Categories: Physics, Quantum mechanics.

Interaction picture

The interaction picture or Dirac picture is an alternative formulation of quantum mechanics, equivalent to both the Schrödinger picture and the Heisenberg picture.

Recall that in the Schrödinger picture, the states ∣ψS(t)⟩\Ket{\psi_S(t)} evolve in time, but time-independent operators L^S\hat{L}_S are fixed. Meanwhile in the Heisenberg picture, the states ∣ψH⟩\Ket{\psi_H} are constant, and all time dependence is on the operators L^H(t)\hat{L}_H(t) instead.

In the interaction picture, both the states ∣ψI(t)⟩\Ket{\psi_I(t)} and the operators L^I(t)\hat{L}_I(t) evolve in tt. This may seem unnecessarily complicated, but it turns out to be convenient when considering a system with a time-dependent “perturbation” H^1,S\hat{H}_{1,S} to a time-independent Hamiltonian H^0,S\hat{H}_{0,S}:

H^S(t)=H^0,S+H^1,S(t)\begin{aligned} \hat{H}_S(t) = \hat{H}_{0,S} + \hat{H}_{1,S}(t) \end{aligned}

Despite being called a perturbation, H^1,S\hat{H}_{1, S} need not be weak compared to H^0,S\hat{H}_{0, S}. Basically, any way of splitting H^S\hat{H}_S is valid as long as H^0,S\hat{H}_{0, S} is time-independent, but only a few ways are useful.

We now define the unitary conversion operator U^0(t)\hat{U}_0(t) as shown below. Note its similarity to the time-evolution operator K^S(t)\hat{K}_S(t):

U^0(t)≡exp⁡ ⁣( ⁣− ⁣iℏH^0,St)\begin{aligned} \boxed{ \hat{U}_0(t) \equiv \exp\!\bigg( \!-\! \frac{i}{\hbar} \hat{H}_{0,S} t \bigg) } \end{aligned}

The interaction-picture states ∣ψI(t)⟩\Ket{\psi_I(t)} and operators L^I(t)\hat{L}_I(t) are then defined as follows:

∣ψI(t)⟩≡U^0†(t)∣ψS(t)⟩L^I(t)≡U^0†(t) L^S(t) U^0(t)\begin{aligned} \boxed{ \Ket{\psi_I(t)} \equiv \hat{U}_0^\dagger(t) \Ket{\psi_S(t)} } \qquad\qquad \boxed{ \hat{L}_I(t) \equiv \hat{U}_0^\dagger(t) \: \hat{L}_S(t) \: \hat{U}{}_0(t) } \end{aligned}

Because H^0,S\hat{H}_{0, S} is time-independent, it commutes with U^0\hat{U}_0, so conveniently H^0,I=H^0,S\hat{H}_{0, I} = \hat{H}_{0, S}.

Equations of motion

To find the equation of motion for ∣ψI(t)⟩\Ket{\psi_I(t)}, we differentiate it and multiply by iℏi \hbar:

iℏddt∣ψI⟩=iℏdU^0†dt∣ψS⟩+U^0†(iℏddt∣ψS⟩)\begin{aligned} i \hbar \dv{}{t} \Ket{\psi_I} &= i \hbar \dv{\hat{U}_0^\dagger}{t} \Ket{\psi_S} + \hat{U}_0^\dagger \bigg( i \hbar \dv{}{t}\Ket{\psi_S} \bigg) \end{aligned}

We insert the definition of U^0\hat{U}_0 in the first term and the Schrödinger equation into the second, and use the fact that [H^0,S,U^0]=0\comm{\hat{H}_{0, S}}{\hat{U}_0} = 0 thanks to the time-independence of H^0,S\hat{H}_{0, S}:

iℏddt∣ψI⟩=−H^0,SU^0†∣ψS⟩+U^0†H^S∣ψS⟩=U^0†( ⁣− ⁣H^0,S+H^S)∣ψS⟩=U^0†H^1,S(U^0U^0†)∣ψS⟩\begin{aligned} i \hbar \dv{}{t} \Ket{\psi_I} &= - \hat{H}_{0,S} \hat{U}_0^\dagger \Ket{\psi_S} + \hat{U}_0^\dagger \hat{H}_S \Ket{\psi_S} \\ &= \hat{U}_0^\dagger \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \Ket{\psi_S} \\ &= \hat{U}_0^\dagger \hat{H}_{1,S} \big( \hat{U}_0 \hat{U}_0^\dagger \big) \Ket{\psi_S} \end{aligned}

Which leads to an analogue of the Schrödinger equation, with H^1,I=U^0†H^1,SU^0\hat{H}_{1,I} = \hat{U}_0^\dagger \hat{H}_{1,S} \hat{U}_0:

iℏddt∣ψI(t)⟩=H^1,I(t)∣ψI(t)⟩\begin{aligned} \boxed{ i \hbar \dv{}{t} \Ket{\psi_I(t)} = \hat{H}_{1,I}(t) \Ket{\psi_I(t)} } \end{aligned}

