rangle = 0
\end{displaymath}

Here $\epsilon$ is the penalty fac­tor; such fac­tors are called “La­grangian mul­ti­pli­ers” af­ter a fa­mous math­e­mati­cian who prob­a­bly watched a lot of soc­cer. For a change in wave func­tion that does not go out of bounds, the sec­ond term is zero, so noth­ing changes. And if the penalty fac­tor is care­fully tuned, the sec­ond term can can­cel any er­ro­neous gain or de­crease in ex­pec­ta­tion en­ergy due to go­ing out of bounds, {D.48}.

You do not, how­ever, have to ex­plic­itly tune the penalty fac­tor your­self. All you need to know is that a proper one ex­ists. In ac­tual ap­pli­ca­tion, all you do in ad­di­tion to en­sur­ing that the pe­nal­ized change in ex­pec­ta­tion en­ergy is zero is en­sure that at least the unchanged wave func­tion is nor­mal­ized. It is re­ally a mat­ter of count­ing equa­tions ver­sus un­knowns. Com­pared to sim­ply set­ting the change in ex­pec­ta­tion en­ergy to zero with no con­straints on the wave func­tion, one ad­di­tional un­known has been added, the penalty fac­tor. And quite gen­er­ally, if you add one more un­known to a sys­tem of equa­tions, you need one more equa­tion to still have a unique so­lu­tion. As the one-more equa­tion, use the nor­mal­iza­tion con­di­tion. With enough equa­tions to solve, you will get the cor­rect so­lu­tion, which means that the im­plied value of the penalty fac­tor should be OK too.

So what does this vari­a­tional state­ment now pro­duce? Writ­ing out the dif­fer­ences ex­plic­itly, you must have

\begin{displaymath}
\Big(\langle\psi+\delta\psi\vert H\vert\psi+\delta\psi\rang...
...psi+\delta\psi\rangle
- \langle\psi\vert\psi\rangle\Big)
= 0
\end{displaymath}

Mul­ti­ply­ing out, can­cel­ing equal terms and ig­nor­ing terms that are qua­drat­i­cally small in $\delta\psi$, you get

\begin{displaymath}
\langle\delta\psi\vert H\vert\psi\rangle
+ \langle\psi\ver...
...vert\psi\rangle
+ \langle\psi\vert\delta\psi\rangle\Big)
= 0
\end{displaymath}

That is not yet good enough to say some­thing spe­cific about. But re­mem­ber that you can ex­change the sides of an in­ner prod­uct if you add a com­plex con­ju­gate, so

\begin{displaymath}
\langle\delta\psi\vert H\vert\psi\rangle
+ \langle\delta\p...
...rt\psi\rangle
- \langle\delta\psi\vert\psi\rangle^*\Big)
= 0
\end{displaymath}

Also re­mem­ber that you can al­low any change $\delta\psi$ you want, in­clud­ing the $\delta\psi$ you are now look­ing at times ${\rm i}$. That means that you also have:

\begin{displaymath}
\langle{\rm i}\delta\psi\vert H\vert\psi\rangle
+ \langle{...
...rangle
+ \langle{\rm i}\delta\psi\vert\psi\rangle^*\Big)
= 0
\end{displaymath}

or us­ing the fact that num­bers come out of the left side of an in­ner prod­uct as com­plex con­ju­gates

\begin{displaymath}
-{\rm i}\langle\delta\psi\vert H\vert\psi\rangle
+{\rm i}\...
...\rangle
+{\rm i}\langle\psi\vert\delta\psi\rangle^*\Big)
= 0
\end{displaymath}

If you di­vide out a $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm i}$ and then av­er­age with the orig­i­nal equa­tion, you get rid of the com­plex con­ju­gates:

\begin{displaymath}
\langle\delta\psi\vert H\vert\psi\rangle
- \epsilon\langle\delta\psi\vert\psi\rangle
= 0
\end{displaymath}

You can now com­bine them into one in­ner prod­uct with $\delta\psi$ on the left:

\begin{displaymath}
\langle\delta\psi\vert H\psi-\epsilon\psi\rangle
= 0
\end{displaymath}

If this is to be zero for any change $\delta\psi$, then the right hand side of the in­ner prod­uct must un­avoid­ably be zero. For ex­am­ple, just take $\delta\psi$ equal to a small num­ber $\varepsilon$ times the right hand side, you will get $\varepsilon$ times the square norm of the right hand side, and that can only be zero if the right hand side is. So $H\psi-\epsilon\psi$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, or

\begin{displaymath}
H\psi=\epsilon\psi.
\end{displaymath}

So you see that you have re­cov­ered the Hamil­ton­ian eigen­value prob­lem from the re­quire­ment that the vari­a­tion of the ex­pec­ta­tion en­ergy is zero. Un­avoid­ably then, $\epsilon$ will have to be an en­ergy eigen­value $E$. It of­ten hap­pens that La­grangian mul­ti­pli­ers have a phys­i­cal mean­ing be­yond be­ing merely penalty fac­tors. But note that there is no re­quire­ment for this to be the ground state. Any en­ergy eigen­state would sat­isfy the equa­tion; the vari­a­tional prin­ci­ple works for them all.

In­deed, you may re­mem­ber from cal­cu­lus that the de­riv­a­tives of a func­tion may be zero at more than one point. For ex­am­ple, a func­tion might also have a max­i­mum, or lo­cal min­ima and max­ima, or sta­tion­ary points where the func­tion is nei­ther a max­i­mum nor a min­i­mum, but the de­riv­a­tives are zero any­way. This sort of thing hap­pens here too: the ground state is the state of low­est pos­si­ble en­ergy, but there will be other states for which $\delta\left\langle{E}\right\rangle $ is zero, and these will cor­re­spond to en­ergy eigen­states of higher en­ergy, {D.49}.