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---
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.
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