Dyson Series
- Quantum Mechanics 1.12
· 8 min 
from 「Modern Quantum Mechanics」: Sakurai, J. J.
The position and momentum operators.
Time Propagator
As we have seen here, the time evolution of a state is given by the time propagator if the hamiltonian of the system is time-independent. The time propagator is given by
\[ \hat{\mathcal{U}}(t,t_0) = \exp\left[ -\frac{i(t-t_0)}{\hbar} \hat{H} \right] \]
If the hamiltonian is time-dependent but still commutes with itself at different times, the following equation holds.
\[ \begin{gather*} \hat{G}(t,t_0) \coloneqq -\frac{i}{\hbar} \int_{t_0}^t \hat{H}(t\rq) \d{t\rq} \nt \comm{\hat{H}(t)}{\hat{G}(t)} = 0 \;\Rightarrow\; \Psi(t) = \exp\left( \hat{G}(t,t_0) \right) \Psi(t_0) \end{gather*} \]
This can be proved easily. However, the equation is not always true because $\hat{H}(t)$ and $\hat{G}(t)$ do not always commute. Therefore, we need to use the time-ordering operator $\mathcal{T}$ to make the equation true, which is known as the Dyson series.
Dyson Series
The Dyson series is given by the following equation.
\[ \hat{\mathcal{U}}(t,t_0) = \sum_{n=0}^\infty \left( -\frac{i}{\hbar} \right)^n \int_{t_0}^t \d{t_1} \int_{t_0}^{t_1} \d{t_2} \cdots \int_{t_0}^{t_{n-1}} \d{t_n} \prod_{i=1}^n \hat{H}(t_i) \]
From the equation above, we can know that $t_i$ are ordered. Specifically, $t\ge t_1\ge t_2\ge \cdots $. Let’s see whether the series works well as a time propagator.
\[ \begin{align*} i\hbar\pdv{ }{t} \Psi(t) &= i\hbar\pdv{ }{t} \left[ \sum_{n=0}^\infty \left( -\frac{i}{\hbar} \right)^n \int_{t_0}^t \d{t_1} \int_{t_0}^{t_1} \d{t_2} \cdots \int_{t_0}^{t_{n-1}} \d{t_n} \prod_{i=1}^n \hat{H}(t_i) \right] \Psi(t_0) \nl &= \hat{H}(t) \left[ \sum_{n=1}^\infty \left( -\frac{i}{\hbar} \right)^{n-1} \int_{t_0}^t \d{t_2} \cdots \int_{t_0}^{t_{n-1}} \d{t_n} \prod_{i=2}^n \hat{H}(t_i) \right] \Psi(t_0) \nl &= \hat{H}(t) \left[ \sum_{n=0}^\infty \left( -\frac{i}{\hbar} \right)^n \int_{t_0}^t \d{t_1} \int_{t_0}^{t_1} \d{t_2} \cdots \int_{t_0}^{t_{n-1}} \d{t_n} \prod_{i=1}^n \hat{H}(t_i) \right] \Psi(t_0) \nl &= \hat{H}(t) \Psi(t) \end{align*} \]
The Schrödinger equation holds, so we have proved that we can write a time propagator in the form of the Dyson series.
Time-Ordering Operator
The time-ordering operator $\mathcal{T}$ is an operator that orders the operators in the order of time. It isn’t actually an operator, since it just rearranges the sequence of the product. However, it is often called a meta-operator. The time-ordering operator is defined as follows.
\[ \mathcal{T} \left[ \prod_{i=1}^n \hat{H}(t_i) \right] = \prod_{i=1}^n \hat{H}( t_{\sigma(i)} ) \quad ( t_{\sigma(1)} \ge \cdots \ge t_{\sigma(n)}) \]
where $\sigma$ is a permutation. While, we can express the dyson series simply using the time-ordering operator. Let’s first discover some interesting properties.
\[ \begin{align*} \mathcal{T} \left[ \hat{G}(t,t_0)^n \right] &= \left( -\frac{i}{\hbar} \right)^n \mathcal{T} \left[ \prod_{i=1}^n \int_{t_0}^t \hat{H}(t_i) \d{t_i} \right] \nl &= \left( -\frac{i}{\hbar} \right)^n \mathcal{T} \left[ \int_{t_0}^t \d{t_1} \cdots \int_{t_0}^t \d{t_n} \prod_{i=1}^n \hat{H}(t_i) \right] \nl &= n! \left( -\frac{i}{\hbar} \right)^n \int_{t_0}^t \d{t_1} \int_{t_0}^{t_1} \d{t_2} \cdots \int_{t_0}^{t_{n-1}} \d{t_n} \prod_{i=1}^n \hat{H}(t_i) \end{align*} \]
This is exactly the same as the $n$-th term of the Dyson series without the factor $n!$. Therefore, we can write the Dyson series as follows.
\[ \begin{align*} \hat{\mathcal{U}}(t,t_0) &= \sum_{n=0}^\infty \frac{1}{n!} \mathcal{T} \left[ \hat{G}(t,t_0)^n \right] \nl &= \mathcal{T} \left[ \sum_{n=0}^\infty \frac{1}{n!} \hat{G}(t,t_0)^n \right] \nl &= \mathcal{T} \exp\left( \hat{G}(t,t_0) \right) \end{align*} \]
Then we finally get the general form of the time propagator.
\[ \hat{\mathcal{U}}(t,t_0) = \mathcal{T} \exp\left( -\frac{i}{\hbar} \int_{t_0}^t \hat{H}(t\rq) \d{t\rq} \right) \]
