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+---
+title: "Time evolution operator"
+sort_title: "Time evolution operator"
+date: 2024-10-15
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+In general, given a system whose governing equation is known,
+the **time evolution operator** $$\hat{U}(t, t_0)$$
+transforms the state at time $$t_0$$ to the one at time $$t$$.
+Although not specific to it,
+this is most often used in quantum mechanics,
+as governed by the Schrödinger equation:
+
+$$\begin{aligned}
+    i \hbar \dv{}{t} \ket{\psi(t)}
+    = \hat{H}(t) \ket{\psi(t)}
+\end{aligned}$$
+
+Such that the definition of $$\hat{U}(t)$$ is as follows,
+where we have set $$t_0 = 0$$:
+
+$$\begin{aligned}
+    \ket{\psi(t)}
+    = \hat{U}(t) \ket{\psi(0)}
+\end{aligned}$$
+
+Clearly, $$\hat{U}(t)$$ must be unitary.
+The goal is to find an expression that satisfies this relation.
+
+
+
+## Time-independent Hamiltonian
+
+We start by inserting the definition of $$\hat{U}(t)$$
+into the Schrödinger equation:
+
+$$\begin{aligned}
+    \dv{}{t} \hat{U}(t) \ket{\psi(0)}
+    = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t) \ket{\psi(0)}
+\end{aligned}$$
+
+If we hide the state $$\ket{\psi(0)}$$,
+then $$\hat{U}(t)$$ can be said to satisfy the equation in its own right:
+
+$$\begin{aligned}
+    \dv{}{t} \hat{U}(t)
+    = - \frac{i}{\hbar} \hat{H} \: \hat{U}(t)
+\end{aligned}$$
+
+If the Hamiltonian $$\hat{H}$$ is time-independent,
+this is straightforward to integrate, yielding:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{U}(t)
+        = \exp\!\bigg( \!-\! \frac{i}{\hbar} t \hat{H} \bigg)
+    }
+\end{aligned}$$
+
+And the generalization to $$t_0 \neq 0$$ is trivial,
+since we can just shift the time axis:
+
+$$\begin{aligned}
+    \hat{U}(t, t_0)
+    = \exp\!\bigg( \!-\! \frac{i}{\hbar} (t - t_0) \hat{H} \bigg)
+\end{aligned}$$
+
+
+
+## Time-dependent Hamiltonian
+
+Even when $$\hat{H}$$ is time-dependent,
+$$\hat{U}(t)$$ can be said to satisfy the Schrödinger equation:
+
+$$\begin{aligned}
+    \dv{}{t} \hat{U}(t)
+    = - \frac{i}{\hbar} \hat{H}(t) \: \hat{U}(t)
+\end{aligned}$$
+
+Integrating from $$0$$ to $$t$$,
+and using $$\hat{U}(0) = 1$$ (which should be clear from its definition):
+
+$$\begin{aligned}
+    \hat{U}(t)
+    = 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \: \hat{U}(\tau_1) \dd{\tau_1}
+\end{aligned}$$
+
+This is a self-consistent equation for $$\hat{U}(t)$$.
+We can recursively insert it into itself, yielding:
+
+$$\begin{aligned}
+    \hat{U}(t)
+    &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1)
+    \bigg( 1 + \frac{1}{i \hbar} \int_0^{\tau_1} \hat{H}(\tau_2) \: \hat{U}(\tau_2) \dd{\tau_2} \bigg) \dd{\tau_1}
+    \\
+    &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1}
+    + \frac{1}{(i \hbar)^2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \: \hat{U}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    \\
+    &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1}
+    + \frac{1}{(i \hbar)^2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    + \frac{1}{(i \hbar)^3} \int_0^t \cdots \: \dd{\tau_1}
+\end{aligned}$$
+
+And so on.
+Let us take a closer look at the third (i.e. second-order) term in this series,
+noting that the integrals are ordered such that $$\tau_2 < \tau_1$$ always.
