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Quantum Mechanics

Picture Transformations

A picture transformation is a time-dependent unitary change of variables that moves time dependence between states, observables, and Hamiltonians without changing physical predictions.

This page uses the convention

∣ψP(t)⟩=R†(t)∣ψS(t)⟩,AP(t)=R†(t)AS(t)R(t),\lvert\psi_P(t)\rangle = R^\dagger(t)\lvert\psi_S(t)\rangle, \qquad A_P(t) = R^\dagger(t)A_S(t)R(t),

where R(t)R(t) is unitary. With this convention, the transformed Hamiltonian is

HP(t)=R†(t)HS(t)R(t)−iℏR†(t)R˙(t).H_P(t) = R^\dagger(t)H_S(t)R(t) - i\hbar R^\dagger(t)\dot R(t).

The second term is the common source of sign mistakes. It appears whenever the transformation itself depends on time.

General Change of Picture

Section titled “General Change of Picture”

The differential calculations below apply directly to differentiable unitary matrices. For unbounded operators, use a common invariant domain preserved by the evolution and R(t)R(t), with the displayed strong derivatives defined there; see domains in Hamiltonian evolution.

Let the Schrödinger-picture state obey

iℏddt∣ψS(t)⟩=HS(t)∣ψS(t)⟩.i\hbar\frac{d}{dt}\lvert\psi_S(t)\rangle = H_S(t)\lvert\psi_S(t)\rangle.

Choose a unitary R(t)R(t) and define the transformed state by

∣ψP(t)⟩=R†(t)∣ψS(t)⟩.\lvert\psi_P(t)\rangle = R^\dagger(t)\lvert\psi_S(t)\rangle.

The transformed observable is chosen so that expectation values are unchanged:

AP(t)=R†(t)AS(t)R(t).A_P(t)=R^\dagger(t)A_S(t)R(t).

Then

⟨ψP(t)∣AP(t)∣ψP(t)⟩=⟨ψS(t)∣(RR†)AS(t)(RR†)∣ψS(t)⟩=⟨ψS(t)∣AS(t)∣ψS(t)⟩.\begin{aligned} \langle\psi_P(t)\rvert A_P(t)\lvert\psi_P(t)\rangle &= \langle\psi_S(t)\rvert (RR^\dagger)A_S(t)(RR^\dagger) \lvert\psi_S(t)\rangle \\ &= \langle\psi_S(t)\rvert A_S(t)\lvert\psi_S(t)\rangle. \end{aligned}

The transformed propagator also has two frame factors:

UP(t,t0)=R†(t)U(t,t0)R(t0).U_P(t,t_0)=R^\dagger(t)U(t,t_0)R(t_0).

The initial factor cannot be dropped unless R(t0)=IR(t_0)=I; it ensures UP(t0,t0)=IU_P(t_0,t_0)=I.

Hamiltonian Transformation

Section titled “Hamiltonian Transformation”

Differentiate the transformed state:

ddt∣ψP⟩=R˙†∣ψS⟩+R†ddt∣ψS⟩.\frac{d}{dt}\lvert\psi_P\rangle = \dot R^\dagger\lvert\psi_S\rangle + R^\dagger\frac{d}{dt}\lvert\psi_S\rangle.

Substitute ∣ψS⟩=R∣ψP⟩\lvert\psi_S\rangle=R\lvert\psi_P\rangle and the Schrödinger equation:

iℏddt∣ψP⟩=(iℏR˙†R+R†HSR)∣ψP⟩.i\hbar\frac{d}{dt}\lvert\psi_P\rangle = \left( i\hbar\dot R^\dagger R + R^\dagger H_SR \right) \lvert\psi_P\rangle.

Unitarity gives

R˙†R=−R†R˙,\dot R^\dagger R = -R^\dagger\dot R,

so

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

The energy observable is EP=R†HSRE_P=R^\dagger H_SR. Define the frame generator K=iℏR†R˙K=i\hbar R^\dagger\dot R, so that HP=EP−KH_P=E_P-K. The state generator HPH_P and the physical energy observable EPE_P are different objects in a moving frame.

Heisenberg Picture as a Special Case

Section titled “Heisenberg Picture as a Special Case”

Set

R(t)=U(t,t0),R(t)=U(t,t_0),

where U(t,t0)U(t,t_0) is the full time-evolution operator. Then

∣ψH⟩=U†(t,t0)∣ψS(t)⟩=∣ψS(t0)⟩,\lvert\psi_H\rangle = U^\dagger(t,t_0)\lvert\psi_S(t)\rangle = \lvert\psi_S(t_0)\rangle,

and

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 transformed Hamiltonian vanishes for the state equation:

HHstate=U†HU−iℏU†U˙=0,H_H^{\mathrm{state}} = U^\dagger HU - i\hbar U^\dagger\dot U =0,

because iℏU˙=HUi\hbar\dot U=HU. The energy observable HH=U†HUH_H=U^\dagger HU remains nonzero in general. This is why a fixed Heisenberg state can have nonzero energy and time-dependent predictions.

