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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.
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