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---
title: "Interaction picture"
firstLetter: "I"
publishDate: 2021-09-13
categories:
- Physics
- Quantum mechanics
date: 2021-09-09T21:15:37+02:00
draft: false
markup: pandoc
---
# 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](/know/concept/heisenberg-picture/).
Recall that Schrödinger lets states $\ket{\psi_S(t)}$ evolve in time,
but keeps operators $\hat{L}_S$ fixed (except for explicit time dependence).
Meanwhile, Heisenberg keeps states $\ket{\psi_H}$ fixed,
and puts all time dependence on the operators $\hat{L}_H(t)$.
However, in the interaction picture,
both the states $\ket{\psi_I(t)}$ and the operators $\hat{L}_I(t)$
evolve in $t$.
This might seem unnecessarily complicated,
but it turns out be convenient when considering
a time-dependent "perturbation" $\hat{H}_{1,S}$
to a time-independent Hamiltonian $\hat{H}_{0,S}$:
$$\begin{aligned}
\hat{H}_S(t)
= \hat{H}_{0,S} + \hat{H}_{1,S}(t)
\end{aligned}$$
With $\hat{H}_S(t)$ the full Schrödinger Hamiltonian.
We define the unitary conversion operator:
$$\begin{aligned}
\hat{U}(t)
\equiv \exp\!\Big( i \frac{\hat{H}_{0,S} t}{\hbar} \Big)
\end{aligned}$$
The interaction-picture states $\ket{\psi_I(t)}$ and operators $\hat{L}_I(t)$
are then defined to be:
$$\begin{aligned}
\boxed{
\ket{\psi_I(t)}
\equiv \hat{U}(t) \ket{\psi_S(t)}
\qquad
\hat{L}_I(t)
\equiv \hat{U}(t) \: \hat{L}_S(t) \: \hat{U}{}^\dagger(t)
}
\end{aligned}$$
## Equations of motion
To find the equation of motion for $\ket{\psi_I(t)}$,
we differentiate it and multiply by $i \hbar$:
$$\begin{aligned}
i \hbar \dv{t} \ket{\psi_I}
&= i \hbar \Big( \dv{\hat{U}}{t} \ket{\psi_S} + \hat{U} \dv{t} \ket{\psi_S} \Big)
\\
&= i \hbar \Big( i \frac{\hat{H}_{0,S}}{\hbar} \Big) \hat{U} \ket{\psi_S} + \hat{U} \Big( i \hbar \dv{t} \ket{\psi_S} \Big)
\end{aligned}$$
We insert the Schrödinger equation into the second term,
and use $\comm*{\hat{U}}{\hat{H}_{0,S}} = 0$:
$$\begin{aligned}
i \hbar \dv{t} \ket{\psi_I}
&= - \hat{H}_{0,S} \hat{U} \ket{\psi_S} + \hat{U} \hat{H}_S \ket{\psi_S}
\\
&= \hat{U} \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \ket{\psi_S}
\\
&= \hat{U} \big( \hat{H}_{1,S} \big) \hat{U}{}^\dagger \hat{U} \ket{\psi_S}
\end{aligned}$$
Which leads to an analogue of the Schrödinger equation,
with $\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$:
$$\begin{aligned}
\boxed{
i \hbar \dv{t} \ket{\psi_I(t)}
= \hat{H}_{1,I}(t) \ket{\psi_I(t)}
}
\end{aligned}$$
Next, we do the same with an operator $\hat{L}_I$
to find a description of its evolution in time:
$$\begin{aligned}
\dv{t} \hat{L}_I
&= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t} + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger
\\
&= \frac{i}{\hbar} \hat{U} \hat{H}_{0,S} \big( \hat{U}{}^\dagger \hat{U} \big) \hat{L}_S \hat{U}{}^\dagger
- \frac{i}{\hbar} \hat{U} \hat{L}_S \big( \hat{U}{}^\dagger \hat{U} \big) \hat{H}_{0,S} \hat{U}{}^\dagger
+ \Big( \dv{\hat{L}_S}{t} \Big)_I
\\
&= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I
- \frac{i}{\hbar} \hat{L}_I \hat{H}_{0,I}
+ \Big( \dv{\hat{L}_S}{t} \Big)_I
= \frac{i}{\hbar} \comm*{\hat{H}_{0,I}}{\hat{L}_I} + \Big( \dv{\hat{L}_S}{t} \Big)_I
\end{aligned}$$
