From 6fb3b28a2ce91b8f12683692cbe32d9a4d35fc9d Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 21 Feb 2021 10:46:28 +0100
Subject: Add "Slater determinant" + improvements to knowledge base

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+% Slater determinant
+
+
+# Slater determinant
+
+In quantum mechanics, the **Slater determinant** is a trick to create an
+antisymmetric wave function for a system of $N$ fermions.
+
+Given an orthogonal set of individual states $\psi_n(x)$, we write
+$\psi_n(x_n)$ to say that particle $x_n$ is in state $\psi_n$. Now the
+goal is to find an expression for an overall many-particle wave
+function $\Psi(x_1, ..., x_N)$ that satisfies the
+[Pauli exclusion principle](/know/concept/pauli-exclusion-principle/).
+Enter the Slater determinant:
+
+$$\begin{aligned}
+    \boxed{
+        \Psi(x_1, ..., x_N)
+        = \frac{1}{\sqrt{N!}} \det\!
+        \begin{bmatrix}
+            \psi_1(x_1) & \cdots & \psi_N(x_1) \\
+            \vdots & \ddots & \vdots \\
+            \psi_1(x_N) & \cdots & \psi_N(x_N)
+        \end{bmatrix}
+    }\end{aligned}$$
+
+Swapping the state of two particles corresponds to exchanging two rows
+in the matrix, which flips the sign of the determinant. Similarly,
+exchanging two columns means swapping two states, which also results in
+a sign change. Finally, putting two particles into the same state makes
+$\Psi$ vanish.
+
+Note that not all valid many-particle fermionic wave functions can be
+written as a single Slater determinant; a linear combination of multiple
+may be needed. Nevertheless, an appropriate choice of the input set
+$\psi_n(x)$ can optimize how well a single determinant approximates a
+given $\Psi$.
+
+In fact, there exists a similar trick for bosons, where the goal is to
+create a symmetric wave function which allows multiple particles to
+occupy the same state. In this case, one needs to take the **Slater
+permanent** of the same matrix, which is simply the determinant, but with
+all minuses replaced by pluses.
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