How a density matrix evolves · Quantum Mechanics

Intuition

Each member of an ensemble obeys Schrödinger’s equation, so the density matrix obeys $i\hbar\,d\hat{\rho}/dt=[\hat{H},\hat{\rho}]$ , the von Neumann equation. In the energy basis the populations stay as they are and the coherences turn at the Bohr frequencies. The evolution is unitary, $\hat{\rho}(t)=\hat{U}\hat{\rho}(0)\hat{U}^{\dagger}$ , so the eigenvalues of $\hat{\rho}$ , its purity and every measure of how mixed it is stay the same: an isolated system never becomes more or less mixed. A density matrix that does not change commutes with $\hat{H}$ . The thermal state, which describes a system in equilibrium with surroundings at temperature $T$ , is one: its populations fall off as $e^{-E/k_{B}T}$ , and it has no coherences between energy levels.

A troupe of dancers following one choreography: the pattern on the floor turns and shifts, but how the dancers are spread among the steps never changes. Isolated evolution moves a mixture without mixing it further.

A spin in a field along $z$ , started in $\ket{+x}$ , with time in units of $1/\omega$ . The population $\rho_{11}$ of the upper level stays at , while the real part of the coherence turns at the Bohr frequency : the Bloch vector precesses about the field.

The von Neumann equation

For a system with Hamiltonian $\hat{H}$ and evolution operator $\hat{U}(t)$ :

\begin{aligned}i\hbar\frac{d\hat{\rho}}{dt}&=\bigl[\hat{H},\hat{\rho}\bigr]\\ \hat{\rho}(t)&=\hat{U}(t)\,\hat{\rho}(0)\,\hat{U}(t)^{\dagger}\end{aligned}

Properties

In the eigenbasis of $\hat{H}$ : $\rho_{nn}$ is constant and $ρ_{m n} (t) = ρ_{m n} (0) e^{- i_{}}$ , with .

The von Neumann equation

i\hbar\frac{d}{dt}\sum_{k}p_{k}\ket{\psi_{k}}\bra{\psi_{k}}=\Bigl[\hat{H},\ \sum_{k}p_{k}\ket{\psi_{k}}\bra{\psi_{k}}\Bigr]

Differentiate the projector of one member by the product rule. Schrödinger’s equation gives the derivative of the ket, and its adjoint the derivative of the bra, with the opposite sign. Together they make the commutator with $\hat{H}$ , and the weights $p_{k}$ , which do not change, carry it to the whole ensemble.

Proof steps

\frac{d}{dt}\ket{\psi}\bra{\psi}=\Bigl(\frac{d}{dt}\ket{\psi}\Bigr)\bra{\psi}+\ket{\psi}\Bigl(\frac{d}{dt}\bra{\psi}\Bigr)

The product rule.

i\hbar\frac{d}{dt}\ket{\psi}=\hat{H}\ket{\psi},\qquad -i\hbar\frac{d}{dt}\bra{\psi}=\bra{\psi}\hat{H}

Schrödinger’s equation and its adjoint; $\hat{H}$ is Hermitian.

i\hbar\frac{d}{dt}\ket{\psi}\bra{\psi}=\hat{H}\ket{\psi}\bra{\psi}-\ket{\psi}\bra{\psi}\hat{H}

Multiply the product rule by $i\hbar$ and substitute.

i\hbar\frac{d}{dt}\ket{\psi}\bra{\psi}=\bigl[\hat{H},\ket{\psi}\bra{\psi}\bigr]

Collect the two terms.

i\hbar\frac{d\hat{\rho}}{dt}=\bigl[\hat{H},\hat{\rho}\bigr]

The weights $p_{k}$ are constant and the commutator is linear.

Applications

Practice

The von Neumann Equation

Every member of an ensemble obeys Schrödinger’s equation, and together they make the density matrix change by its commutator with the Hamiltonian.

i\hbar\frac{d\hat{\rho}}{dt}=\bigl[\hat{H},\hat{\rho}\bigr]

Try it

Which equation does a density matrix obey under a Hamiltonian $\hat{H}$ ?