---
title: "Pauli exclusion principle"
date: 2021-02-22
categories:
- Quantum mechanics
- Physics
layout: "concept"
---

In quantum mechanics, the **Pauli exclusion principle** is a theorem with
profound consequences for how the world works.

Suppose we have a composite state
$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$, where the two
identical particles $x_1$ and $x_2$ each can occupy the same two allowed
states $a$ and $b$. We then define the permutation operator $\hat{P}$ as
follows:

$$\begin{aligned}
    \hat{P} \Ket{a}\Ket{b} = \Ket{b}\Ket{a}
\end{aligned}$$

That is, it swaps the states of the particles. Obviously, swapping the
states twice simply gives the original configuration again, so:

$$\begin{aligned}
    \hat{P}^2 \Ket{a}\Ket{b} = \Ket{a}\Ket{b}
\end{aligned}$$

Therefore, $\Ket{a}\Ket{b}$ is an eigenvector of $\hat{P}^2$ with
eigenvalue $1$. Since $[\hat{P}, \hat{P}^2] = 0$, $\Ket{a}\Ket{b}$
must also be an eigenket of $\hat{P}$ with eigenvalue $\lambda$,
satisfying $\lambda^2 = 1$, so we know that $\lambda = 1$ or $\lambda = -1$:

$$\begin{aligned}
    \hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b}
\end{aligned}$$

As it turns out, in nature, each class of particle has a single
associated permutation eigenvalue $\lambda$, or in other words: whether
$\lambda$ is $-1$ or $1$ depends on the type of particle that $x_1$
and $x_2$ are. Particles with $\lambda = -1$ are called
**fermions**, and those with $\lambda = 1$ are known as **bosons**. We
define $\hat{P}_f$ with $\lambda = -1$ and $\hat{P}_b$ with
$\lambda = 1$, such that:

$$\begin{aligned}
    \hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b}
    \qquad
    \hat{P}_b \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = \Ket{a}\Ket{b}
\end{aligned}$$

Another fundamental fact of nature is that identical particles cannot be
distinguished by any observation. Therefore it is impossible to tell
apart $\Ket{a}\Ket{b}$ and the permuted state $\Ket{b}\Ket{a}$,
regardless of the eigenvalue $\lambda$. There is no physical difference!

But this does not mean that $\hat{P}$ is useless: despite not having any
observable effect, the resulting difference between fermions and bosons
is absolutely fundamental. Consider the following superposition state,
where $\alpha$ and $\beta$ are unknown:

$$\begin{aligned}
    \Ket{\Psi(a, b)}
    = \alpha \Ket{a}\Ket{b} + \beta \Ket{b}\Ket{a}
\end{aligned}$$

When we apply $\hat{P}$, we can "choose" between two "intepretations" of
its action, both shown below. Obviously, since the left-hand sides are
equal, the right-hand sides must be equal too:

$$\begin{aligned}
    \hat{P} \Ket{\Psi(a, b)}
    &= \lambda \alpha \Ket{a}\Ket{b} + \lambda \beta \Ket{b}\Ket{a}
    \\
    \hat{P} \Ket{\Psi(a, b)}
    &= \alpha \Ket{b}\Ket{a} + \beta \Ket{a}\Ket{b}
\end{aligned}$$

This gives us the equations $\lambda \alpha = \beta$ and
$\lambda \beta = \alpha$. In fact, just from this we could have deduced
that $\lambda$ can be either $-1$ or $1$. In any case, for bosons
($\lambda = 1$), we thus find that $\alpha = \beta$:

$$\begin{aligned}
    \Ket{\Psi(a, b)}_b = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big)
\end{aligned}$$

Where $C$ is a normalization constant. As expected, this state is
**symmetric**: switching $a$ and $b$ gives the same result. Meanwhile, for
fermions ($\lambda = -1$), we find that $\alpha = -\beta$:

$$\begin{aligned}
    \Ket{\Psi(a, b)}_f = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big)
\end{aligned}$$

This state is called **antisymmetric** under exchange: switching $a$ and $b$
causes a sign change, as we would expect for fermions.

Now, what if the particles $x_1$ and $x_2$ are in the same state $a$?
For bosons, we just need to update the normalization constant $C$:

$$\begin{aligned}
    \Ket{\Psi(a, a)}_b
    = C \Ket{a}\Ket{a}
\end{aligned}$$

However, for fermions, the state is unnormalizable and thus unphysical:

$$\begin{aligned}
    \Ket{\Psi(a, a)}_f
    = C \big( \Ket{a}\Ket{a} - \Ket{a}\Ket{a} \big)
    = 0
\end{aligned}$$

And this is the Pauli exclusion principle: **fermions may never
occupy the same quantum state**. One of the many notable consequences of
this is that the shells of atoms only fit a limited number of
electrons (which are fermions), since each must have a different quantum number.