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% Pauli exclusion principle
# Pauli exclusion principle
In quantum mechanics, the *Pauli exclusion principle* is a theorem that
has 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
$\lambda = -1$.
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 species of particle that $x_1$
and $x_2$ represent. 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 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:
$$\begin{aligned}
\ket{\Psi(a, a)}_f
= C \big( \ket{a}\ket{a} - \ket{a}\ket{a} \big)
= 0
\end{aligned}$$
At last, 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 an atom only fit a limited number of
electrons, since each must have a different quantum number.
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