The result is analogous to the equation of motion in the Heisenberg picture:
$$\begin{aligned}
\boxed{
\dv{t} \hat{L}_I(t)
= \frac{i}{\hbar} \comm*{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \Big( \dv{t} \hat{L}_S(t) \Big)_I
}
\end{aligned}$$
## Time evolution operator
Recall that an alternative form of the Schrödinger equation is as follows,
where a **time evolution operator** or
**generator of translations in time** $K_S(t, t_0)$
brings $\ket{\psi_S}$ from time $t_0$ to $t$:
$$\begin{aligned}
\ket{\psi_S(t)}
= \hat{K}_S(t, t_0) \ket{\psi_S(t_0)}
\qquad \quad
\hat{K}_S(t, t_0)
\equiv \exp\!\Big( \!-\! i \frac{\hat{H}_S (t - t_0)}{\hbar} \Big)
\end{aligned}$$
We want to find an analogous operator in the interaction picture, satisfying:
$$\begin{aligned}
\ket{\psi_I(t)}
\equiv \hat{K}_I(t, t_0) \ket{\psi_I(t_0)}
\end{aligned}$$
Inserting this definition into the equation of motion for $\ket{\psi_I}$ yields
an equation for $\hat{K}_I$, with the logical boundary condition $\hat{K}_I(t_0, t_0) = 1$:
$$\begin{aligned}
i \hbar \dv{t} \Big( \hat{K}_I(t, t_0) \ket{\psi_I(t_0)} \Big)
&= \hat{H}_{1,I}(t) \Big( \hat{K}_I(t, t_0) \ket{\psi_I(t_0)} \Big)
\\
i \hbar \dv{t} \hat{K}_I(t, t_0)
&= \hat{H}_{1,I}(t) \hat{K}_I(t, t_0)
\end{aligned}$$
We turn this into an integral equation
by integrating both sides from $t_0$ to $t$:
$$\begin{aligned}
i \hbar \int_{t_0}^t \dv{t'} K_I(t', t_0) \dd{t'}
= \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'}
\end{aligned}$$
After evaluating the left integral,
we see an expression for $\hat{K}_I$ as a function of $\hat{K}_I$ itself:
$$\begin{aligned}
K_I(t, t_0)
= 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'}
\end{aligned}$$
By recursively inserting $\hat{K}_I$ once, we get a longer expression,
still with $\hat{K}_I$ on both sides:
$$\begin{aligned}
K_I(t, t_0)
= 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'}
+ \frac{1}{(i \hbar)^2} \int_{t_0}^t \hat{H}_{1,I}(t') \int_{t_0}^{t'} \hat{H}_{1,I}(t'') \hat{K}_I(t'', t_0) \dd{t''} \dd{t'}
\end{aligned}$$
And so on. Note the ordering of the integrals and integrands:
upon closer inspection, we see that the $n$th term is
a [time-ordered product](/know/concept/time-ordered-product/) $\mathcal{T}$
of $n$ factors $\hat{H}_{1,I}$:
$$\begin{aligned}
\hat{K}_I(t, t_0)
&= 1 + \int_{t_0}^t \hat{H}_{1,I}(t_1) \dd{t_1}
+ \frac{1}{2} \int_{t_0}^{t} \int_{t_0}^{t_1} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \hat{H}_{1,I}(t_2) \Big\} \dd{t_1} \dd{t_2}
+ \: ...
\\
&= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
\int_{t_0}^{t} \cdots \int_{t_0}^{t_n} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \cdots \hat{H}_{1,I}(t_n) \Big\} \dd{t_1} \cdots \dd{t_n}
\\
&= \sum_{n = 0}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
\mathcal{T} \bigg\{ \bigg( \int_{t_0}^{t} \hat{H}_{1,I}(t') \dd{t'} \bigg)^n \bigg\}
\end{aligned}$$
Here, we recognize the Taylor expansion of $\exp$,
leading us to a final expression for $\hat{K}_I$:
$$\begin{aligned}
\boxed{
\hat{K}_I(t, t_0)
= \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\}
}
\end{aligned}$$
## References
1. H. Bruus, K. Flensberg,
*Many-body quantum theory in condensed matter physics*,
2016, Oxford.
|
