path: root/source/know/concept/interaction-picture/index.md

---
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 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:

$$\begin{aligned}
    \boxed{
        \hat{U}(t)
        \equiv \exp\!\bigg( i \frac{\hat{H}_{0,S} t}{\hbar} \bigg)
    }
\end{aligned}$$

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 $$\hat{H}_{1,I}$$:

$$\begin{aligned}
    \hat{K}_I(t, t_0)
    &= 1 + \int_{t_0}^t \hat{H}_{1,I}(t_1) \dd{t_1}
    + \frac{1}{2} \int_{t_0}^{t} \int_{t_0}^{t_1} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \hat{H}_{1,I}(t_2) \Big\} \dd{t_1} \dd{t_2}
    + \: ...
    \\
    &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
    \int_{t_0}^{t} \cdots \int_{t_0}^{t_n} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \cdots \hat{H}_{1,I}(t_n) \Big\} \dd{t_1} \cdots \dd{t_n}
    \\
    &= \sum_{n = 0}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
    \mathcal{T} \bigg\{ \bigg( \int_{t_0}^{t} \hat{H}_{1,I}(t') \dd{t'} \bigg)^n \bigg\}
\end{aligned}$$

This construction is occasionally called the **Dyson series**.
We recognize the well-known Taylor expansion of $$\exp(x)$$,
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.