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

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

and consists of four single-particle functions:

$$\begin{eqnarray*} \pp1/{\skew0\vec r}//z/ = \psi_{\rm {l}}({\skew0\vec r}){\up... ...0\vec r}//z/ = \psi_{\rm {r}}({\skew0\vec r}){\downarrow}(S_z). \end{eqnarray*}$$

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

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

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:

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

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:

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

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:

$$\Psi = \sum_{n_1=1}^N \sum_{n_2=1}^N \ldots \sum_{n_I=1}^N... ...kew0\vec r}_2//z2/ \ldots \pp{n_I}/{\skew0\vec r}_I//zI/ %$$ (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

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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