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
An example is provided by the approximate ground state of the hydrogen
molecule from the previous section,
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:
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
and are swapped. To understand what
that means, the various terms need to be arranged in groups:
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 , , , and 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; , , ... So only six coefficients , , ...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:
To find the actual hydrogen molecule ground state from the above expression, additional physical requirements have to be imposed. For example, the coefficients and 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 and must be zero since their Slater determinants produce the excited antisymmetric spatial state times the , respectively spin states. Finally, the coefficients and must be opposite in order that their Slater determinants combine into the lowest-energy symmetric spatial state times the and spin states. That leaves the single coefficient 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 through 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; , , ... 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
, instead of 4, single-particle functions
, , ...,
are used to describe , instead of 2, particles. Then the
most general possible wave function assumes the form:
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 terms: there are possibilities for the function
number of particle 1, times possibilities for the function
number of particle 2, ... In general then, a corresponding
total of unknown coefficients 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
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 consists of the single term
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 identical fermions with anything less than different single-particle functions . This critically important observation is known as the “Pauli exclusion principle:” fermions occupying single-particle functions exclude a -th fermion from simply entering the same 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 different single-particle functions
, , ... reduces
under the antisymmetrization requirement to a single Slater
determinant of the form
In the case that the bare minimum of functions is used to describe identical fermions, only one Slater determinant can be formed. Then the antisymmetrization requirement reduces the unknown coefficients to just one, ; obviously a tremendous reduction.
At the other extreme, when the number of functions is very large,
much larger than to be precise, most terms have all indices
different and the reduction is
only from to
about terms. The latter would also be true for identical
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 100 functions are used to describe
an arsenic atom, with 33 electrons, there would be a
prohibitive 10 terms in the sum (5.30). Even
after reduction to Slater determinants, there would still be a
prohibitive 3 10 or so unknown coefficients left. The
precise expression for the number of Slater determinants is called
choose ; it is given by
Hartree-Fock approach, discussed in chapter
9.3, goes to the extreme in reducing the number of functions:
it uses the very minimum of 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 more
determinants; adding another function gives another 2 more
- Wave functions for multiple-particle systems can be formed using sums of products of single-particle wave functions.
- The coefficients of these products are constrained by the symmetrization requirements.
- In particular, for identical fermions such as electrons, the single-particle wave functions must combine into Slater determinants.
- Systems of identical fermions require at least as many single-particle states as there are particles. This is known as the Pauli exclusion principle.
- If more single-particle states are used to describe a system, the problem size increases rapidly.
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?
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?