Suppose we have a system of several identical particles. In classical mechanics the identity of the particles leads to no special consequences. For example, consider identical billiard balls rolling on a billiard table. We can follow the motion of any individual ball, say by taking a motion picture of the system. We can say that ball number one is moving along a certain path, ball two is on another definite path, and so on, the paths being determined by Newton's laws of motion. Thus, although the balls are identical, we can distinguish among them by specifying the path each takes. The identity of the balls has no special effect on their motions.
In quantum mechanics the uncertainty principle tells us that we cannot follow the exact path taken by a microscopic "particle." If the microscopic particles of the system all have different masses or charges or spins, we can use one of these properties to distinguish the particles from one another. But if they are all identical, then the one way we had in classical mechanics of distinguishing them, namely by specifying their paths, is lost in quantum mechanics because of the uncertainty principle. Therefore, the wave function of a system of interacting identical particles must not distinguish among the particles. For example, in the perturbation treatment of the helium-atom excited states in Chapter 9, we saw that the function $1 s(1) 2 s(2)$, which says that electron 1 is in the $1 s$ orbital and electron 2 is in the $2 s$ orbital, was not a correct zeroth-order wave function.
Rather, we had to use the functions $2^{-1 / 2}[1 s(1) 2 s(2) \pm 1 s(2) 2 s(1)]$, which do not specify which electron is in which orbital. (If the identical particles are well separated from one another so that their wave functions do not overlap, they may be regarded as distinguishable.)
We now derive the restrictions on the wave function due to the requirement of indistinguishability of identical particles in quantum mechanics. The wave function of a system of $n$ identical microscopic particles depends on the space and spin variables of the particles. For particle 1 , these variables are $x{1}, y{1}, z{1}, m{s 1}$. Let $q{1}$ stand for all four of these variables. Thus $\psi=\psi\left(q{1}, q{2}, \ldots, q{n}\right)$.
We define the exchange or permutation operator $\hat{P}_{12}$ as the operator that interchanges all the coordinates of particles 1 and 2 :
\(
\begin{equation}
\hat{P}{12} f\left(q{1}, q{2}, q{3}, \ldots, q{n}\right)=f\left(q{2}, q{1}, q{3}, \ldots, q_{n}\right) \tag{10.14}
\end{equation}
\)
For example, the effect of $\hat{P}_{12}$ on the function that has electron 1 in a $1 s$ orbital with spin up and electron 2 in a $3 s$ orbital with spin down is
\(
\begin{equation}
\hat{P}_{12}[1 s(1) \alpha(1) 3 s(2) \beta(2)]=1 s(2) \alpha(2) 3 s(1) \beta(1) \tag{10.15}
\end{equation}
\)
What are the eigenvalues of $\hat{P}{12}$ ? Applying $\hat{P}{12}$ twice has no net effect:
\(
\hat{P}{12} \hat{P}{12} f\left(q{1}, q{2}, \ldots, q{n}\right)=\hat{P}{12} f\left(q{2}, q{1}, \ldots, q{n}\right)=f\left(q{1}, q{2}, \ldots, q{n}\right)
\)
Therefore, $\hat{P}{12}^{2}=\hat{1}$. Let $w{i}$ and $c{i}$ denote the eigenfunctions and eigenvalues of $\hat{P}{12}$. We have $\hat{P}{12} w{i}=c{i} w{i}$. Application of $\hat{P}{12}$ to this equation gives $\hat{P}{12}^{2} w{i}=c{i} P{12} w{i}$. Substitution of $\hat{P}{12}^{2}=\hat{1}$ and $\hat{P}{12} w{i}=c{i} w{i}$ in $\hat{P}{12}^{2} w{i}=c{i} \hat{P}{12} w{i}$ gives $w{i}=c{i}^{2} w{i}$. Since zero is not allowed as an eigenfunction, we can divide by $w{i}$ to get $1=c{i}^{2}$ and $c{i}= \pm 1$. The eigenvalues of $\hat{P}_{12}$ (and of any linear operator whose square is the unit operator) are +1 and -1 .