Interaction Picture as a Special Case

Section titled “Interaction Picture as a Special Case”

For a split

HS(t)=H0+V(t),H_S(t)=H_0+V(t),

take H0H_0 time independent and choose

R(t)=U0(t,t0)=e−iH0(t−t0)/ℏ.R(t)=U_0(t,t_0)=e^{-iH_0(t-t_0)/\hbar}.

Then

∣ψI(t)⟩=U0†(t,t0)∣ψS(t)⟩,AI(t)=U0†(t,t0)ASU0(t,t0).\lvert\psi_I(t)\rangle = U_0^\dagger(t,t_0)\lvert\psi_S(t)\rangle, \qquad A_I(t) = U_0^\dagger(t,t_0)A_SU_0(t,t_0).

The transformed Hamiltonian is

HI(t)=U0†(H0+V)U0−iℏU0†U˙0=U0†(t,t0)V(t)U0(t,t0)=VI(t).\begin{aligned} H_I(t) &= U_0^\dagger(H_0+V)U_0-i\hbar U_0^\dagger\dot U_0 \\ &= U_0^\dagger(t,t_0)V(t)U_0(t,t_0) = V_I(t). \end{aligned}

The H0H_0 part cancels against the extra generator term. This cancellation is the algebraic heart of the interaction picture.

Rotating Frames

Section titled “Rotating Frames”

A rotating frame is another time-dependent picture. Let

R(t)=e−iωtG/ℏ,R(t)=e^{-i\omega tG/\hbar},

where GG generates the rotation. Since

R˙(t)=−iωℏGR(t),\dot R(t) = -\frac{i\omega}{\hbar}GR(t),

the extra term is

−iℏR†R˙=−ωG.-i\hbar R^\dagger\dot R = -\omega G.

If HSH_S commutes with GG, then

HP=HS−ωG.H_P=H_S-\omega G.

For a spin with HS=ω0SzH_S=\omega_0S_z and G=SzG=S_z, a frame rotating about zz has

HP=(ω0−ω)Sz.H_P=(\omega_0-\omega)S_z.

This subtraction exposes detunings in driven rotating frames. The transformation itself is exact.

Operator Equations Under a Picture Change

Section titled “Operator Equations Under a Picture Change”

The transformed operator is

AP(t)=R†(t)AS(t)R(t).A_P(t)=R^\dagger(t)A_S(t)R(t).

Differentiating gives both explicit time dependence and picture-induced time dependence:

dAPdt=R†∂AS∂tR+R˙†ASR+R†ASR˙.\frac{dA_P}{dt} = R^\dagger\frac{\partial A_S}{\partial t}R + \dot R^\dagger A_SR + R^\dagger A_S\dot R.

With the frame generator K=iℏR†R˙K=i\hbar R^\dagger\dot R, this is

dAPdt=iℏ[K,AP]+R†∂AS∂tR.\frac{dA_P}{dt}=\frac{i}{\hbar}[K,A_P] +R^\dagger\frac{\partial A_S}{\partial t}R.

The operator’s picture-induced derivative involves KK, not the state generator HPH_P. State evolution and operator evolution combine in an expectation value: HP+K=EPH_P+K=E_P recovers the original physical dynamics. For R=UR=U, K=HHK=H_H, giving the Heisenberg equation.

Convention Warning

Section titled “Convention Warning”

Some books define the transformed state as R(t)∣ψS(t)⟩R(t)\lvert\psi_S(t)\rangle instead of R†(t)∣ψS(t)⟩R^\dagger(t)\lvert\psi_S(t)\rangle. That convention is equally valid, but the transformed operator and Hamiltonian formulas change:

∣ψP⟩=R∣ψS⟩⟹HP=RHSR†+iℏR˙R†.\lvert\psi_P\rangle=R\lvert\psi_S\rangle \quad \Longrightarrow \quad H_P=RH_SR^\dagger+i\hbar\dot R R^\dagger.

Before comparing formulas across sources, identify which side carries the dagger.

