From bcf2e9b649425d2df16b64752c4396a07face7ea Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 3 Mar 2021 18:03:22 +0100
Subject: Expand knowledge base

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+---
+title: "Density operator"
+firstLetter: "D"
+publishDate: 2021-03-03
+categories:
+- Physics
+- Quantum mechanics
+
+date: 2021-03-03T09:07:51+01:00
+draft: false
+markup: pandoc
+---
+
+# Density operator
+
+In quantum mechanics, the expectation value of an observable
+$\expval*{\hat{L}}$ represents the average result from measuring
+$\hat{L}$ on a large number of systems (an **ensemble**)
+prepared in the same state $\ket{\Psi}$,
+known as a **pure ensemble** or (somewhat confusingly) **pure state**.
+
+But what if the systems of the ensemble are not all in the same state?
+To work with such a **mixed ensemble** or **mixed state**,
+the **density operator** $\hat{\rho}$ or **density matrix** (in a basis) is useful.
+It is defined as follows, where $p_n$ is the probability
+that the system is in state $\ket{\Psi_n}$,
+i.e. the proportion of systems in the ensemble that are
+in state $\ket{\Psi_n}$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\rho}
+        = \sum_{n} p_n \ket{\Psi_n} \bra{\Psi_n}
+    }
+\end{aligned}$$
+
+Do not let is this form fool you into thinking that $\hat{\rho}$ is diagonal:
+$\ket{\Psi_n}$ need not be basis vectors.
+Instead, the matrix elements of $\hat{\rho}$ are found as usual,
+where $\ket{j}$ and $\ket{k}$ are basis vectors:
+
+$$\begin{aligned}
+    \matrixel{j}{\hat{\rho}}{k}
+    = \sum_{n} p_n \braket{j}{\Psi_n} \braket{\Psi_n}{k}
+\end{aligned}$$
+
+However, from the special case where $\ket{\Psi_n}$ are indeed basis vectors,
+we can conclude that $\hat{\rho}$ is Hermitian,
+and that its trace (i.e. the total probability) is 100%:
+
+$$\begin{gathered}
+    \boxed{
+        \hat{\rho}^\dagger = \hat{\rho}
+    }
+    \qquad \qquad
+    \boxed{
+        \mathrm{Tr}(\hat{\rho}) = 1
+    }
+\end{gathered}$$
+
+These properties are preserved by all changes of basis.
+If the ensemble is purely $\ket{\Psi}$,
+then $\hat{\rho}$ is given by a single state vector:
+
+$$\begin{aligned}
+    \hat{\rho} = \ket{\Psi} \bra{\Psi}
+\end{aligned}$$
+
+From the special case where $\ket{\Psi}$ is a basis vector,
+we can conclude that for a pure ensemble,
+$\hat{\rho}$ is idempotent, which means that:
+
+$$\begin{aligned}
+    \hat{\rho}^2 = \hat{\rho}
+\end{aligned}$$
+
+This can be used to find out whether a given $\hat{\rho}$
+represents a pure or mixed ensemble.
+
+Next, we define the ensemble average $\expval*{\expval*{\hat{L}}}$
+as the mean of the expectation values for states in the ensemble,
+which can be calculated like so:
+
+$$\begin{aligned}
+    \boxed{
+        \expval*{\expval*{\hat{L}}}
+        = \sum_{n} p_n \matrixel{\Psi_n}{\hat{L}}{\Psi_n}
+        = \mathrm{Tr}(\hat{L} \hat{\rho})
+    }
+\end{aligned}$$
+
+To prove the latter,
+we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so:
+
+$$\begin{aligned}
+    \mathrm{Tr}(\hat{L} \hat{\rho})
+    &= \sum_{j} \matrixel{j}{\hat{L} \hat{\rho}}{j}
+    = \sum_{j} \sum_{n} p_n \matrixel{j}{\hat{L}}{\Psi_n} \braket{\Psi_n}{j}
+    \\
+    &= \sum_{n} \sum_{j} p_n \braket{\Psi_n}{j} \matrixel{j}{\hat{L}}{\Psi_n}
+    = \sum_{n} p_n \matrixel{\Psi_n}{\hat{I} \hat{L}}{\Psi_n}
+    = \expval*{\expval*{\hat{L}}}
+\end{aligned}$$
+
+In both the pure and mixed cases,
+if the state probabilities $p_n$ are constant with respect to time,
+then the evolution of the ensemble obeys the **Von Neumann equation**:
+
+$$\begin{aligned}
+    \boxed{
+        i \hbar \dv{\hat{\rho}}{t} = [\hat{H}, \hat{\rho}]
+    }
+\end{aligned}$$
+
+This equivalent to the Schrödinger equation:
+one can be derived from the other.
+We differentiate $\hat{\rho}$ with the product rule,
+and then substitute the opposite side of the Schrödinger equation:
+
+$$\begin{aligned}
+    i \hbar \dv{\hat{\rho}}{t}
+    &= i \hbar \dv{t} \sum_n p_n \ket{\Psi_n} \bra{\Psi_n}
+    \\
+    &= \sum_n p_n \Big( i \hbar \dv{t} \ket{\Psi_n} \Big) \bra{\Psi_n} + \sum_n p_n \ket{\Psi_n} \Big( i \hbar \dv{t} \bra{\Psi_n} \Big)
+    \\
+    &= \sum_n p_n \ket*{\hat{H} n} \bra{n} - \sum_n p_n \ket{n} \bra*{\hat{H} n}
+    = \hat{H} \hat{\rho} - \hat{\rho} \hat{H}
+    = [\hat{H}, \hat{\rho}]
+\end{aligned}$$
+
+
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