5.7 Ways to Sym­metrize the Wave Func­tion

This sec­tion dis­cusses ways in which the sym­metriza­tion re­quire­ments for wave func­tions of sys­tems of iden­ti­cal par­ti­cles can be achieved in gen­eral. This is a key is­sue in the nu­mer­i­cal so­lu­tion of any non­triv­ial quan­tum sys­tem, so this sec­tion will ex­am­ine it in some de­tail.

It will be as­sumed that the ap­prox­i­mate de­scrip­tion of the wave func­tion is done us­ing a set of cho­sen sin­gle-par­ti­cle func­tions, or states,

$$\pp1/{\skew0\vec r}//z/, \pp2/{\skew0\vec r}//z/, \ldots$$

An ex­am­ple is pro­vided by the ap­prox­i­mate ground state of the hy­dro­gen mol­e­cule from the pre­vi­ous sec­tion,

$$a \left[ \psi_{\rm {l}}({\skew0\vec r}_1)\psi_{\rm {r}}({\... ...\frac{{\uparrow}{\downarrow}- {\downarrow}{\uparrow}}{\sqrt2}.$$

This can be mul­ti­plied out to be

$$\begin{eqnarray*} && \frac{a}{\sqrt2} \Big[ \psi_{\rm {l}}({\skew0\vec r}_1)... ...S_{z1})\psi_{\rm {l}}({\skew0\vec r}_2){\uparrow}(S_{z2}) \Big] \end{eqnarray*}$$

and con­sists of four sin­gle-par­ti­cle func­tions:

$$\begin{eqnarray*} \pp1/{\skew0\vec r}//z/ = \psi_{\rm {l}}({\skew0\vec r}){\upa... ...w0\vec r}//z/ = \psi_{\rm {r}}({\skew0\vec r}){\downarrow}(S_z). \end{eqnarray*}$$

The first of the four func­tions rep­re­sents a sin­gle elec­tron in the ground state around the left pro­ton with spin up, the sec­ond a sin­gle elec­tron in the same spa­tial state with spin down, etcetera. For bet­ter ac­cu­racy, more sin­gle-par­ti­cle func­tions could be in­cluded, say ex­cited atomic states in ad­di­tion to the ground states. In terms of the above four func­tions, the ex­pres­sion for the hy­dro­gen mol­e­cule ground state is

$$\begin{eqnarray*} && \phantom{{} + {}} \frac{a}{\sqrt2} \pp1/{\skew0\vec r}_1... ...a}{\sqrt2} \pp4/{\skew0\vec r}_1//z1/ \pp1/{\skew0\vec r}_2//z2/ \end{eqnarray*}$$

The is­sue in this sec­tion is that the above hy­dro­gen ground state is just one spe­cial case of the most gen­eral wave func­tion for the two par­ti­cles that can be formed from four sin­gle-par­ti­cle states:

$$\begin{eqnarray*} \lefteqn{\Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2}... ... + a_{44} \pp4/{\skew0\vec r}_1//z1/ \pp4/{\skew0\vec r}_2//z2/ \end{eqnarray*}$$

This can be writ­ten much more con­cisely us­ing sum­ma­tion in­dices as

$$\Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t) = ... ... \pp{n_1}/{\skew0\vec r}_1//z1/ \pp{n_2}/{\skew0\vec r}_2//z2/$$

How­ever, the in­di­vid­ual terms will be fully writ­ten out for now to re­duce the math­e­mat­i­cal ab­strac­tion. The in­di­vid­ual terms are some­times called “Hartree prod­ucts.”

The an­ti­sym­metriza­tion re­quire­ment says that the wave func­tion must be an­ti­sym­met­ric un­der ex­change of the two elec­trons. More con­cretely, it must turn into its neg­a­tive when the ar­gu­ments ${\skew0\vec r}_1,S_{z1}$ and ${\skew0\vec r}_2,S_{z2}$ are swapped. To un­der­stand what that means, the var­i­ous terms need to be arranged in groups:

$$\begin{array}{rl} {\rm I}: & \quad a_{11} \pp1/{\skew0\ve... ...p4/{\skew0\vec r}_1//z1/\pp3/{\skew0\vec r}_2//z2/ \end{array}$$

Within each group, all terms in­volve the same com­bi­na­tion of func­tions, but in a dif­fer­ent or­der. Dif­fer­ent groups have a dif­fer­ent com­bi­na­tion of func­tions.

