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Density Operators in Different Pictures

Density operators transform between pictures in the same spirit as state vectors: the mathematical representative changes, but expectation values do not.

The invariant quantity is the trace rule:

⟨A⟩=Tr⁡(ρA).\langle A\rangle = \operatorname{Tr}(\rho A).

If states and observables are transformed consistently, every picture gives the same number.

In the Schrödinger picture, the density operator carries the state time dependence. For closed evolution,

ρS(t)=U(t,t0)ρS(t0)U†(t,t0).\rho_S(t) = U(t,t_0)\rho_S(t_0)U^\dagger(t,t_0).

It obeys the Liouville–von Neumann equation:

iℏdρSdt=[HS(t),ρS(t)].i\hbar\frac{d\rho_S}{dt} = [H_S(t),\rho_S(t)].

For an observable AS(t)A_S(t),

⟨A⟩t=Tr⁡(ρS(t)AS(t)).\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_S(t)A_S(t)\bigr).

This is the density-operator version of Schrödinger-picture state evolution.

In the Heisenberg picture, the density operator is fixed at the reference time:

ρH=ρS(t0).\rho_H = \rho_S(t_0).

Observables carry the time dependence:

AH(t)=U†(t,t0)AS(t)U(t,t0).A_H(t) = U^\dagger(t,t_0)A_S(t)U(t,t_0).

The expectation value is

⟨A⟩t=Tr⁡(ρHAH(t)).\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_HA_H(t)\bigr).

This equals the Schrödinger-picture trace:

Tr⁡(ρHAH(t))=Tr⁡(ρS(t0)U†ASU)=Tr⁡(UρS(t0)U†AS)=Tr⁡(ρS(t)AS).\begin{aligned} \operatorname{Tr}\bigl(\rho_HA_H(t)\bigr) &= \operatorname{Tr}\bigl( \rho_S(t_0)U^\dagger A_SU \bigr) \\ &= \operatorname{Tr}\bigl( U\rho_S(t_0)U^\dagger A_S \bigr) \\ &= \operatorname{Tr}\bigl(\rho_S(t)A_S\bigr). \end{aligned}

The second line uses cyclicity of the trace. The density operator is fixed, but predictions can still be time dependent because AH(t)A_H(t) is time dependent.

For a split

H(t)=H0+V(t),H(t)=H_0+V(t),

define

U0(t,t0)=e−iH0(t−t0)/ℏU_0(t,t_0)=e^{-iH_0(t-t_0)/\hbar}

when H0H_0 is time independent. The interaction-picture density operator is

ρI(t)=U0†(t,t0)ρS(t)U0(t,t0).\rho_I(t) = U_0^\dagger(t,t_0)\rho_S(t)U_0(t,t_0).

The interaction-picture observable is

AI(t)=U0†(t,t0)AS(t)U0(t,t0).A_I(t) = U_0^\dagger(t,t_0)A_S(t)U_0(t,t_0).

The expectation value remains

⟨A⟩t=Tr⁡(ρI(t)AI(t)).\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_I(t)A_I(t)\bigr).

The interaction-picture density operator evolves under the transformed interaction:

iℏdρIdt=[VI(t),ρI(t)],i\hbar\frac{d\rho_I}{dt} = [V_I(t),\rho_I(t)],

where

VI(t)=U0†(t,t0)V(t)U0(t,t0).V_I(t) = U_0^\dagger(t,t_0)V(t)U_0(t,t_0).

Thus the density operator version of the interaction picture is exact before any perturbative expansion is made.

For a unitary picture transformation R(t)R(t) using the convention

∣ψP⟩=R†∣ψS⟩,\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle,

the density operator and observable transform as

ρP(t)=R†(t)ρS(t)R(t),AP(t)=R†(t)AS(t)R(t).\rho_P(t)=R^\dagger(t)\rho_S(t)R(t), \qquad A_P(t)=R^\dagger(t)A_S(t)R(t).

Then

Tr⁡(ρPAP)=Tr⁡(ρSAS).\operatorname{Tr}\bigl(\rho_P A_P\bigr) = \operatorname{Tr}\bigl(\rho_S A_S\bigr).

The transformed Hamiltonian is the same one used for state vectors:

HP=R†HSR−iℏR†R˙.H_P = R^\dagger H_SR - i\hbar R^\dagger\dot R.

The density operator satisfies

iℏdρPdt=[HP,ρP].i\hbar\frac{d\rho_P}{dt} = [H_P,\rho_P].

For the Heisenberg picture, R=UR=U and HP=0H_P=0 for the density-operator equation, so ρH\rho_H is fixed. For the interaction picture, R=U0R=U_0 and HP=VIH_P=V_I.

For a time-independent Hamiltonian with eigenstates H∣n⟩=En∣n⟩H\lvert n\rangle=E_n\lvert n\rangle, Schrödinger-picture density-matrix elements evolve as

ρmn(t)=e−i(Em−En)(t−t0)/ℏρmn(t0).\rho_{mn}(t) = e^{-i(E_m-E_n)(t-t_0)/\hbar} \rho_{mn}(t_0).

Diagonal energy populations are constant. Off-diagonal energy coherences rotate at Bohr frequencies. In the Heisenberg picture, this same phase information appears in the time dependence of observables rather than in ρH\rho_H.

The terminology “population” and “coherence” is basis-dependent; see Density Matrix Conventions.