+We can exploit this fact to introduce several
+[Heaviside step functions](/know/concept/heaviside-step-function/) $$\Theta(t)$$:
+
+$$\begin{aligned}
+    &\quad \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    \\
+    &= \frac{1}{2} \int_0^t \hat{H}(\tau_1) \int_0^{\tau_1} \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    + \frac{1}{2} \int_0^t \hat{H}(\tau_2) \int_0^{\tau_2} \hat{H}(\tau_1) \dd{\tau_1} \dd{\tau_2}
+    \\
+    &= \frac{1}{2} \int_0^t \! \hat{H}(\tau_1)
+    \int_0^{\tau_1} \! \Theta(\tau_1 \!-\! \tau_2) \hat{H}(\tau_2) \dd{\tau_2} \dd{\tau_1}
+    + \frac{1}{2} \int_0^t \! \hat{H}(\tau_2)
+    \int_0^{\tau_2} \! \Theta(\tau_2 \!-\! \tau_1) \hat{H}(\tau_1) \dd{\tau_1} \dd{\tau_2}
+    \\
+    &= \frac{1}{2} \int_0^t \int_0^t
+    \bigg( \Theta(\tau_1 \!-\! \tau_2) \: \hat{H}(\tau_1) \: \hat{H}(\tau_2)
+    + \Theta(\tau_2 \!-\! \tau_1) \: \hat{H}(\tau_1) \: \hat{H}(\tau_2) \bigg) \dd{\tau_1} \dd{\tau_2}
+    \\
+    &= \frac{1}{2} \int_0^t \int_0^t \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \Big\} \dd{\tau_1} \dd{\tau_2}
+\end{aligned}$$
+
+Where we have recognized the
+[time-ordering meta-operator](/know/concept/time-ordered-product/) $$\mathcal{T}$$.
+The above procedure is easy to generalize to the higher-order terms,
+so we arrive at the following expression for $$\hat{U}(t)$$:
+
+$$\begin{aligned}
+    \hat{U}(t)
+    &= 1 + \frac{1}{i \hbar} \int_0^t \hat{H}(\tau_1) \dd{\tau_1}
+    + \frac{1}{2} \frac{1}{(i \hbar)^2} \iint_0^t \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \Big\} \dd{\tau_2} \dd{\tau_1}
+    \\
+    &\qquad+ \frac{1}{6} \frac{1}{(i \hbar)^3} \iiint_0^t
+    \mathcal{T} \Big\{ \hat{H}(\tau_1) \: \hat{H}(\tau_2) \: \hat{H}(\tau_3) \Big\} \dd{\tau_3} \dd{\tau_2} \dd{\tau_1}
+    + \: ...
+    \\
+    &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n} \int_0^t \!\cdots\! \int_0^t
+    \mathcal{T} \Big\{ \hat{H}(\tau_1) \cdots \hat{H}(\tau_n) \Big\} \dd{\tau_n} \cdots \dd{\tau_1}
+\end{aligned}$$
+
+This result is sometimes called a **Dyson series**.
+Convention allows us to write it as follows,
+despite such a use of $$\mathcal{T}$$ looking a bit strange:
+
+$$\begin{aligned}
+    \hat{U}(t)
+    &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
+    \mathcal{T} \bigg\{ \bigg( \int_0^t \hat{H}(\tau) \dd{\tau} \bigg)^n \bigg\}
+\end{aligned}$$
+
+Here, we recognize the Taylor expansion of $$\exp(x)$$,
+leading us to the desired result:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{U}(t)
+        = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{i}{\hbar} \int_0^t \hat{H}(\tau) \dd{\tau} \bigg) \bigg\}
+    }
+\end{aligned}$$
+
+Where once again $$\mathcal{T}$$ is being used according to convention.
+Finally, the time axis can be shifted arbitrarily,
+so many authors write the evolution operator from $$t_0$$ to $$t$$ as $$\hat{U}(t, t_0)$$:
+
+$$\begin{aligned}
+    \hat{U}(t, t_0)
+    = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{i}{\hbar} \int_{t_0}^t \hat{H}(\tau) \dd{\tau} \bigg) \bigg\}
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.