Now if the elec­trons are ex­changed, it turns the terms in groups I through IV back into them­selves. Since the wave func­tion must change sign in the ex­change, and some­thing can only be its own neg­a­tive if it is zero, the an­ti­sym­metriza­tion re­quire­ment re­quires that the co­ef­fi­cients $a_{11}$, $a_{22}$, $a_{33}$, and $a_{44}$ must all be zero. Four co­ef­fi­cients have been elim­i­nated from the list of un­known quan­ti­ties.

Fur­ther, in each of the groups V through X with two dif­fer­ent states, ex­change of the two elec­trons turn the terms into each other, ex­cept for their co­ef­fi­cients. If that is to achieve a change of sign, the co­ef­fi­cients must be each other’s neg­a­tives; $a_{21}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-a_{12}$, $a_{31}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-a_{13}$, ... So only six co­ef­fi­cients $a_{12}$, $a_{13}$, ...still need to be found from other phys­i­cal re­quire­ments, such as en­ergy min­i­miza­tion for a ground state. Less than half of the orig­i­nal six­teen un­knowns sur­vive the an­ti­sym­metriza­tion re­quire­ment, sig­nif­i­cantly re­duc­ing the prob­lem size.

There is a very neat way of writ­ing the an­ti­sym­metrized wave func­tion of sys­tems of fermi­ons, which is es­pe­cially con­ve­nient for larger num­bers of par­ti­cles. It is done us­ing de­ter­mi­nants. The an­ti­sym­met­ric wave func­tion for the above ex­am­ple is:

$$\begin{eqnarray*} \Psi & = & a_{12} \left\vert \begin{array}{cc} \pp1/{\ske... ...ec r}_2//z2/&\pp4/{\skew0\vec r}_2//z2/ \end{array} \right\vert \end{eqnarray*}$$

These de­ter­mi­nants are called “Slater de­ter­mi­nants”.

To find the ac­tual hy­dro­gen mol­e­cule ground state from the above ex­pres­sion, ad­di­tional phys­i­cal re­quire­ments have to be im­posed. For ex­am­ple, the co­ef­fi­cients $a_{12}$ and $a_{34}$ can rea­son­ably be ig­nored for the ground state, be­cause ac­cord­ing to the given de­f­i­n­i­tion of the states, their Slater de­ter­mi­nants have the elec­trons around the same nu­cleus, and that pro­duces el­e­vated en­ergy due to the mu­tual re­pul­sion of the elec­trons. Also, fol­low­ing the ar­gu­ments of sec­tion 5.2, the co­ef­fi­cients $a_{13}$ and $a_{24}$ must be zero since their Slater de­ter­mi­nants pro­duce the ex­cited an­ti­sym­met­ric spa­tial state $\psi_{\rm {l}}\psi_{\rm {r}}-\psi_{\rm {r}}\psi_{\rm {l}}$ times the ${\uparrow}{\uparrow}$, re­spec­tively ${\downarrow}{\downarrow}$ spin states. Fi­nally, the co­ef­fi­cients $a_{14}$ and $a_{23}$ must be op­po­site in or­der that their Slater de­ter­mi­nants com­bine into the low­est-en­ergy sym­met­ric spa­tial state $\psi_{\rm {l}}\psi_{\rm {r}}+\psi_{\rm {r}}\psi_{\rm {l}}$ times the ${\uparrow}{\downarrow}$ and ${\downarrow}{\uparrow}$ spin states. That leaves the sin­gle co­ef­fi­cient $a_{14}$ that can be found from the nor­mal­iza­tion re­quire­ment, tak­ing it real and pos­i­tive for con­ve­nience.

