---
title: "Pauli exclusion principle"
sort_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.