---
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:

$$\begin{aligned}
    i \hbar \dv{\hat{\rho}}{t}
    &= i \hbar \dv{}{t}\sum_n p_n \Ket{\Psi_n} \Bra{\Psi_n}
    \\
    &= \sum_n p_n \Big( i \hbar \dv{}{t}\Ket{\Psi_n} \Big) \Bra{\Psi_n} + \sum_n p_n \Ket{\Psi_n} \Big( i \hbar \dv{}{t}\Bra{\Psi_n} \Big)
    \\
    &= \sum_n p_n \ket{\hat{H} n} \Bra{n} - \sum_n p_n \Ket{n} \bra{\hat{H} n}
    = \hat{H} \hat{\rho} - \hat{\rho} \hat{H}
    = \comm{\hat{H}}{\hat{\rho}}
\end{aligned}$$