But the is­sue in this sec­tion is what the sym­metriza­tion re­quire­ments say about wave func­tions in gen­eral, whether they are some ground state or not. And for four sin­gle-par­ti­cle states for two iden­ti­cal fermi­ons, the con­clu­sion is that the wave func­tion must be some com­bi­na­tion of the six Slater de­ter­mi­nants, re­gard­less of what other physics may be rel­e­vant.

The next ques­tion is how that con­clu­sion changes if the two par­ti­cles in­volved are not fermi­ons, but iden­ti­cal bosons. The sym­metriza­tion re­quire­ment is then that ex­chang­ing the par­ti­cles must leave the wave func­tion un­changed. Since the terms in groups I through IV do re­main the same un­der par­ti­cle ex­change, their co­ef­fi­cients $a_{11}$ through $a_{44}$ can have any nonzero value. This is the sense in which the an­ti­sym­metriza­tion re­quire­ment for fermi­ons is much more re­stric­tive than the one for bosons: groups in­volv­ing a du­pli­cated state must be zero for fermi­ons, but not for bosons.

In groups V through X, where par­ti­cle ex­change turns each of the two terms into the other one, the co­ef­fi­cients must now be equal in­stead of neg­a­tives; $a_{21}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_{12}$, $a_{31}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $a_{13}$, ... That elim­i­nates six co­ef­fi­cients from the orig­i­nal six­teen un­knowns, leav­ing ten co­ef­fi­cients that must be de­ter­mined by other phys­i­cal re­quire­ments on the wave func­tion.

(The equiv­a­lent of Slater de­ter­mi­nants for bosons are “per­ma­nents,” ba­si­cally de­ter­mi­nants with all mi­nus signs in their de­f­i­n­i­tion re­placed by plus signs. Un­for­tu­nately, many of the help­ful prop­er­ties of de­ter­mi­nants do not ap­ply to per­ma­nents.)

All of the above ar­gu­ments can be ex­tended to the gen­eral case that $N$, in­stead of 4, sin­gle-par­ti­cle func­tions $\pp1/{\skew0\vec r}//z/$, $\pp2/{\skew0\vec r}//z/$, ..., $\pp{N}/{\skew0\vec r}//z/$ are used to de­scribe $I$, in­stead of 2, par­ti­cles. Then the most gen­eral pos­si­ble wave func­tion as­sumes the form:

$$\Psi = \sum_{n_1=1}^N \sum_{n_2=1}^N \ldots \sum_{n_I=1}^N ... ...{\skew0\vec r}_2//z2/ \ldots \pp{n_I}/{\skew0\vec r}_I//zI/ %$$ (5.30)

where the $a_{n_1n_2\ldots{n}_I}$ are nu­mer­i­cal co­ef­fi­cients that are to be cho­sen to sat­isfy the phys­i­cal con­straints on the wave func­tion, in­clud­ing the (anti) sym­metriza­tion re­quire­ments.

This sum­ma­tion is again the every pos­si­ble com­bi­na­tion idea of com­bin­ing every pos­si­ble state for par­ti­cle 1 with every pos­si­ble state for par­ti­cle 2, etcetera. So the to­tal sum above con­tains $N^I$ terms: there are $N$ pos­si­bil­i­ties for the func­tion num­ber $n_1$ of par­ti­cle 1, times $N$ pos­si­bil­i­ties for the func­tion num­ber $n_2$ of par­ti­cle 2, ... In gen­eral then, a cor­re­spond­ing to­tal of $N^I$ un­known co­ef­fi­cients $a_{n_1n_2\ldots{n}_I}$ must be de­ter­mined to find out the pre­cise wave func­tion.

