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\begin{document}

\section{Multiple Particles}

\subsection{Nonidentical Particles}

So far we have dealt exclusively with the quantum mechanics of single
particles.  When multiple particles are involved, we seek a single
state vector that describes all particles in the system.  As an
example, imagine two particles moving along the $x$-axis.  A wave
function describing these two particles would take the form $\psi (
x_1 , x_2 )$, which is the amplitude to find simultaneously particle 1
at $x_1$ {\em and} particle 2 at $x_2$.

Decomposing the state vector in energy space rather than position
space, we could also represent the state of the system by $A ( E_i ,
E_j )$, which is the amplitude for particle 1 to have energy $E_i$ and
particle 2 to have energy $E_j$.

Normalization of the state vector in these two cases would require
\begin{equation}
	\int dx_1 \int dx_2 | \psi (x_1 , x_2 ) |^2 = 1
\label{norm}
\end{equation}
for the space representation and
\begin{equation}
	\sum_{i,j} | A ( E_i , E_j ) |^2 = 1
\end{equation}
in the energy representation.

If the two particles do not interact, then the state vector can be
broken down into a sum of products of individual state vectors.  For
instance, if $\psi_{1i} (x_1 )$ and $\psi_{2j} (x_2 )$ are the energy
eigenvectors for the first and second particles, then the wave
function for the system of two particles can be expressed as
\begin{equation}
	\psi (x_1 ,x_2 ) = \sum_{i,j} A (E_i , E_j ) \psi_{1i} (x_1 )
		\psi_{2j} (x_2 ) .
\end{equation}
The time dependence of each $A$ takes the form $\exp (-iE_i t) \exp
(-iE_j t) = \exp [-i(E_i + E_j )t]$.  If the two particles {\em do}
interact, say, by some force between them, then the wave function
still takes the above form, but the $A$'s have a more complex time
dependence.

Time independent perturbation theory can be used to obtain the
approximate energies of systems of particles which interact via some
weak potential energy, say, $U( x_2 - x_1 )$.  In the case of no
interactions, the unperturbed eigenvectors are simply $\psi^{(0)}_{ij}
(x_1 ,x_2 ) = \psi_{1i} (x_1 ) \psi_{2j} (x_2 )$ and the corresponding
energy eigenvalues are just $E^{(0)}_{ij} = E_i + E_j$.  Perturbation
theory then says that the perturbed energies are given by
\begin{equation}
	E_{ij} = E^{(0)}_{ij} + \int dx_1 \int dx_2
		\psi_{1i} (x_1 )' \psi_{2j} (x_2 )' U( x_2 - x_1 )
		\psi_{1i} (x_1 ) \psi_{2j} (x_2 )
\label{jointe}
\end{equation}
to first order, with corresponding expressions for the corrected wave
functions.

The above scheme shows how to construct the eigenstates for a two
particle system in the case where the particles either interact weakly
or don't interact at all.  First, construct the energy eigenvectors,
$\psi_i (x)$, and eigenvalues, $E_i$, for the individual particles.
Second, construct the combined eigenvectors and eigenvalues,
$\psi_{ij} (x_1 , x_2 ) = \psi_{1i} (x_1 ) \psi_{2j} (x_2 )$ and
$E_{ij} = E_i + E_j$ for each value of $i$ and $j$.  If the particles
don't interact, we are done.  However, if they do, the corrected
energy eigenvalues can be obtained using equation (\ref{jointe}).

\subsection{Identical Particles}

The above development must be modified if the two particles being
considered are identical, e. g., two electrons, two photons, etc.  In
quantum mechanics, two identical particles are truly indistinguishable,
which means that exchanging the two particles cannot change their
joint probability distribution.  This turns out to have far-reaching
consequences.

Suppose that $X$ is an operator that exchanges the particles in a
two particle system.  Then,
\begin{equation}
	X \psi (x_1 , x_2 ) = \psi (x_2 , x_1 ) .
\end{equation}
If we apply $X$ twice to a two particle system, then the system is
returned to its original state, i. e.,
\begin{equation}
	X^2 \psi (x_1 , x_2 ) = \psi (x_1 , x_2 ) .
\end{equation}
Thus, $X^2$ is the identity operator, which has all eigenvalues
equal to unity.  The operator $X$ must therefore have eigenvalues
$\pm 1$.  An eigenstate of $X$ must therefore be either symmetric
or antisymmetric under the exchange of particles.

The Hamiltonian, $H$, shouldn't change if identical particles are
interchanged.  Thus, $[X,H] = 0$, and $X$ and $H$ share eigenvalues.
Furthermore, once a state is either symmetric or antisymmetric to
the interchange of identical particles, it remains that way forever.