Common Mistakes

Section titled “Common Mistakes”
  • Transforming states but leaving observables in the old picture.
  • Forgetting the extra term −iℏR†R˙-i\hbar R^\dagger\dot R for time-dependent R(t)R(t).
  • Using the right formula with the opposite convention for R(t)R(t).
  • Treating the interaction picture as approximate before any expansion is truncated.
  • Confusing explicit time dependence in AS(t)A_S(t) with time dependence induced by R(t)R(t).
  • Calling a nonunitary similarity transformation a change of picture without checking the inner product and expectation values.
  • Dropping picture labels in a calculation where several transformed objects appear.

References

Section titled “References”
  • C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
  • P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
  • H. Goldstein, C. Poole, and J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley, 2002.
  • 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.

Exercises

Section titled “Exercises”
  1. Derive the transformed Hamiltonian for the convention ∣ψP⟩=R†∣ψS⟩\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle.
Solution

Differentiate:

ddt∣ψP⟩=R˙†∣ψS⟩+R†ddt∣ψS⟩.\frac{d}{dt}\lvert\psi_P\rangle = \dot R^\dagger\lvert\psi_S\rangle + R^\dagger\frac{d}{dt}\lvert\psi_S\rangle.

Use ∣ψS⟩=R∣ψP⟩\lvert\psi_S\rangle=R\lvert\psi_P\rangle and

iℏddt∣ψS⟩=HS∣ψS⟩.i\hbar\frac{d}{dt}\lvert\psi_S\rangle=H_S\lvert\psi_S\rangle.

Then

iℏddt∣ψP⟩=(iℏR˙†R+R†HSR)∣ψP⟩.i\hbar\frac{d}{dt}\lvert\psi_P\rangle = \left( i\hbar\dot R^\dagger R + R^\dagger H_SR \right) \lvert\psi_P\rangle.

Since R˙†R=−R†R˙\dot R^\dagger R=-R^\dagger\dot R,

HP=R†HSR−iℏR†R˙.H_P=R^\dagger H_SR-i\hbar R^\dagger\dot R.
  1. Show that choosing R(t)=U(t,t0)R(t)=U(t,t_0) makes the Heisenberg state time independent.
Solution

With R=UR=U,

∣ψP(t)⟩=U†(t,t0)∣ψS(t)⟩.\lvert\psi_P(t)\rangle = U^\dagger(t,t_0)\lvert\psi_S(t)\rangle.

But ∣ψS(t)⟩=U(t,t0)∣ψS(t0)⟩\lvert\psi_S(t)\rangle=U(t,t_0)\lvert\psi_S(t_0)\rangle, so

∣ψP(t)⟩=U†(t,t0)U(t,t0)∣ψS(t0)⟩=∣ψS(t0)⟩.\lvert\psi_P(t)\rangle = U^\dagger(t,t_0)U(t,t_0)\lvert\psi_S(t_0)\rangle = \lvert\psi_S(t_0)\rangle.

This fixed state is the Heisenberg state.

  1. Let R(t)=e−iωtG/ℏR(t)=e^{-i\omega tG/\hbar}. Compute the extra term in the transformed Hamiltonian.
Solution

Differentiate:

R˙(t)=−iωℏGR(t).\dot R(t) = -\frac{i\omega}{\hbar}GR(t).

Then

R†R˙=−iωℏG,R^\dagger\dot R = -\frac{i\omega}{\hbar}G,

assuming GG commutes with its own exponential. Therefore

−iℏR†R˙=−ωG.-i\hbar R^\dagger\dot R = -\omega G.
  1. Under a unitary picture transformation, why must observables transform as AP=R†ASRA_P=R^\dagger A_SR if states transform as ∣ψP⟩=R†∣ψS⟩\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle?
Solution

The expectation value should be invariant:

⟨ψP∣AP∣ψP⟩=⟨ψS∣AS∣ψS⟩.\langle\psi_P\rvert A_P\lvert\psi_P\rangle = \langle\psi_S\rvert A_S\lvert\psi_S\rangle.

Since ∣ψP⟩=R†∣ψS⟩\lvert\psi_P\rangle=R^\dagger\lvert\psi_S\rangle, one has ⟨ψP∣=⟨ψS∣R\langle\psi_P\rvert=\langle\psi_S\rvert R. Choosing

AP=R†ASRA_P=R^\dagger A_SR

gives

⟨ψP∣AP∣ψP⟩=⟨ψS∣RR†ASRR†∣ψS⟩=⟨ψS∣AS∣ψS⟩.\langle\psi_P\rvert A_P\lvert\psi_P\rangle = \langle\psi_S\rvert RR^\dagger A_SRR^\dagger\lvert\psi_S\rangle = \langle\psi_S\rvert A_S\lvert\psi_S\rangle.