Categories: Physics, Quantum mechanics.

# 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 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 \braket{j}{\Psi_n} \matrixel{\Psi_n}{\hat{O}}{j} \\ &= \sum_{n} \sum_{j} p_n\matrixel{\Psi_n}{\hat{O}}{j} \braket{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} = [\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}