But for iden­ti­cal par­ti­cles, the num­ber that must be de­ter­mined is much less. That num­ber can again be de­ter­mined by di­vid­ing the terms into groups in which the terms all in­volve the same com­bi­na­tion of $I$ sin­gle-par­ti­cle func­tions, just in a dif­fer­ent or­der. The sim­plest groups are those that in­volve just a sin­gle sin­gle-par­ti­cle func­tion, gen­er­al­iz­ing the groups I through IV in the ear­lier ex­am­ple. Such groups con­sist of only a sin­gle term; for ex­am­ple, the group that only in­volves $\pp1////$ con­sists of the sin­gle term

$$a_{11\ldots1}\pp1/{\skew0\vec r}_1//z1/\pp1/{\skew0\vec r}_2//z2/\ldots\pp1/{\skew0\vec r}_I//zI/.$$

At the other ex­treme, groups in which every sin­gle-par­ti­cle func­tion is dif­fer­ent have as many as $I!$ terms, since $I!$ is the num­ber of ways that $I$ dif­fer­ent items can be or­dered. In the ear­lier ex­am­ple, that were groups V through X, each hav­ing 2! = 2 terms. If there are more than two par­ti­cles, there will also be groups in which some states are the same and some are dif­fer­ent.

For iden­ti­cal bosons, the sym­metriza­tion re­quire­ment says that all the co­ef­fi­cients within a group must be equal. Any term in a group can be turned into any other by par­ti­cle ex­changes; so, if they would not all have the same co­ef­fi­cients, the wave func­tion could be changed by par­ti­cle ex­changes. As a re­sult, for iden­ti­cal bosons the num­ber of un­known co­ef­fi­cients re­duces to the num­ber of groups.

For iden­ti­cal fermi­ons, only groups in which all sin­gle-par­ti­cle func­tions are dif­fer­ent can be nonzero. That fol­lows be­cause if a term has a du­pli­cated sin­gle-par­ti­cle func­tion, it turns into it­self with­out the re­quired sign change un­der an ex­change of the par­ti­cles of the du­pli­cated func­tion.

So there is no way to de­scribe a sys­tem of $I$ iden­ti­cal fermi­ons with any­thing less than $I$ dif­fer­ent sin­gle-par­ti­cle func­tions $\pp{n}////$. This crit­i­cally im­por­tant ob­ser­va­tion is known as the “Pauli ex­clu­sion prin­ci­ple:” $I-1$ fermi­ons oc­cu­py­ing $I-1$ sin­gle-par­ti­cle func­tions ex­clude a $I$-​th fermion from sim­ply en­ter­ing the same $I-1$ func­tions; a new func­tion must be added to the mix for each ad­di­tional fermion. The more iden­ti­cal fermi­ons there are in a sys­tem, the more dif­fer­ent sin­gle-par­ti­cle func­tions are re­quired to de­scribe it.

Each group in­volv­ing $I$ dif­fer­ent sin­gle-par­ti­cle func­tions $\pp{n_1}////$, $\pp{n_2}////$, ...$\pp{n_I}////$ re­duces un­der the an­ti­sym­metriza­tion re­quire­ment to a sin­gle Slater de­ter­mi­nant of the form

$$\frac{1}{\sqrt{I!}} \left\vert \begin{array}{ccccc} \pp{... ...ots & \pp{n_I}/{\skew0\vec r}_I//zI/ \end{array} \right\vert$$ (5.31)

mul­ti­plied by a sin­gle un­known co­ef­fi­cient. The nor­mal­iza­tion fac­tor 1$\raisebox{.5pt}{$/$}$​$\sqrt{I!}$ has been thrown in merely to en­sure that if the func­tions $\pp{n}////$ are or­tho­nor­mal, then so are the Slater de­ter­mi­nants. Us­ing Slater de­ter­mi­nants en­sures the re­quired sign changes of fermion sys­tems au­to­mat­i­cally, be­cause de­ter­mi­nants change sign if two rows are ex­changed.

In the case that the bare min­i­mum of $I$ func­tions is used to de­scribe $I$ iden­ti­cal fermi­ons, only one Slater de­ter­mi­nant can be formed. Then the an­ti­sym­metriza­tion re­quire­ment re­duces the $I^I$ un­known co­ef­fi­cients $a_{n_1n_2\ldots{n}_I}$ to just one, $a_{12\ldots{I}}$; ob­vi­ously a tremen­dous re­duc­tion.

