*Michael Fowler, University of Virginia*

So far, we have used Schrödinger’s equation to see how a single particle, usually an electron, behaves in a variety of potentials. If we are going to think about atoms other than hydrogen, it is necessary to extend the Schrödinger equation so that it describes more than one particle.

As a simple example of a two particle system, let us consider two electrons confined to the same one-dimensional infinite square well,

$\begin{array}{c}V\left(x\right)=\infty ,\text{\hspace{1em}}x<-L/2,\\ V\left(x\right)=0,\text{\hspace{1em}}-L/2\le x\le L/2,\\ V\left(x\right)=\infty ,\text{\hspace{1em}}L/2<x.\end{array}$

To make things even simpler, let us assume that the electrons do not interact with each other$\u2014$we switch off their electrostatic repulsion. Then by analogy with our construction of the Schrödinger equation for a single electron, we write

$$-\frac{{\hslash}^{2}}{2m}\frac{{\partial}^{2}\psi \left({x}_{1},{x}_{2},t\right)}{\partial {x}_{1}^{2}}-\frac{{\hslash}^{2}}{2m}\frac{{\partial}^{2}\psi \left({x}_{1},{x}_{2},t\right)}{\partial {x}_{2}^{2}}=E\psi \left({x}_{1},{x}_{2},t\right)$$

inside the well, with the wave function going to zero for ${x}_{1}$ or ${x}_{2}$ equal to $\pm L/2.$

On looking at this equation, we see it is *the same* as the Schrödinger equation for
a *single* electron in a *two* dimensional square well, and so can
be solved in the same way, by separation of variables. For example, the wave function we plotted for
the two dimensional rectangular well is in the square case:

${\psi}_{\left(2,3\right)}\left({x}_{1},{x}_{2},t\right)=A\mathrm{sin}\left(2\pi {x}_{1}/L\right)\mathrm{cos}\left(3\pi {x}_{2}/L\right){e}^{-iEt/\hslash}$

has the same energy

$$E=\frac{{\hslash}^{2}}{2m}\frac{{\pi}^{2}}{{L}^{2}}\left({2}^{2}+{3}^{2}\right)$$

as the physically distinct wavefunction:

${\psi}_{\left(3,2\right)}\left({x}_{1},{x}_{2},t\right)=A\mathrm{sin}\left(2\pi {x}_{2}/L\right)\mathrm{cos}\left(3\pi {x}_{1}/L\right){e}^{-iEt/\hslash}.$

We have already discussed how the above wavefunctions are to be interpreted if they are regarded as two-dimensional wavefunctions for a single electron: ${\left|\psi \left({x}_{1},{x}_{2}\right)\right|}^{2}d{x}_{1}d{x}_{2}$ is the probability of finding the electron in a small area $d{x}_{1}d{x}_{2}$ at $\left({x}_{1},\text{\hspace{0.17em}}{x}_{2}\right).$

The natural extension of this interpretation to two
electrons is to assume now ${\left|\psi \left({x}_{1},{x}_{2}\right)\right|}^{2}d{x}_{1}d{x}_{2}$ is the*
joint* probability of finding electron 1 in a small line length $d{x}_{1}$ at
position ${x}_{1},$ and at the same time finding electron 2 in a
small length $d{x}_{2}$ at ${x}_{2}.$

This, however, leads to a real problem. Consider the wavefunction

${\psi}_{\left(2,3\right)}\left({x}_{1},{x}_{2},t\right)=A\mathrm{sin}\left(2\pi {x}_{1}/L\right)\mathrm{cos}\left(3\pi {x}_{2}/L\right){e}^{-iEt/\hslash}$

again. Let us now take specific points, ${x}_{1}=0,\text{\hspace{0.17em}}{x}_{}=L/4.$ Then
the probability of finding electron 1 in an infinitesimal interval at ${x}_{1}$ and electron 2 similarly at ${x}_{2}$ is zero, because $\psi $ is zero at ${x}_{1}=0.$ On the
other hand, the probability of finding electron 2 at ${x}_{1}\text{\hspace{0.17em}}\left(=0\right)$ and
electron 1 at ${x}_{2}\text{\hspace{0.17em}}\left(=L/4\right)$ is* not*
zero.

The problem is this: the electrons are *identical* (we assume their spins point the same way). We can’t tell which is which, and nobody else
can either. The indistinguishability of
elementary particles is not like that of apparently identical macroscopic
objects, where one could always place some tiny mark. There is no way to mark an electron. *This
means, though, that the best we can do is to talk about the probability of
finding one electron at **${x}_{1}$** and
another at **${x}_{2},$** we cannot specify which electron we find
where. *Therefore, any alleged wave
function that gives different probabilities for finding electron 1 at ${x}_{1},$ 2 at ${x}_{2}$ and finding 2 at ${x}_{1},$ 1 at ${x}_{2}$ is not physically meaningful.

