Categories: Physics, Quantum mechanics.

Density operator

In quantum mechanics, the expectation value of an observable ⟨L^⟩\expval{\hat{L}} represents the average result from measuring L^\hat{L} on a large number of systems (an ensemble) prepared in the same state ∣Ψ⟩\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 ρ^\hat{\rho} or density matrix (in a basis) is useful. It is defined as follows, where pnp_n is the probability that the system is in state ∣Ψn⟩\Ket{\Psi_n}, i.e. the proportion of systems in the ensemble that are in state ∣Ψn⟩\Ket{\Psi_n}:

ρ^=∑npn∣Ψn⟩⟨Ψn∣\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 ρ^\hat{\rho} is diagonal: ∣Ψn⟩\Ket{\Psi_n} need not be basis vectors. Instead, the matrix elements of ρ^\hat{\rho} are found as usual, where ∣j⟩\Ket{j} and ∣k⟩\Ket{k} are basis vectors:

⟨j∣ρ^∣k⟩=∑npn⟨j | Ψn⟩⟨Ψn | k⟩\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 ∣Ψn⟩\Ket{\Psi_n} are indeed basis vectors, we can conclude that ρ^\hat{\rho} is positive semidefinite and Hermitian, and that its trace (i.e. the total probability) is 100%:

ρ^≥0ρ^†=ρ^Tr(ρ^)=1\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 ∣Ψ⟩\Ket{\Psi}, then ρ^\hat{\rho} is given by a single state vector:

ρ^=∣Ψ⟩⟨Ψ∣\begin{aligned} \hat{\rho} = \Ket{\Psi} \Bra{\Psi} \end{aligned}

From the special case where ∣Ψ⟩\Ket{\Psi} is a basis vector, we can conclude that for a pure ensemble, ρ^\hat{\rho} is idempotent, which means that:

ρ^2=ρ^\begin{aligned} \hat{\rho}^2 = \hat{\rho} \end{aligned}

This can be used to find out whether a given ρ^\hat{\rho} represents a pure or mixed ensemble.

Next, we define the ensemble average ⟨O^⟩\expval{\hat{O}} as the mean of the expectation values of O^\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:

⟨O^⟩=∑npn⟨Ψn∣O^∣Ψn⟩=Tr(ρ^O^)\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 Tr\mathrm{Tr} as the sum of the diagonal elements, so:

Tr(ρ^O^)=∑j⟨j∣ρ^O^∣j⟩=∑j∑npn⟨j | Ψn⟩⟨Ψn∣O^∣j⟩=∑n∑jpn⟨Ψn∣O^∣j⟩⟨j | Ψn⟩=∑npn⟨Ψn∣O^I^∣Ψn⟩=⟨O^⟩\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 pnp_n are constant with respect to time, then the evolution of the ensemble obeys the Von Neumann equation:

iℏdρ^dt=[H^,ρ^]\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 ρ^\hat{\rho} with the product rule, and then substitute the opposite side of the Schrödinger equation:

iℏdρ^dt=iℏddt∑npn∣Ψn⟩⟨Ψn∣=∑npn(iℏddt∣Ψn⟩)⟨Ψn∣+∑npn∣Ψn⟩(iℏddt⟨Ψn∣)=∑npn∣H^n⟩⟨n∣−∑npn∣n⟩⟨H^n∣=H^ρ^−ρ^H^=[H^,ρ^]\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}