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+---
+title: "Density operator"
+date: 2021-03-03
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+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:
+
+$$\begin{aligned}
+    \matrixel{j}{\hat{\rho}}{k}
+    = \sum_{n} p_n \Inprod{j}{\Psi_n} \Inprod{\Psi_n}{k}
+\end{aligned}$$
+
+However, from the special case where $\Ket{\Psi_n}$ are indeed basis vectors,
+we can conclude that $\hat{\rho}$ is positive semidefinite and Hermitian,
+and that its trace (i.e. the total probability) is 100%:
+
+$$\begin{gathered}
+    \boxed{
+        \hat{\rho} \ge 0
+    }
+    \qquad \qquad
+    \boxed{
+        \hat{\rho}^\dagger = \hat{\rho}
+    }
+    \qquad \qquad
+    \boxed{
+        \mathrm{Tr}(\hat{\rho}) = 1
+    }
+\end{gathered}$$
+
+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 $\expval{\hat{O}}$
+as the mean of the expectation values of $\hat{O}$ for states in the ensemble.
+We use the same notation as for the pure expectation value,
+since this is only a small extension of the concept to mixed ensembles.
+It is calculated like so:
+
+$$\begin{aligned}
+    \boxed{
+        \expval{\hat{O}}
+        = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O}}{\Psi_n}
+        = \mathrm{Tr}(\hat{\rho} \hat{O})
+    }
+\end{aligned}$$
+
+To prove the latter,
+we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so:
+
+$$\begin{aligned}
+    \mathrm{Tr}(\hat{\rho} \hat{O})
+    &= \sum_{j} \matrixel{j}{\hat{\rho} \hat{O}}{j}
+    = \sum_{j} \sum_{n} p_n \Inprod{j}{\Psi_n} \matrixel{\Psi_n}{\hat{O}}{j}
+    \\
+    &= \sum_{n} \sum_{j} p_n\matrixel{\Psi_n}{\hat{O}}{j} \Inprod{j}{\Psi_n}
+    = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O} \hat{I}}{\Psi_n}
+    = \expval{\hat{O}}
+\end{aligned}$$
+
+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**:
+
+$$\begin{aligned}
+    \boxed{
+        i \hbar \dv{\hat{\rho}}{t} = \comm{\hat{H}}{\hat{\rho}}
+    }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    i \hbar \dv{\hat{\rho}}{t}
+    &= i \hbar \dv{}{t}\sum_n p_n \Ket{\Psi_n} \Bra{\Psi_n}
+    \\
+    &= \sum_n p_n \Big( i \hbar \dv{}{t}\Ket{\Psi_n} \Big) \Bra{\Psi_n} + \sum_n p_n \Ket{\Psi_n} \Big( i \hbar \dv{}{t}\Bra{\Psi_n} \Big)
+    \\
+    &= \sum_n p_n \ket{\hat{H} n} \Bra{n} - \sum_n p_n \Ket{n} \bra{\hat{H} n}
+    = \hat{H} \hat{\rho} - \hat{\rho} \hat{H}
+    = \comm{\hat{H}}{\hat{\rho}}
+\end{aligned}$$
+
+
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