That is to say, a wave function describing two identical particles must have a symmetric probability distribution

${\left|\psi \left({x}_{1},{x}_{2}\right)\right|}^{2}={\left|\psi \left({x}_{2},{x}_{1}\right)\right|}^{2}.$

This is definitely not the case with our function ${\psi}_{\left(2,3\right)}\left({x}_{1},{x}_{2}\right),$, so although it is a solution to the two particle Schrödinger equation, it is not a physically meaningful wave function for two particles in a one-dimensional box.

In fact, this is not difficult to fix$\u2014$recall that the function ${\psi}_{\left(3,2\right)}\left({x}_{1},{x}_{2}\right)$ has the same energy, and in fact just corresponds to the two particles being switched around, that is to say, ${\psi}_{\left(3,2\right)}\left({x}_{1},{x}_{2}\right)={\psi}_{\left(2,3\right)}\left({x}_{2},{x}_{1}\right).$

It follows that the symmetric function

${\psi}_{\left(2,3\right)}^{S}\left({x}_{1},{x}_{2}\right)=\frac{1}{\sqrt{2}}\left({\psi}_{\left(2,3\right)}\left({x}_{1},{x}_{2}\right)+{\psi}_{\left(2,3\right)}\left({x}_{2},{x}_{1}\right)\right)$

and the antisymmetric function

${\psi}_{\left(2,3\right)}^{S}\left({x}_{1},{x}_{2}\right)=\frac{1}{\sqrt{2}}\left({\psi}_{\left(2,3\right)}\left({x}_{1},{x}_{2}\right)-{\psi}_{\left(2,3\right)}\left({x}_{2},{x}_{1}\right)\right)$

are both solutions to Schrödinger’s equation for the energy $E,$ and both satisfy the requirement ${\left|\psi \left({x}_{1},{x}_{2}\right)\right|}^{2}={\left|\psi \left({x}_{2},{x}_{1}\right)\right|}^{2}$ necessary for identical particles, so these are the appropriate candidate wave functions for the two particles in the one-dimensional box.

It turns out that both symmetric and antisymmetric
wavefunctions arise in nature in describing identical particles. In fact, all
elementary particles are either *fermions*,
which have antisymmetric multiparticle wavefunctions, or *bosons*, which have symmetric wave functions. Electrons, protons and neutrons are fermions;
photons, $\alpha $ -particles and helium atoms are bosons.

It is important to realize that this requirement of symmetry of the probability distribution, arising from the true indistinguishability of the particles, has a large effect on the probability distribution, and, furthermore, the effect is very different for fermions and bosons. The simplest way to see this is just to plot ${\left|\psi \left({x}_{1},{x}_{2}\right)\right|}^{2}$ for the symmetric and for the antisymmetric wave functions, and compare them.

For the *symmetric *case,
the probability distribution (unnormalized) looks like:

(The graytone figure is a contour map.)

The main point to note here is that the highest joint probability is that of finding the two particles close together, near (1,1) and (4,4). (These actual points depend on the wave function we chose initially, of course).

Now look at the *antisymmetric*
case:

The difference is dramatic.
For the antisymmetric wave function, the particles are most likely to be
found *far away* from each other. In fact, there is *zero* probability that they will be found at the same spot, because
if $\psi \left({x}_{1},{x}_{2}\right)=-\psi \left({x}_{2},{x}_{1}\right),$ obviously $\psi \left(x,x\right)=0.$ A more general statement is that *two fermions cannot be in the same quantum
state*, because if they were, the wave function would be of the symmetric
form $f\left({x}_{1}\right)f\left({x}_{2}\right),$ and
could not be antisymmetrized.

This is the **Pauli Exclusion Principle**$\u2014$it is the
basis of the periodic table, and consequently of almost everything else.

We should perhaps emphasize that these wave functions were
calculated with the electrostatic repulsion between the electrons switched off. That is*
not* what is keeping the electrons apart, although it will increase the
separation if it is included. We should
also warn against a simple classical picture of the Pauli principle, the
thought that two things can’t be in the same place, after all, so perhaps it’s
no surprise two electrons can’t be in the same state. Two electrons *can* be in the same identical *space* wave function provided that their
spins point in opposite ways.
Furthermore, two bosons can be in the same state, and although that is
perhaps reasonable sounding for photons, it is equally true for heavy atoms. In
Bose-Einstein Condensation, a large number of atoms occupy the same quantum
state. This happens in liquid Helium^{4}
below about two kelvin, it becomes a superfluid and flows without
friction. Furthermore, BE condensation of
laser cooled collections of large atoms in a trap is now fairly routine.