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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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