1 files changed, 112 insertions, 0 deletions
diff --git a/latex/know/concept/pauli-exclusion-principle/source.md b/latex/know/concept/pauli-exclusion-principle/source.md
new file mode 100644
index 0000000..0a35869
--- /dev/null
+++ b/latex/know/concept/pauli-exclusion-principle/source.md
@@ -0,0 +1,112 @@
+% 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 have 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.