--- title: "Interaction picture" sort_title: "Interaction picture" date: 2021-09-13 categories: - Physics - Quantum mechanics layout: "concept" --- The **interaction picture** or **Dirac picture** is an alternative formulation of quantum mechanics, equivalent to both the Schrödinger picture and the [Heisenberg picture](/know/concept/heisenberg-picture/). Recall that in the Schrödinger picture, the states $$\Ket{\psi_S(t)}$$ evolve in time, but time-independent operators $$\hat{L}_S$$ are fixed. Meanwhile in the Heisenberg picture, the states $$\Ket{\psi_H}$$ are constant, and all time dependence is on the operators $$\hat{L}_H(t)$$ instead. In the interaction picture, both the states $$\Ket{\psi_I(t)}$$ and the operators $$\hat{L}_I(t)$$ evolve in $$t$$. This may seem unnecessarily complicated, but it turns out to be convenient when considering a system with a time-dependent "perturbation" $$\hat{H}_{1,S}$$ to a time-independent Hamiltonian $$\hat{H}_{0,S}$$: $$\begin{aligned} \hat{H}_S(t) = \hat{H}_{0,S} + \hat{H}_{1,S}(t) \end{aligned}$$ Despite being called a perturbation, $$\hat{H}_{1, S}$$ need not be weak compared to $$\hat{H}_{0, S}$$. Basically, any way of splitting $$\hat{H}_S$$ is valid as long as $$\hat{H}_{0, S}$$ is time-independent, but only a few ways are useful. We now define the unitary conversion operator $$\hat{U}_0(t)$$ as shown below. Note its similarity to the [time-evolution operator](/know/concept/time-evolution-operator/) $$\hat{K}_S(t)$$: $$\begin{aligned} \boxed{ \hat{U}_0(t) \equiv \exp\!\bigg( \!-\! \frac{i}{\hbar} \hat{H}_{0,S} t \bigg) } \end{aligned}$$ The interaction-picture states $$\Ket{\psi_I(t)}$$ and operators $$\hat{L}_I(t)$$ are then defined as follows: $$\begin{aligned} \boxed{ \Ket{\psi_I(t)} \equiv \hat{U}_0^\dagger(t) \Ket{\psi_S(t)} } \qquad\qquad \boxed{ \hat{L}_I(t) \equiv \hat{U}_0^\dagger(t) \: \hat{L}_S(t) \: \hat{U}{}_0(t) } \end{aligned}$$ Because $$\hat{H}_{0, S}$$ is time-independent, it commutes with $$\hat{U}_0$$, so conveniently $$\hat{H}_{0, I} = \hat{H}_{0, S}$$. ## Equations of motion To find the equation of motion for $$\Ket{\psi_I(t)}$$, we differentiate it and multiply by $$i \hbar$$: $$\begin{aligned} i \hbar \dv{}{t} \Ket{\psi_I} &= i \hbar \dv{\hat{U}_0^\dagger}{t} \Ket{\psi_S} + \hat{U}_0^\dagger \bigg( i \hbar \dv{}{t}\Ket{\psi_S} \bigg) \end{aligned}$$ We insert the definition of $$\hat{U}_0$$ in the first term and the Schrödinger equation into the second, and use the fact that $$\comm{\hat{H}_{0, S}}{\hat{U}_0} = 0$$ thanks to the time-independence of $$\hat{H}_{0, S}$$: $$\begin{aligned} i \hbar \dv{}{t} \Ket{\psi_I} &= - \hat{H}_{0,S} \hat{U}_0^\dagger \Ket{\psi_S} + \hat{U}_0^\dagger \hat{H}_S \Ket{\psi_S} \\ &= \hat{U}_0^\dagger \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \Ket{\psi_S} \\ &= \hat{U}_0^\dagger \hat{H}_{1,S} \big( \hat{U}_0 \hat{U}_0^\dagger \big) \Ket{\psi_S} \end{aligned}$$ Which leads to an analogue of the Schrödinger equation, with $$\hat{H}_{1,I} = \hat{U}_0^\dagger \hat{H}_{1,S} \hat{U}_0$$: $$\begin{aligned} \boxed{ i \hbar \dv{}{t} \Ket{\psi_I(t)} = \hat{H}_{1,I}(t) \Ket{\psi_I(t)} } \end{aligned}$$ Next, we do the same with an operator $$\hat{L}_I$$ in order to describe its evolution in time: $$\begin{aligned} \dv{\hat{L}_I}{t} &= \dv{\hat{U}_0^\dagger}{t} \hat{L}_S \hat{U}_0 + \hat{U}_0^\dagger \hat{L}_S \dv{\hat{U}_0}{t} + \hat{U}_0^\dagger \dv{\hat{L}_S}{t} \hat{U}_0 \\ &= \frac{i}{\hbar} \hat{U}_0^\dagger \hat{H}_{0,S} \big( \hat{U}_0 \hat{U}_0^\dagger \big) \hat{L}_S \hat{U}_0 - \frac{i}{\hbar} \hat{U}_0^\dagger \hat{L}_S \big( \hat{U}_0 \hat{U}_0^\dagger \big) \hat{H}_{0,S} \hat{U}_0 + \bigg( \dv{\hat{L}_S}{t} \bigg)_I \\ &= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I - \frac{i}{\hbar} \hat{L}_I \hat{H}_{0,I} + \bigg( \dv{\hat{L}_S}{t} \bigg)_I \end{aligned}$$ The result is analogous to the equation of motion in the Heisenberg picture: $$\begin{aligned} \boxed{ \dv{}{t} \hat{L}_I(t) = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \bigg( \dv{}{t}\hat{L}_S(t) \bigg)_I } \end{aligned}$$ In other words, in the interaction picture, the "simple" time-dependence (from $$\hat{H}_{0, S}$$) is given to the operators, and the "complicated" dependence (from $$\hat{H}_{1, S}$$) 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 $$\hat{K}_S(t)$$? Its interaction version $$\hat{K}_I(t)$$ is unsurprisingly obtained by the standard transform $$\hat{K}_I = \hat{U}_0^\dagger \hat{K}_S \hat{U}_0$$: $$\begin{aligned} \Ket{\psi_I(t)} &= \hat{U}_0^\dagger(t) \Ket{\psi_S(t)} \\ &= \hat{U}_0^\dagger(t) \: \hat{K}_S(t) \Ket{\psi_S(0)} \\ &= \hat{U}_0^\dagger(t) \: \hat{K}_S(t) \: \hat{U}_0(t) \: \hat{U}_0^\dagger(t) \Ket{\psi_S(0)} \\ &\equiv \hat{K}_I(t) \Ket{\psi_I(0)} \end{aligned}$$ But we can do better. By inserting this definition of $$\hat{K}_I$$ into the interaction picture's analogue of Schrödinger's equation, we get the following relation for $$\hat{K}_I$$: $$\begin{aligned} i \hbar \dv{}{t} \hat{K}_I(t) &= \hat{H}_{1,I}(t) \: \hat{K}_I(t) \end{aligned}$$ In other words, $$\hat{K}_I$$ can be said to also obey the standard equation of motion for states, despite being an operator. We integrate both sides and use $$\hat{K}_I(0) = 1$$: $$\begin{aligned} K_I(t) = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \: \hat{K}_I(\tau) \dd{\tau} \end{aligned}$$ This equation can be recursively inserted into itself forever. We recognize the resulting so called *Dyson series* from the derivation of $$\hat{K}_S(t)$$ for time-dependent Hamiltonians in the Schrödinger picture ([given here](/know/concept/time-evolution-operator/)), so we know that the result is given by: $$\begin{aligned} \boxed{ \hat{K}_I(t) = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_0^t \hat{H}_{1,I}(\tau) \dd{\tau} \bigg) \bigg\} } \end{aligned}$$ Where $$\mathcal{T}$$ is the [time-ordering meta-operator](/know/concept/time-ordered-product/), which is conventionally written in this way to say that it applies to the terms of a Taylor expansion of $$\exp(x)$$. This means that the evolution of a quantum state in the interaction picture is determined by the perturbation $$\hat{H}_{1, I}$$. ## References 1. H. Bruus, K. Flensberg, *Many-body quantum theory in condensed matter physics*, 2016, Oxford.