If $w{+}$is an eigenfunction of $\hat{P}{12}$ with eigenvalue +1 , then
\(
\begin{gathered}
\hat{P}{12} w{+}\left(q{1}, q{2}, \ldots, q{n}\right)=(+1) w{+}\left(q{1}, q{2}, \ldots, q{n}\right) \
w{+}\left(q{2}, q{1}, \ldots, q{n}\right)=w{+}\left(q{1}, q{2}, \ldots, q_{n}\right)
\end{gathered}
\)
(10.16)
A function such as $w_{+}$that has the property (10.16) of being unchanged when particles 1 and 2 are interchanged is said to be symmetric with respect to interchange of particles 1 and 2 . For eigenvalue -1 , we have
\(
\begin{equation}
w{-}\left(q{2}, q{1}, \ldots, q{n}\right)=-w{-}\left(q{1}, q{2}, \ldots, q{n}\right) \tag{10.17}
\end{equation}
\)
The function $w{-}$in (10.17) is antisymmetric with respect to interchange of particles 1 and 2 , meaning that this interchange multiplies $w{-}$by -1 . There is no necessity for an arbitrary function $f\left(q{1}, q{2}, \ldots, q_{n}\right)$ to be either symmetric or antisymmetric with respect to interchange of 1 and 2 .
Do not confuse the property of being symmetric or antisymmetric with respect to particle interchange with the property of being even or odd with respect to inversion in space. The function $x{1}+x{2}$ is symmetric with respect to interchange of 1 and 2 and is an odd function of $x{1}$ and $x{2}$. The function $x{1}^{2}+x{2}^{2}$ is symmetric with respect to interchange of 1 and 2 and is an even function of $x{1}$ and $x{2}$.
The operator $\hat{P}_{i k}$ is defined by
\(
\begin{equation}
\hat{P}{i k} f\left(q{1}, \ldots, q{i}, \ldots, q{k}, \ldots, q{n}\right)=f\left(q{1}, \ldots, q{k}, \ldots, q{i}, \ldots, q_{n}\right) \tag{10.18}
\end{equation}
\)
The eigenvalues of $\hat{P}{i k}$ are, like those of $\hat{P}{12},+1$ and -1 .
We now consider the wave function of a system of $n$ identical microscopic particles. Since the particles are indistinguishable, the way we label them cannot affect the state of the system. Thus the two wave functions
\(
\psi\left(q{1}, \ldots, q{i}, \ldots, q{k}, \ldots, q{n}\right) \quad \text { and } \psi\left(q{1}, \ldots, q{k}, \ldots, q{i}, \ldots, q{n}\right)
\)
must correspond to the same state of the system. Two wave functions that correspond to the same state can differ at most by a multiplicative constant. Hence
\(
\begin{aligned}
\psi\left(q{1}, \ldots, q{k}, \ldots, q{i}, \ldots, q{n}\right) & =c \psi\left(q{1}, \ldots, q{i}, \ldots, q{k}, \ldots, q{n}\right) \
\hat{P}{i k} \psi\left(q{1}, \ldots, q{i}, \ldots, q{k}, \ldots, q{n}\right) & =c \psi\left(q{1}, \ldots, q{i}, \ldots, q{k}, \ldots, q_{n}\right)
\end{aligned}
\)
The last equation states that $\psi$ is an eigenfunction of $\hat{P}{i k}$. But we know that the only possible eigenvalues of $\hat{P}{i k}$ are 1 and -1 . We conclude that the wave function for a system of $n$ identical particles must be symmetric or antisymmetric with respect to interchange of any two of the identical particles, $i$ and $k$. Since the $n$ particles are all identical, we could not have the wave function symmetric with respect to some interchanges and antisymmetric with respect to other interchanges. Thus the wave function of $n$ identical particles must be either symmetric with respect to every possible interchange or antisymmetric with respect to every possible interchange of two particles. (The argument just given is not rigorous. The statement that the wave function of a system of identical particles must be either completely symmetric or completely antisymmetric with respect to interchange of two particles is called the symmetrization postulate.)
We have seen that there are two possible cases for the wave function of a system of identical particles, the symmetric and the antisymmetric cases. Experimental evidence (such as the periodic table of the elements to be discussed later) shows that for electrons only the antisymmetric case occurs. Thus we have an additional postulate of quantum mechanics, which states that the wave function of a system of electrons must be antisymmetric with respect to interchange of any two electrons.
