Categories: Physics, Quantum mechanics.

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:

\[\begin{aligned} i \hbar \dv{\hat{\rho}}{t} &= i \hbar \dv{t} \sum_n p_n \ket{\Psi_n} \bra{\Psi_n} \\ &= \sum_n p_n \Big( i \hbar \dv{t} \ket{\Psi_n} \Big) \bra{\Psi_n} + \sum_n p_n \ket{\Psi_n} \Big( i \hbar \dv{t} \bra{\Psi_n} \Big) \\ &= \sum_n p_n \ket*{\hat{H} n} \bra{n} - \sum_n p_n \ket{n} \bra*{\hat{H} n} = \hat{H} \hat{\rho} - \hat{\rho} \hat{H} = [\hat{H}, \hat{\rho}] \end{aligned}\]

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