At the other ex­treme, when the num­ber of func­tions $N$ is very large, much larger than $I^2$ to be pre­cise, most terms have all in­dices dif­fer­ent and the re­duc­tion is only from $N^I$ to about $N^I$$\raisebox{.5pt}{$/$}$​$I!$ terms. The lat­ter would also be true for iden­ti­cal bosons.

The func­tions bet­ter be cho­sen to pro­duce a good ap­prox­i­ma­tion to the wave func­tion with a small num­ber of terms. As an ar­bi­trary ex­am­ple to fo­cus the thoughts, if $N$ $\vphantom0\raisebox{1.5pt}{$=$}$ 100 func­tions are used to de­scribe an ar­senic atom, with $I$ $\vphantom0\raisebox{1.5pt}{$=$}$ 33 elec­trons, there would be a pro­hib­i­tive 10$\POW9,{66}$ terms in the sum (5.30). Even af­ter re­duc­tion to Slater de­ter­mi­nants, there would still be a pro­hib­i­tive 3 10$\POW9,{26}$ or so un­known co­ef­fi­cients left. The pre­cise ex­pres­sion for the num­ber of Slater de­ter­mi­nants is called “$N$ choose $I$;” it is given by

$$\left(\begin{array}{c}N\ I\end{array}\right) = \frac{N!}{(N-I)!I!} = \frac{N(N-1)(N-2)\ldots(N-I+1)}{I!},$$

since the top gives the to­tal num­ber of terms that have all func­tions dif­fer­ent, ($N$ pos­si­ble func­tions for par­ti­cle 1, times $N-1$ pos­si­ble func­tions left for par­ti­cle 2, etcetera,) and the bot­tom re­flects that it takes $I!$ of them to form a sin­gle Slater de­ter­mi­nant. {D.25}.

The ba­sic Hartree-Fock ap­proach, dis­cussed in chap­ter 9.3, goes to the ex­treme in re­duc­ing the num­ber of func­tions: it uses the very min­i­mum of $I$ sin­gle-par­ti­cle func­tions. How­ever, rather than choos­ing these func­tions a pri­ori, they are ad­justed to give the best ap­prox­i­ma­tion that is pos­si­ble with a sin­gle Slater de­ter­mi­nant. Un­for­tu­nately, if a sin­gle de­ter­mi­nant still turns out to be not ac­cu­rate enough, adding a few more func­tions quickly blows up in your face. Adding just one more func­tion gives $I$ more de­ter­mi­nants; adding an­other func­tion gives an­other $I(I+1)$$\raisebox{.5pt}{$/$}$​2 more de­ter­mi­nants, etcetera.

Key Points

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Wave func­tions for mul­ti­ple-par­ti­cle sys­tems can be formed us­ing sums of prod­ucts of sin­gle-par­ti­cle wave func­tions.

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The co­ef­fi­cients of these prod­ucts are con­strained by the sym­metriza­tion re­quire­ments.

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In par­tic­u­lar, for iden­ti­cal fermi­ons such as elec­trons, the sin­gle-par­ti­cle wave func­tions must com­bine into Slater de­ter­mi­nants.

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Sys­tems of iden­ti­cal fermi­ons re­quire at least as many sin­gle-par­ti­cle states as there are par­ti­cles. This is known as the Pauli ex­clu­sion prin­ci­ple.

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If more sin­gle-par­ti­cle states are used to de­scribe a sys­tem, the prob­lem size in­creases rapidly.

5.7 Re­view Ques­tions

1.

How many sin­gle-par­ti­cle states would a ba­sic Hartree-Fock ap­prox­i­ma­tion use to com­pute the elec­tron struc­ture of an ar­senic atom? How many Slater de­ter­mi­nants would that in­volve?

So­lu­tion symways-a

2.

If two more sin­gle-par­ti­cle states would be used to im­prove the ac­cu­racy for the ar­senic atom, (one more nor­mally does not help), how many Slater de­ter­mi­nants could be formed with those states?

So­lu­tion symways-b