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---
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 type 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:
$$\begin{aligned}
\hat{P}_f \Ket{a}\Ket{b}
&= \Ket{b}\Ket{a}
= - \Ket{a}\Ket{b}
\\
\hat{P}_b \Ket{a}\Ket{b}
&= \Ket{b}\Ket{a}
= \Ket{a}\Ket{b}
\end{aligned}$$
Another fundamental fact 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 profound.
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 therefore 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.
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