In 1926, Dirac concluded (based on theoretical work and experimental data) that electrons require antisymmetric wave functions and photons require symmetric wave functions. However, Dirac and other physicists erroneously believed in 1926 that all material particles required antisymmetric wave functions. In 1930, experimental data indicated that $\alpha$ particles (which have $s=0$ ) require symmetric wave functions; physicists eventually realized that what determines whether a system of identical particles requires symmetric or antisymmetric wave functions is the spin of the particle. Particles with half-integral spin ( $s=\frac{1}{2}, \frac{3}{2}$, and so on) require antisymmetric wave functions, while particles with integral spin ( $s=0,1$, and so on) require symmetric wave functions. In 1940, the physicist Wolfgang Pauli used relativistic quantum field theory to prove this result. Particles requiring antisymmetric wave functions, such as electrons, are called fermions (after E. Fermi), whereas particles requiring symmetric wave functions, such as pions, are called bosons (after S. N. Bose). In nonrelativistic quantum mechanics, we must postulate that the wave function of a system of identical particles must be antisymmetric with respect to interchange of any two particles if the particles have half-integral spin and must be symmetric with respect to interchange if the particles have integral spin. This statement is called the spin-statistics theorem (since the statistical mechanics of a system of bosons differs from that of a system of fermions).
Many proofs of varying validity have been offered for the spin-statistics theorem; see I. Duck and E. C. G. Sudurshan, Pauli and the Spin-Statistics Theorem, World
Scientific, 1997; Am. J. Phys., 66, 284 (1998); Sudurshan and Duck, Pramana-J. Phys., 61, 645 (2003) (available at www.ias.ac.in/pramana/v61/p645/fulltext.pdf). Several experiments have confirmed the validity of the spin-statistics theorem to extremely high accuracy; see G. M. Tino, Fortschr. Phys., 48, 537 (2000) (available at arxiv.org/ abs/quant-ph/9907028).
The spin-statistics theorem has an important consequence for a system of identical fermions. The antisymmetry requirement means that
\(
\begin{equation}
\psi\left(q{1}, q{2}, q{3}, \ldots, q{n}\right)=-\psi\left(q{2}, q{1}, q{3}, \ldots, q{n}\right) \tag{10.19}
\end{equation}
\)
Consider the value of $\psi$ when electrons 1 and 2 have the same coordinates, that is, when $x{1}=x{2}, y{1}=y{2}, z{1}=z{2}$, and $m{s 1}=m{s 2}$. Putting $q{2}=q{1}$ in (10.19), we have
\(
\begin{align}
\psi\left(q{1}, q{1}, q{3}, \ldots, q{n}\right) & =-\psi\left(q{1}, q{1}, q{3}, \ldots, q{n}\right) \
2 \psi & =0 \
\psi\left(q{1}, q{1}, q{3}, \ldots, q{n}\right) & =0 \tag{10.20}
\end{align}
\)
Thus, two electrons with the same spin have zero probability of being found at the same point in three-dimensional space. (By "the same spin," we mean the same value of $m_{s}$ ). Since $\psi$ is a continuous function, Eq. (10.20) means that the probability of finding two electrons with the same spin close to each other in space is quite small. Thus the antisymmetry requirement forces electrons of like spin to keep apart from one another. To describe this, one often speaks of a Pauli repulsion between such electrons. This "repulsion" is not a real physical force, but a reflection of the fact that the electronic wave function must be antisymmetric with respect to exchange.
The requirement for symmetric or antisymmetric wave functions also applies to a system containing two or more identical composite particles. Consider, for example, an ${ }^{16} \mathrm{O}{2}$ molecule. The ${ }^{16} \mathrm{O}$ nucleus has 8 protons and 8 neutrons. Each proton and each neutron has $s=\frac{1}{2}$ and is a fermion. Therefore, interchange of the two ${ }^{16} \mathrm{O}$ nuclei interchanges 16 fermions and must multiply the molecular wave function by $(-1)^{16}=1$. Thus the ${ }^{16} \mathrm{O}{2}$ molecular wave function must be symmetric with respect to interchange of the nuclear coordinates. The requirement for symmetry or antisymmetry with respect to interchange of identical nuclei affects the degeneracy of molecular wave functions and leads to the symmetry number in the rotational partition function [see McQuarrie (2000), pp. 104-105].
For interchange of two identical composite particles containing $m$ identical bosons and $n$ identical fermions, the wave function is multiplied by $(+1)^{m}(-1)^{n}=(-1)^{n}$.
A composite particle is thus a fermion if it contains an odd number of fermions and is a boson otherwise.
When the variational principle (Section 8.1) is used to get approximate electronic wave functions of atoms and molecules, the requirement that the trial variation function be well-behaved includes the requirement that it be antisymmetric.