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:

$$\langle A\rangle = \operatorname{Tr}(\rho A).$$

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

Schrödinger Picture Density Operator

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

$$\rho_S(t) = U(t,t_0)\rho_S(t_0)U^\dagger(t,t_0).$$

It obeys the Liouville–von Neumann equation:

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

For an observable $A_S(t)$,

$$\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_S(t)A_S(t)\bigr).$$

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

Heisenberg Picture Density Operator

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

$$\rho_H = \rho_S(t_0).$$

Observables carry the time dependence:

$$A_H(t) = U^\dagger(t,t_0)A_S(t)U(t,t_0).$$

The expectation value is

$$\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_HA_H(t)\bigr).$$

This equals the Schrödinger-picture trace:

$$\begin{aligned} \operatorname{Tr}\bigl(\rho_HA_H(t)\bigr) &= \operatorname{Tr}\bigl( \rho_S(t_0)U^\dagger A_SU \bigr) \\ &= \operatorname{Tr}\bigl( U\rho_S(t_0)U^\dagger A_S \bigr) \\ &= \operatorname{Tr}\bigl(\rho_S(t)A_S\bigr). \end{aligned}$$

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

Interaction Picture Density Operator

For a split

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

define

$$U_0(t,t_0)=e^{-iH_0(t-t_0)/\hbar}$$

when $H_0$ is time independent. The interaction-picture density operator is

$$\rho_I(t) = U_0^\dagger(t,t_0)\rho_S(t)U_0(t,t_0).$$

The interaction-picture observable is

$$A_I(t) = U_0^\dagger(t,t_0)A_S(t)U_0(t,t_0).$$

The expectation value remains

$$\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_I(t)A_I(t)\bigr).$$

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

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

where

$$V_I(t) = U_0^\dagger(t,t_0)V(t)U_0(t,t_0).$$

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

General Picture Transformation

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

$$\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle,$$

the density operator and observable transform as

$$\rho_P(t)=R^\dagger(t)\rho_S(t)R(t), \qquad A_P(t)=R^\dagger(t)A_S(t)R(t).$$

Then

$$\operatorname{Tr}\bigl(\rho_P A_P\bigr) = \operatorname{Tr}\bigl(\rho_S A_S\bigr).$$

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

$$H_P = R^\dagger H_SR - i\hbar R^\dagger\dot R.$$

The density operator satisfies

$$i\hbar\frac{d\rho_P}{dt} = [H_P,\rho_P].$$

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

Matrix Elements and Coherences

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

$$\rho_{mn}(t) = e^{-i(E_m-E_n)(t-t_0)/\hbar} \rho_{mn}(t_0).$$

Diagonal energy populations are constant. Coherences between distinct energies rotate at Bohr frequencies; off-diagonal entries within a degenerate energy subspace stay fixed. In the Heisenberg picture, this same phase information appears in the time dependence of observables rather than in $\rho_H$.

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

Explicit Time-Dependent Observables

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

$$\langle A\rangle_t = \operatorname{Tr}\bigl(\rho_S(t)A_S(t)\bigr) = \operatorname{Tr}\bigl(\rho_HA_H(t)\bigr).$$

The derivative of the expectation value includes the explicit term:

$$\frac{d}{dt}\langle A\rangle = \frac{i}{\hbar}\operatorname{Tr}\bigl(\rho[H,A]\bigr) + \operatorname{Tr}\left( \rho\frac{\partial A}{\partial t} \right).$$

Operators with Explicit Time Dependence explains that term through rotating measurement axes.

Reduced states and changes of picture

A subsystem coupled to an environment generally has nonunitary reduced evolution after the environment is traced out. A unitary change of picture remains possible for the subsystem, but cannot turn a genuine dissipator into a Hamiltonian commutator. For a Lindblad equation, the jump operators transform as $L_{\mu,P}=R^\dagger L_\mu R$, while the Hamiltonian acquires the same frame term as above.

Unitary conjugation preserves every eigenvalue of $\rho$, hence its purity $\operatorname{Tr}\rho^2$ and entropy. A change in those quantities under reduced dynamics is therefore a physical effect, not a change of picture. See quantum operations for the reduced dynamical description.

Common Mistakes

• Thinking the Heisenberg density operator being fixed means mixed-state predictions cannot change.
• Transforming observables but forgetting to transform $\rho$ in a general picture.
• Reversing the order in $\rho_S(t)=U\rho_S(t_0)U^\dagger$.
• 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.

References

• 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.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.

Exercises

1. Show that Schrödinger and Heisenberg density-operator expectation values agree.

Solution

Use

$$\rho_S(t)=U\rho_S(t_0)U^\dagger, \qquad A_H(t)=U^\dagger A_SU, \qquad \rho_H=\rho_S(t_0).$$

Then

$$\begin{aligned} \operatorname{Tr}(\rho_HA_H) &= \operatorname{Tr}\bigl(\rho_S(t_0)U^\dagger A_SU\bigr) \\ &= \operatorname{Tr}\bigl(U\rho_S(t_0)U^\dagger A_S\bigr) \\ &= \operatorname{Tr}\bigl(\rho_S(t)A_S\bigr). \end{aligned}$$

2. Derive the interaction-picture density-operator equation $i\hbar\dot\rho_I=[V_I,\rho_I]$ for time-independent $H_0$.

Solution

Start with

$$\rho_I=U_0^\dagger\rho_SU_0, \qquad i\hbar\dot\rho_S=[H_0+V,\rho_S].$$

Differentiate $\rho_I$. The terms involving $H_0$ from $\dot U_0$ and $\dot U_0^\dagger$ cancel the commutator with $H_0$, leaving

$$i\hbar\dot\rho_I = [U_0^\dagger VU_0,\rho_I] = [V_I,\rho_I].$$

3. For a time-independent Hamiltonian with $H\lvert n\rangle=E_n\lvert n\rangle$, derive the phase evolution of $\rho_{mn}(t)$.

Solution

The Liouville–von Neumann equation gives

$$i\hbar\dot\rho_{mn} = (E_m-E_n)\rho_{mn}.$$

Therefore

$$\rho_{mn}(t) = e^{-i(E_m-E_n)(t-t_0)/\hbar} \rho_{mn}(t_0).$$

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

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

Solution

A picture transformation is a unitary change of representation. Applied to closed-system dynamics, it preserves the form of closed-system dynamics as unitary conjugation and gives

$$i\hbar\dot\rho=[H,\rho]$$

in the transformed picture with the transformed Hamiltonian.

A Lindblad equation contains extra dissipative terms such as

$$L_\mu\rho L_\mu^\dagger - \frac12\{L_\mu^\dagger L_\mu,\rho\},$$

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