It turns out that particles with half-integral spins, i. e., fermions,
are antisymmetric with respect to interchange, while particles with
integral spins, i. e., bosons, are symmetric.  There is a proof of
this buried deep in quantum field theory, but at this level we simply
accept it as an empirical result.

\begin{figure}
\psfig{figure=ident.eps}
\caption{Joint space probability distributions for two particles in a
square well, one in the ground state, the other in the first excited
state.  Left: Non-identical particles.  Middle: Identical fermions.
Right: Identical bosons.}
\label{ident}
\end{figure}

Joint states of two identical particles may be constructed as was
done above if one additional step is taken, namely the enforcement of
the symmetry or antisymmetry of the joint state.  This is easily
done as follows.  We write the joint state of two non-interacting
bosons or fermions as
\begin{equation}
	\psi (x_1, x_2 ) = [ \psi_{1} (x_1 ) \psi_{2} (x_2 ) \pm
		\psi_{1} (x_2 ) \psi_{2} (x_1 ) ]/N ,
\end{equation}
where the plus sign applies to bosons and the minus sign applies to
fermions.  The number $N$ is a normalization factor which enforces
equation (\ref{norm}).  An equivalent construction for energy space is
\begin{equation}
	A_J (E_i , E_j ) = [A_1 (E_i ) A_2 (E_j ) \pm
		 A_1 (E_j ) A_2 (E_i)]/N ,
\label{energy}
\end{equation}
where $A_1 (E)$ and $A_2 (E)$ are the amplitudes for particle 1 and
particle 2 to have energy $E$ and $A_J$ is the joint amplitude.

\subsection{Pauli Exclusion Principle}

A consequence of equation (\ref{energy}) is that non-interacting
identical fermions cannot occupy the same single particle state.  If
the fermions interact via some potential, then the actual states are
modified from the single particle states, but the same result holds
--- no more than a single fermion may occupy a perturbed single
particle state.  This result has enormous consequences, for instance,
in the structure of matter and the nature of the chemical elements.

Analogous behavior occurs in position space.  Figure \ref{ident}
shows $|\psi (x_1 , x_2 )|^2$ for two particles in an infinite square
well, with one particle in the ground state and the other in the first
excited state.  Three different cases are shown, (1) two non-identical
particles, (2) two identical fermions, and (3) two identical bosons.
Notice that the fermions are ``antisocial'', i. e., they try to remain
as far from each other as possible.  In particular, the probability of
finding the two particles in the same location (i. e., where $x_1 =
x_2$) is zero.  If one of the fermions inhabits the left end of the
box, the other is most likely to be found at the right end.

\subsection{Bose-Einstein Condensation}

Figure \ref{ident} shows that, unlike fermions, bosons behave like
``sheep'', in that they like to stick together.  In particular, if one
particle is found at a particular location, the other is most likely
to be found there as well.  This is behavior which is accentuated as
more particles are added to the system.  If many bosons are involved,
a macroscopically observable quantum state may be created in which
multiple bosons move in lockstep.  Superconductivity and superfluidity
in liquid Helium are examples of such behavior.  In the former case,
the bosons are electron pairs in which the total spin adds to zero.
In the later case the bosons are $\mbox{He}^4$ atoms.

\subsection{Entwinement of States}

Figure \ref{ident} graphically indicates that identical particles
cannot be considered individually, but must be treated collectively.
This can even be true for identical particles separated by large
distances.

For free photons the spin is either aligned with the direction of
motion, resulting in right circular polarization, or antialigned,
resulting in left circular polarization.  (Though the photon has spin
one and thus should have a third spin orientation in which the
component of angular momentum along the direction of motion is zero,
this orientation does not occur for free photons.)

Suppose two photons are emitted in opposite directions by a piece of
apparatus as shown in figure \ref{photons} such that the spins of the
photons are not individually known, but are known to be oppositely
aligned, so that the total spin angular momentum is zero.  Since the
photons are moving in opposite directions, they both have either left
circular polarization or right circular polarization.  In this
situation the polarization of each individual photon is not known a
priori, as the state of each photon is not an eigenstate of that
photon's angular momentum.  It is only known that the polarizations
are the same.  However, if the polarization of {\em one} of the
photons is {\em measured}, then the polarization of the {\em other}
photon immediately becomes known.  This is true even if the two photons
are, say, 10 light years apart!

\begin{figure}
\psfig{figure=photons.eps}
\caption{Sketch of two possible joint circular polarization states of
two photons emitted in opposite directions with total spin zero.}
\label{photons}
\end{figure}

The states of the two photons are said to be {\em entwined}.  This
really means that there is a joint state in which particular
polarization or spin amplitudes for different particles are correlated
with each other.  This aspect of quantum mechanics bothered Einstein
so much that he put a great deal of effort into trying to disprove the
theory.  However, experiments have shown that Einstein's ``spooky
action at a distance'', which results from the above-described
entwinement of states, actually occurs.

\end{document}