If the observable itself depends explicitly on time, the trace rule still works:

⟨A⟩t=Tr⁡(ρS(t)AS(t))=Tr⁡(ρHAH(t)).\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_S(t)A_S(t)\bigr) = \operatorname{Tr}\bigl(\rho_HA_H(t)\bigr).

The derivative of the expectation value includes the explicit term:

ddt⟨A⟩=iℏTr⁡(ρ[H,A])+Tr⁡(ρ∂A∂t).\frac{d}{dt}\langle A\rangle = \frac{i}{\hbar}\operatorname{Tr}\bigl(\rho[H,A]\bigr) + \operatorname{Tr}\left( \rho\frac{\partial A}{\partial t} \right).

The page Operators with Explicit Time Dependence owns the detailed interpretation of that term.

This page describes closed-system picture transformations. In a closed system, density operators evolve by unitary conjugation and obey the Liouville–von Neumann equation.

If a subsystem is coupled to an environment and the environment is traced out, the reduced density operator generally does not evolve by unitary conjugation. Its dynamics may require quantum operations, master equations, or non-Markovian maps. Those belong to Measurement and Open Quantum Systems.

A common Markovian reference equation is the Lindblad equation:

ρ˙=−iℏ[H,ρ]+∑μ(LμρLμ†−12{Lμ†Lμ,ρ}).\dot\rho = -\frac{i}{\hbar}[H,\rho] + \sum_\mu \left( L_\mu\rho L_\mu^\dagger - \frac12\{L_\mu^\dagger L_\mu,\rho\} \right).

Only the first term is the closed-system Liouville–von Neumann term.

  • Thinking the Heisenberg density operator being fixed means mixed-state predictions cannot change.
  • Transforming observables but forgetting to transform ρ\rho in a general picture.
  • Reversing the order in ρS(t)=UρS(t0)U†\rho_S(t)=U\rho_S(t_0)U^\dagger.
  • Treating a reduced open-system density operator as if it obeyed closed-system unitary conjugation.
  • Confusing a basis-dependent density matrix with the abstract density operator.
  • Calling a time-dependent ensemble update a picture transformation.
  • Forgetting that trace cyclicity requires well-defined trace-class products in infinite-dimensional settings.
  • J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
  • P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
  • R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
  • J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
  • L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
  • H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002.
  1. Show that Schrödinger and Heisenberg density-operator expectation values agree.
Solution

Use

ρS(t)=UρS(t0)U†,AH(t)=U†ASU,ρH=ρS(t0).\rho_S(t)=U\rho_S(t_0)U^\dagger, \qquad A_H(t)=U^\dagger A_SU, \qquad \rho_H=\rho_S(t_0).

Then

Tr⁡(ρHAH)=Tr⁡(ρS(t0)U†ASU)=Tr⁡(UρS(t0)U†AS)=Tr⁡(ρS(t)AS).\begin{aligned} \operatorname{Tr}(\rho_HA_H) &= \operatorname{Tr}\bigl(\rho_S(t_0)U^\dagger A_SU\bigr) \\ &= \operatorname{Tr}\bigl(U\rho_S(t_0)U^\dagger A_S\bigr) \\ &= \operatorname{Tr}\bigl(\rho_S(t)A_S\bigr). \end{aligned}
  1. Derive the interaction-picture density-operator equation iℏρ˙I=[VI,ρI]i\hbar\dot\rho_I=[V_I,\rho_I] for time-independent H0H_0.
Solution

Start with

ρI=U0†ρSU0,iℏρ˙S=[H0+V,ρS].\rho_I=U_0^\dagger\rho_SU_0, \qquad i\hbar\dot\rho_S=[H_0+V,\rho_S].

Differentiate ρI\rho_I. The terms involving H0H_0 from U˙0\dot U_0 and U˙0†\dot U_0^\dagger cancel the commutator with H0H_0, leaving

iℏρ˙I=[U0†VU0,ρI]=[VI,ρI].i\hbar\dot\rho_I = [U_0^\dagger VU_0,\rho_I] = [V_I,\rho_I].
  1. For a time-independent Hamiltonian with H∣n⟩=En∣n⟩H\lvert n\rangle=E_n\lvert n\rangle, derive the phase evolution of ρmn(t)\rho_{mn}(t).
Solution

The Liouville–von Neumann equation gives

iℏρ˙mn=(Em−En)ρmn.i\hbar\dot\rho_{mn} = (E_m-E_n)\rho_{mn}.

Therefore

ρmn(t)=e−i(Em−En)(t−t0)/ℏρmn(t0).\rho_{mn}(t) = e^{-i(E_m-E_n)(t-t_0)/\hbar} \rho_{mn}(t_0).

When m=nm=n, the phase is one, so energy-basis populations are constant.

  1. Why is a Lindblad equation not just the Schrödinger-picture density operator written in another picture?
Solution

A picture transformation is unitary bookkeeping for a closed system. It preserves the form of closed-system dynamics as unitary conjugation and gives

iℏρ˙=[H,ρ]i\hbar\dot\rho=[H,\rho]

in the transformed picture with the transformed Hamiltonian.

A Lindblad equation contains extra dissipative terms such as

LμρLμ†−12{Lμ†Lμ,ρ},L_\mu\rho L_\mu^\dagger - \frac12\{L_\mu^\dagger L_\mu,\rho\},

which represent environmental or coarse-grained effects. They are not produced by a unitary change of picture on the same closed-system Hilbert space.