Next, we do the same with an operator L^I\hat{L}_I in order to describe its evolution in time:

dL^Idt=dU^0†dtL^SU^0+U^0†L^SdU^0dt+U^0†dL^SdtU^0=iℏU^0†H^0,S(U^0U^0†)L^SU^0−iℏU^0†L^S(U^0U^0†)H^0,SU^0+(dL^Sdt)I=iℏH^0,IL^I−iℏL^IH^0,I+(dL^Sdt)I\begin{aligned} \dv{\hat{L}_I}{t} &= \dv{\hat{U}_0^\dagger}{t} \hat{L}_S \hat{U}_0 + \hat{U}_0^\dagger \hat{L}_S \dv{\hat{U}_0}{t} + \hat{U}_0^\dagger \dv{\hat{L}_S}{t} \hat{U}_0 \\ &= \frac{i}{\hbar} \hat{U}_0^\dagger \hat{H}_{0,S} \big( \hat{U}_0 \hat{U}_0^\dagger \big) \hat{L}_S \hat{U}_0 - \frac{i}{\hbar} \hat{U}_0^\dagger \hat{L}_S \big( \hat{U}_0 \hat{U}_0^\dagger \big) \hat{H}_{0,S} \hat{U}_0 + \bigg( \dv{\hat{L}_S}{t} \bigg)_I \\ &= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I - \frac{i}{\hbar} \hat{L}_I \hat{H}_{0,I} + \bigg( \dv{\hat{L}_S}{t} \bigg)_I \end{aligned}

The result is analogous to the equation of motion in the Heisenberg picture:

ddtL^I(t)=iℏ[H^0,I(t),L^I(t)]+(ddtL^S(t))I\begin{aligned} \boxed{ \dv{}{t} \hat{L}_I(t) = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_I } \end{aligned}

In other words, in the interaction picture, the “simple” time-dependence (from H^0,S\hat{H}_{0, S}) is given to the operators, and the “complicated” dependence (from H^1,S\hat{H}_{1, S}) to the states. This means that the difficult part of a problem can be solved in isolation in a kind of Schrödinger picture.

Time evolution operator

What about the time evolution operator K^S(t)\hat{K}_S(t)? Its interaction version K^I(t)\hat{K}_I(t) is unsurprisingly obtained by the standard transform K^I=U^0†K^SU^0\hat{K}_I = \hat{U}_0^\dagger \hat{K}_S \hat{U}_0:

∣ψI(t)⟩=U^0†(t)∣ψS(t)⟩=U^0†(t) K^S(t)∣ψS(0)⟩=U^0†(t) K^S(t) U^0(t) U^0†(t)∣ψS(0)⟩≡K^I(t)∣ψI(0)⟩\begin{aligned} \Ket{\psi_I(t)} &= \hat{U}_0^\dagger(t) \Ket{\psi_S(t)} \\ &= \hat{U}_0^\dagger(t) \: \hat{K}_S(t) \Ket{\psi_S(0)} \\ &= \hat{U}_0^\dagger(t) \: \hat{K}_S(t) \: \hat{U}_0(t) \: \hat{U}_0^\dagger(t) \Ket{\psi_S(0)} \\ &\equiv \hat{K}_I(t) \Ket{\psi_I(0)} \end{aligned}

But we can do better. By inserting this definition of K^I\hat{K}_I into the interaction picture’s analogue of Schrödinger’s equation, we get the following relation for K^I\hat{K}_I:

iℏddtK^I(t)=H^1,I(t) K^I(t)\begin{aligned} i \hbar \dv{}{t} \hat{K}_I(t) &= \hat{H}_{1,I}(t) \: \hat{K}_I(t) \end{aligned}

In other words, K^I\hat{K}_I can be said to also obey the standard equation of motion for states, despite being an operator. We integrate both sides and use K^I(0)=1\hat{K}_I(0) = 1:

KI(t)=1+1iℏ∫0tH^1,I(τ) K^I(τ)dτ\begin{aligned} K_I(t) = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \: \hat{K}_I(\tau) \dd{\tau} \end{aligned}

This equation can be recursively inserted into itself forever. We recognize the resulting so called Dyson series from the derivation of K^S(t)\hat{K}_S(t) for time-dependent Hamiltonians in the Schrödinger picture (given here), so we know that the result is given by:

K^I(t)=T{exp⁡ ⁣(1iℏ∫0tH^1,I(τ)dτ)}\begin{aligned} \boxed{ \hat{K}_I(t) = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \dd{\tau} \bigg) \bigg\} } \end{aligned}

Where T\mathcal{T} is the time-ordering meta-operator, which is conventionally written in this way to say that it applies to the terms of a Taylor expansion of exp⁡(x)\exp(x). This means that the evolution of a quantum state in the interaction picture is determined by the perturbation H^1,I\hat{H}_{1, I}.

References

  1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.