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

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

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:

$$\fbox{$\displaystyle \left\langle{E}\right\rangle \mbox{ is minimal for the ground state wave function.} $} %$$ (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:

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

Fol­low­ing that idea, what you do is ad­just the pa­ra­me­ters val­ues so that you get the low­est pos­si­ble value of the ex­pec­ta­tion en­ergy for your type of ap­prox­i­mate wave func­tion. The true ground state wave func­tion al­ways has the low­est pos­si­ble en­ergy, so the lower you make your ap­prox­i­mate en­ergy, the closer that en­ergy is to the ex­act value.

So this pro­ce­dure gives you the best pos­si­ble ap­prox­i­ma­tion to the true en­ergy, and en­ergy is usu­ally the key quan­tity in quan­tum me­chan­ics. In ad­di­tion you know for sure that the true en­ergy must be lower than your ap­prox­i­ma­tion, which is also of­ten very use­ful in­for­ma­tion.

The vari­a­tional method as de­scribed above has al­ready been used ear­lier in this book to find an ap­prox­i­mate ground state for the hy­dro­gen mol­e­c­u­lar ion, chap­ter 4.6, and for the hy­dro­gen mol­e­cule, chap­ter 5.2. It will also be used to find an ap­prox­i­mate 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.

To be sure, it is not at all ob­vi­ous that get­ting the best en­ergy will also pro­duce the best 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 in this case, your wave func­tion will in fact be close to the ex­act wave func­tion if you man­age to get close enough to the ex­act en­ergy. More pre­cisely, as­sum­ing that the ground state is unique, the closer your en­ergy gets to the ex­act en­ergy, the closer your wave func­tion gets to the ex­act wave func­tion. One way of think­ing about it is to note that your ap­prox­i­mate wave func­tion is al­ways a com­bi­na­tion of the de­sired ex­act ground state plus pol­lut­ing amounts of higher en­ergy states. By min­i­miz­ing the en­ergy, in some sense you min­i­mize the amount of these pol­lut­ing higher en­ergy states. The math­e­mat­ics of that idea is ex­plored in more de­tail in ad­den­dum {A.7}.

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 is not ob­vi­ous. But if us­ing a suit­able 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 the ground state 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 atomic 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 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 might search for­ever.

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 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. Math­e­mati­cians may not like that, since the word small has no rig­or­ous mean­ing. But un­less you mis­use your small quan­ti­ties, you can al­ways con­vert your re­sults us­ing them to rig­or­ous math­e­mat­ics af­ter the fact.

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. But a physi­cists may not like that, 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. There is of­ten more than one pos­si­ble choice for $x$, with none pre­ferred un­der all cir­cum­stances. So a typ­i­cal physi­cist would say that the change ${\rm d}{f}$ in $f$ due to a small change in what­ever vari­able it de­pends on must be zero. It is the same thing, since for a small enough change ${\rm d}{x}$ in the vari­able, ${\rm d}{f}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f'{\rm d}{x}$, so that if $f'$ is zero, then so is ${\rm d}{f}$. (Math­e­mat­i­cally more ac­cu­rately, if ${\rm d}{x}$ be­comes small enough, ${\rm d}{f}$ be­comes zero com­pared to ${\rm d}{x}$.) If there is more than one in­de­pen­dent vari­able that the func­tion de­pends on, then the de­riv­a­tives be­come par­tial de­riv­a­tives, ${\rm d}{f}$ be­comes $\partial{f}$, and spec­i­fy­ing the pre­cise de­riv­a­tives would be­come much messier still.

In vari­a­tional pro­ce­dures, it is com­mon to use $\delta{f}$ in­stead of ${\rm d}{f}$ or $\partial{f}$ for the small change in $f$. This book will do so too.

So in quan­tum me­chan­ics, the fact that the ex­pec­ta­tion en­ergy must be min­i­mal in the ground state can be writ­ten as:

$$\fbox{$\displaystyle \delta \left\langle{E}\right\rangle =... ...mbox{ for all acceptable small changes in wave function} $} %$$ (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 the vari­a­tional for­mu­la­tion of the pre­vi­ous sub­sec­tion can be ap­plied 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 the La­grangian mul­ti­plier method can deal with the con­straints.

The dif­fer­en­tial state­ment is:

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

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:

$$\delta \langle\psi\vert H\vert\psi\rangle = 0 \mbox{ whenever } \delta\langle\psi\vert\psi\rangle = 0.$$

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:

$$\delta \langle\psi\vert H\vert\psi\rangle - \epsilon \delta\langle\psi\vert\psi\rangle = 0$$

Here $\epsilon$ is the penalty fac­tor. Such penalty 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 change does go out of bounds, the sec­ond term will can­cel any re­sult­ing er­ro­neous gain or de­crease in ex­pec­ta­tion en­ergy, {D.48}, as­sum­ing that the penalty fac­tor is care­fully tuned. Note that the penalty fac­tor $\epsilon$ must be real be­cause the other two quan­ti­ties in the equa­tion above are changes in real func­tions.

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

$$\Big(\langle\psi+\delta\psi\vert H\vert\psi+\delta\psi\rang... ...psi+\delta\psi\rangle - \langle\psi\vert\psi\rangle\Big) = 0$$

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

$$\langle\delta\psi\vert H\vert\psi\rangle + \langle\psi\ver... ...vert\psi\rangle + \langle\psi\vert\delta\psi\rangle\Big) = 0$$

Re­mark­ably, you can throw away the sec­ond of each pair of in­ner prod­ucts in the ex­pres­sion above. To see why, re­mem­ber that you can al­low any change $\delta\psi$ you want, in­clud­ing the $\delta\psi$ you are now look­ing at times $-{\rm i}$. If you plug that into the above equa­tion and di­vide the en­tire thing by ${\rm i}$ to get rid of the added fac­tors ${\rm i}$ again, you get

$$\langle\delta\psi\vert H\vert\psi\rangle - \langle\psi\ver... ...vert\psi\rangle - \langle\psi\vert\delta\psi\rangle\Big) = 0$$

The two ad­di­tional mi­nus signs arise be­cause an $-{\rm i}$ comes out of the left side of an in­ner prod­uct as ${\rm i}$, but out of the right side as $-{\rm i}$. Av­er­ag­ing this equa­tion with the orig­i­nal above it has the ef­fect of throw­ing away the sec­ond of each pair of in­ner prod­ucts in the orig­i­nal equa­tion.

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

$$\langle\delta\psi\vert H\psi-\epsilon\psi\rangle = 0$$

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

$$H\psi=\epsilon\psi.$$

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}.