---
address: |
    Department of Mathematics\
    Hong Kong University of Science and Technology\
    Clear Water Bay, Kowloon, Hong Kong
author:
- Guowu Meng
title: Comment on Perturbation theory for Heisenberg Operators
---

\[defn\][Claim]{}

I would like to comment on a recent paper[@F] published in Phys. Rev. A. In that paper, the author has tried to establish the following main conclusion: [*Both the standard perturbation series and the expanded Magnus exponential operator lead to the result of Franson and Donegan [@FD] for Heisenberg operators*]{}. He establishes this conclusion through second order and presents a concrete example to convince the readers that the conclusion must be true in all orders. Here, I would like to point out that the author’s conclusion is indeed true in [*all*]{} orders. To prove this claim, one just need to observe that the formal perturbation series for Heisenberg operators is unique. The uniqueness is due to the fact that the initial value problem for a linear 1st ordinary differential equation has a unique (formal) solution.

Recall that the evolution operator is the unique solution for $$\begin{aligned}
\label{eqn}
\left\{
\begin{matrix}{dU\over dt}&=&\lambda AU\cr
U(0) &= &I\end{matrix} \right.\end{aligned}$$ where $I$ is the identity operator, $\lambda$ is an imaginary parameter $1\over i\hbar$, and $A$ is a hermitian operator (the Hamiltonian in the Schrodinger picture) and could be time dependent.

For any hermitian operator $\hat O$ in the quantum system, we let and . (Both $\tau(\lambda A)$ and $\rho_r(U)$ are operators on the space of hermitian operators of the quantum system.) Recall that (hermitian)operator $\hat O$ in the Schrodinger picture and the corresponding operator $\hat
O_H$ in the Heisenberg picture are related via equation $\hat
O_H=\rho_r(U)(\hat O)$. The initial value problem we mentioned early is $$\begin{aligned}
\label{eqnr}
\left\{
\begin{matrix}{d\over dt}\rho_r(U)&=& \rho_r(U)\tau(\lambda A)\cr
\rho_r(U(0)) &= &I.\end{matrix} \right.\end{aligned}$$ and it has the following [**unique**]{} formal perturbative solution: $$\begin{aligned}
\label{sol}
\rho_r(U(t))=I+\sum_{n\ge 1}\int_{\Delta_n(t)}d^nt\,\tau(\lambda
A(t_1))\cdots \tau(\lambda A(t_n)),\end{aligned}$$ where $\Delta_n(t)\equiv\{(t_1,\ldots,t_n)\in
\mathbb{R}^n\,|\,0\le t_1\le \cdots \le t_n\le t\}$. Consequently, by using the definition of $\tau(\lambda A)$, we have $\hat
O_H\equiv \rho_r(U(t))(\hat O)$ is equal to $$\begin{aligned}
\label{O}
\hskip -15pt & &\hat O+\sum_{n\ge 1}\int_{\Delta_n(t)}\,d^nt\,
[\cdots [\hat O,\lambda A(t_n)],\ldots,\lambda A(t_1)]\cr \hskip
-3pt&=&\hskip -3pt \hat O+\sum_{n\ge
1}\lambda^n\int_{\Delta_n(t)}\,d^nt\, [\cdots [\hat
O,A(t_n)],\ldots,A(t_1)].\end{aligned}$$ (Note that, equation (\[O\]) with $A=H$ and $\lambda={1\over
i\hbar}$, being identical to equation (6) in [@FD], is the main result of Franson and Donegan.)

In view of the uniqueness of the formal expansion for $\rho_r(U(t))$, it is now clear that the conclusion expected by F. M. Fernández is valid in all orders.

. To solve equation (\[eqnr\]), let $$\begin{aligned}
\label{fexp}
\rho_r(U(t))=u_0(t)+\lambda u_1(t)+\cdots=\sum \lambda^n u_n(t)\end{aligned}$$ be the formal perturbative expansion of $\rho_r(U(t))$ in powers of $\lambda$. Plugging (\[fexp\]) into (\[eqnr\]), we have $u_0(t)=I$ and a sequence of initial value problems $$\begin{aligned}
\label{eqns}
{d(\lambda u_n)\over dt}&=& u_{n-1}\tau(\lambda A)\cr \lambda
u_n(0) &=& 0\end{aligned}$$ Solving (\[eqns\]) inductively, the unique solution in equation (\[sol\]) is then obtained. Next, we note that, to arrive at equation (\[eqnr\]) we just need to verify from the definitions that 1) $\rho_r(U(0))(\hat O)=O$, 2)$\left({d\over
dt}\rho_r(U(t))\right)(\hat O)\equiv{d\over dt}\left(U^\dag(t)\hat
O U(t) \right)=\left(\rho_r(U(t))\tau(\lambda A(t))\right) (\hat
O)$. The actual verification is just a simple computation based on definitions and equation (\[eqn\]) and is omitted here due to the restriction of the length of this comment. (For those people who know representation theory, the quickest way to obtain equation (\[eqnr\]) is to apply the right adjoint representation to equation (\[eqn\]). In fact, the quickest way to obtain equation (\[O\]) is to apply the right adjoint representation to the formal perturbative expansion formula for solution of equation (\[eqn\]) — a formula in many standard textbook of quantum mechanics.)

[99]{}

F. M. Fernández, Phys. Rev. A [**67**]{}, 022104 (2003)

J.D. Franson and M.M. Donegan, Phys. Rev. A [**65**]{}, 052107 (2002)
