5.1 Wave Function for Multiple Particles

While a single particle is described by a wave function $\Psi({\skew0\vec r};t)$, a system of two particles, call them 1 and 2, is described by a wave function

\begin{displaymath}
\Psi({\skew0\vec r}_1,{\skew0\vec r}_2;t)
\end{displaymath} (5.1)

depending on both particle positions. The value of $\vert\Psi({\skew0\vec r}_1,{\skew0\vec r}_2;t)\vert^2{\,\rm d}^3{\skew0\vec r}_1{\,\rm d}^3{\skew0\vec r}_2$ gives the probability of simultaneously finding particle 1 within a vicinity ${\,\rm d}^3{\skew0\vec r}_1$ of ${\skew0\vec r}_1$ and particle 2 within a vicinity ${\,\rm d}^3{\skew0\vec r}_2$ of ${\skew0\vec r}_2$.

The wave function must be normalized to express that the electrons must be somewhere:

\begin{displaymath}
\langle \Psi \vert \Psi \rangle_6 =
\mathop{\int\kern-7p...
...t^2{\,\rm d}^3 {\skew0\vec r}_1{\rm d}^3 {\skew0\vec r}_2 = 1
\end{displaymath} (5.2)

where the subscript 6 of the inner product is just a reminder that the integration is over all six scalar position coordinates of $\Psi$.

The underlying idea of increasing system size is “every possible combination:” allow for every possible combination of state for particle 1 and state for particle 2. For example, in one dimension, all possible $x$ positions of particle 1 geometrically form an $x_1$-​axis. Similarly all possible $x$ positions of particle 2 form an $x_2$-​axis. If every possible position $x_1$ is separately combined with every possible position $x_2$, the result is an $x_1,x_2$-​plane of possible positions of the combined system.

Similarly, in three dimensions the three-di­men­sion­al space of positions ${\skew0\vec r}_1$ combines with the three-di­men­sion­al space of positions ${\skew0\vec r}_2$ into a six-di­men­sion­al space having all possible combinations of values for ${\skew0\vec r}_1$ with all possible values for ${\skew0\vec r}_2$.

The increase in the number of dimensions when the system size increases is a major practical problem for quantum mechanics. For example, a single arsenic atom has 33 electrons, and each electron has 3 position coordinates. It follows that the wave function is a function of 99 scalar variables. (Not even counting the nucleus, spin, etcetera.) In a brute-force numerical solution of the wave function, maybe you could restrict each position coordinate to only ten computational values, if no very high accuracy is desired. Even then, $\Psi$ values at 10$\POW9,{99}$ different combined positions must be stored, requiring maybe 10$\POW9,{91}$ Gigabytes of storage. To do a single multiplication on each of those those numbers within a few years would require a computer with a speed of 10$\POW9,{82}$ gigaflops. No need to take any of that arsenic to be long dead before an answer is obtained. (Imagine what it would take to compute a microgram of arsenic instead of an atom.) Obviously, more clever numerical procedures are needed.

Sometimes the problem size can be reduced. In particular, the problem for a two-particle system like the proton-electron hydrogen atom can be reduced to that of a single particle using the concept of reduced mass. That is shown in addendum {A.5}.


Key Points
$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
To describe multiple-particle systems, just keep adding more independent variables to the wave function.

$\begin{picture}(15,5.5)(0,-3)
\put(2,0){\makebox(0,0){\scriptsize\bf0}}
\put(12...
...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}}
\end{picture}$
Unfortunately, this makes many-particle problems impossible to solve by brute force.

5.1 Review Questions
1.

A simple form that a six-di­men­sion­al wave function can take is a product of two three-di­men­sion­al ones, as in $\psi({\skew0\vec r}_1,{\skew0\vec r}_2)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\psi_1({\skew0\vec r}_1)\psi_2({\skew0\vec r}_2)$. Show that if $\psi_1$ and $\psi_2$ are normalized, then so is $\psi$.

Solution complex-a

2.

Show that for a simple product wave function as in the previous question, the relative probabilities of finding particle 1 near a position ${\skew0\vec r}_a$ versus finding it near another position ${\skew0\vec r}_b$ is the same regardless where particle 2 is. (Or rather, where particle 2 is likely to be found.)

Note: This is the reason that a simple product wave function is called uncorrelated. For particles that interact with each other, an uncorrelated wave function is often not a good approximation. For example, two electrons repel each other. All else being the same, the electrons would rather be at positions where the other electron is nowhere close. As a result, it really makes a difference for electron 1 where electron 2 is likely to be and vice-versa. To handle such situations, usually sums of product wave functions are used. However, for some cases, like for the helium atom, a single product wave function is a perfectly acceptable first approximation. Real-life electrons are crowded together around attracting nuclei and learn to live with each other.

Solution complex-b