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+---
+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.
+