I have a question concerning the time ordering operator. Let's suppose we have a time evolution generated by some Hamiltonian $H(t)$ given by $$ U(t)=T_\leftarrow\exp\left(-\mathrm{i}\int_0^t\mathrm{d}s\,H(s)\right)\tag{1}. $$ In Breuer and Petruccione, it's said that if the commutator of the Hamiltonian at some time $t$ with itself at some other time $t^\prime$ is a $c$-number function/ a complex function, i.e. $$\left[H(t),\,H(t^\prime)\right]=f(t,t'),$$ then the time evolution is given by $$ U(t)=\exp\left(-\frac{1}{2}\int_0^t\mathrm{d}s\int_0^t\mathrm{d}s^\prime\,\left[H(s),\,H(s^\prime)\right]\Theta(s-s^\prime)\right)$$ $$\times \exp\left(-\mathrm{i}\int_0^t\mathrm{ds}\,H(s)\right),\tag{2} $$ where $\Theta(s-s^\prime)$ is the Heaviside function. No proof or reference is given and I could not find any explanation anywhere, this is why I ask here. Any help would be very appreciated.


2 Answers 2


You are looking for a mathematical result known as the Magnus expansion. In general, this gives an exact representation of the time-ordered matrix exponential $$V(t) = {\rm T} \exp \left( \int_0^t dt' \, A(t')\right),$$ in terms of an equivalent ordinary exponential $$ V(t) = \exp\left (S(t)\right),$$ where $S(t)$ can be expressed as an infinite series of nested commutators, $S(t) = \sum_{n=1}^\infty S_n(t)$, e.g.\begin{align} S_1 & = \int_0^t dt_1\, A(t_1),\\ S_2 & = \frac{1}{2}\int_0^t dt_1 \int_0^{t_1} dt_2\, [A(t_1),A(t_2)],\\ & \vdots \end{align} The next terms in the expansion involve higher-order commutators like $[A(t_3),[A(t_1),A(t_2)]]$, which obviously vanish when $[A(t_1),A(t_2)]$ is a $c$-number. See Blanes et al., Physics Reports 470 (2009), 151-238 for further details.


OP's formula (2) is a continuum version of

$$\begin{align} &\exp(A_n)\ldots \exp(A_1)\cr &~=~\exp\left(\sum_{i\in\{1, \ldots, n\}} A_i + \frac{1}{2}\sum_{i,j\in\{1, \ldots, n\}}^{i>j} [A_i,A_j]\right),\end{align}\tag{A}$$

or equivalently,

$$\begin{align} & \exp(A_1)\ldots \exp(A_n)\cr &~=~\exp\left(\sum_{i\in\{1, \ldots, n\}} A_i + \frac{1}{2}\sum_{i,j\in\{1, \ldots, n\}}^{i<j} [A_i,A_j]\right),\end{align}\tag{B}$$

which are valid if we assume that

$$ \forall i,j,k~\in~\{1, \ldots, n\}: [[A_i,A_j],A_k]~=~0. \tag{C} $$

Eq. (B) in turn follows by repeated application of the truncated BCH formula:

$$\begin{align} e^Ae^B~=~&e^{A+B+\frac{C}{2}}, \cr \qquad \text{with}\qquad &C~\equiv~[A,B],\cr \qquad \text{if}\qquad & [A,C]~=~0~=~[B,C].\end{align} \tag{D}$$


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