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Next: Constant perturbations Up: Approximation methods Previous: Spin magnetic resonance

The Dyson series

Let us now try to find approximate solutions of Eq. (740) for a general system. It is convenient to work in terms of the time evolution operator, $U(t_0, t)$, which is defined
\begin{displaymath}
\vert A, t_0, t\rangle = U(t_0, t)  \vert A\rangle.
\end{displaymath} (768)

Here, $\vert A, t_0, t\rangle$ is the state ket of the system at time $t$, given that the state ket at the initial time $t_0$ is $\vert A\rangle$. It is easily seen that the time evolution operator satisfies the differential equation
\begin{displaymath}
{\rm i}  \hbar  \frac{\partial U(t_0, t)}{\partial t} = (H_0 + H_1) 
U(t_0, t),
\end{displaymath} (769)

subject to the boundary condition
\begin{displaymath}
U(t_0, t_0 ) = 1.
\end{displaymath} (770)

In the absence of the external perturbation, the time evolution operator reduces to

\begin{displaymath}
U(t_0, t) = \exp[-{\rm i}   H_0(t-t_0)/\hbar].
\end{displaymath} (771)

Let us switch on the perturbation and look for a solution of the form
\begin{displaymath}
U(t_0, t) = \exp[ -{\rm i}   H_0(t-t_0)/\hbar]  U_I(t_0, t).
\end{displaymath} (772)

It is readily demonstrated that $U_I$ satisfies the differential equation
\begin{displaymath}
{\rm i}  \hbar  \frac{\partial U_I(t_0, t)}{\partial t} = H_I(t_0, t) 
U_I(t_0, t),
\end{displaymath} (773)

where
\begin{displaymath}
H_I(t_0,t) = \exp[ +{\rm i}   H_0(t-t_0)/\hbar]   H_1 
\exp[ -{\rm i}   H_0(t-t_0)/\hbar],
\end{displaymath} (774)

subject to the boundary condition
\begin{displaymath}
U_I(t_0, t_0) = 1.
\end{displaymath} (775)

Note that $U_I$ specifies that component of the time evolution operator which is due to the time-dependent perturbation. Thus, we would expect $U_I$ to contain all of the information regarding transitions between different eigenstates of $H_0$ caused by the perturbation.

Suppose that the system starts off at time $t_0$ in the eigenstate $\vert i\rangle$ of the unperturbed Hamiltonian. The subsequent evolution of the state ket is given by Eq. (735),

\begin{displaymath}
\vert i, t_0, t\rangle = \sum_m c_m(t) \exp[ -{\rm i}   E_m(t-t_0)/\hbar] 
\vert m\rangle.
\end{displaymath} (776)

However, we also have
\begin{displaymath}
\vert i, t_0, t\rangle = \exp[-{\rm i}   H_0(t-t_0)/\hbar]  U_I(t_0, t)  \vert i\rangle.
\end{displaymath} (777)

It follows that
\begin{displaymath}
c_n(t) = \langle n\vert U_I(t_0, t) \vert i\rangle,
\end{displaymath} (778)

where use has been made of $\langle n\vert m\rangle
=\delta_{nm}$. Thus, the probability that the system is found in state $\vert n\rangle$ at time $t$, given that it is definitely in state $\vert i\rangle$ at time $t_0$, is simply
\begin{displaymath}
P_{i\rightarrow n} (t_0, t) = \vert\langle n\vert U_I(t_0, t) \vert i\rangle\vert^2.
\end{displaymath} (779)

This quantity is usually termed the transition probability between states $\vert i\rangle$ and $\vert n\rangle$.

Note that the differential equation (773), plus the boundary condition (775), are equivalent to the following integral equation,

\begin{displaymath}
U_I(t_0, t) = 1 - \frac{\rm i}{\hbar} \int_{t_0}^t H_I(t_0, t') 
U_I(t_0, t')  dt'.
\end{displaymath} (780)

We can obtain an approximate solution to this equation by iteration:
$\displaystyle U_I(t_0, t)$ $\textstyle \simeq$ $\displaystyle 1 - \frac{\rm i}{\hbar} \int_{t_0}^t H_I(t_0, t')
\left[ 1 - \frac{\rm i}{\hbar} \int_{t_0}^{t'} H_I(t_0, t'') 
U_I(t_0, t'')\right] dt'$  
  $\textstyle \simeq$ $\displaystyle 1 - \frac{\rm i}{\hbar} \int_{t_0}^t H_I(t_0, t') dt'$  
    $\displaystyle + \left(\frac{-{\rm i}}{\hbar}\right)^2 \int_{t_0}^t dt'
\int_{t_0}^{t'} H_I(t_0, t' ) H_I(t_0, t'' )  dt'' + \cdots.$ (781)

This expansion is known as the Dyson series. Let
\begin{displaymath}
c_n = c_n^{(0)} + c_n^{(1)} + c_n^{(2)} + \cdots,
\end{displaymath} (782)

where the superscript $^{(1)}$ refers to a first-order term in the expansion, etc. It follows from Eqs. (778) and (781) that
$\displaystyle c_n^{(0)}(t)$ $\textstyle =$ $\displaystyle \delta_{in},$ (783)
$\displaystyle c_n^{(1)}(t)$ $\textstyle =$ $\displaystyle -\frac{\rm i}{\hbar} \int_{t_0}^t \langle n \vert H_I(t_0, t')\vert i\rangle 
dt',$ (784)
$\displaystyle c_n^{(2)}(t)$ $\textstyle =$ $\displaystyle \left(\frac{-{\rm i}}{\hbar}\right)^2 \int_{t_0}^t dt'
\int_{t_0}^{t'} \langle n\vert H_I(t_0, t' ) H_I(t_0, t'' )\vert i\rangle dt''.$ (785)

These expressions simplify to
$\displaystyle c_n^{(0)}(t)$ $\textstyle =$ $\displaystyle \delta_{in},$ (786)
$\displaystyle c_n^{(1)}(t)$ $\textstyle =$ $\displaystyle -\frac{\rm i}{\hbar} \int_{t_0}^t \exp[ {\rm i}  \omega_{ni} 
(t'-t_0)]  H_{ni}(t')  dt',$ (787)
$\displaystyle c_n^{(2)}(t)$ $\textstyle =$ $\displaystyle \left(\frac{-{\rm i}}{\hbar}\right)^2
\sum_m \int_{t_0}^t dt'\int_{t_0}^{t' }dt'' \exp[ {\rm i}
 \omega_{nm} (t'-t_0)]$  
    $\displaystyle \times H_{nm}(t')  
\exp[ {\rm i}  \omega_{mi} (t''-t_0)] H_{mi}(t''),$ (788)

where
\begin{displaymath}
\omega_{nm} = \frac{E_n -E_m}{\hbar},
\end{displaymath} (789)

and
\begin{displaymath}
H_{nm} (t) = \langle n\vert H_1(t) \vert m\rangle.
\end{displaymath} (790)

The transition probability between states $i$ and $n$ is simply
\begin{displaymath}
P_{i\rightarrow n} (t_0, t) = \vert c_n^{(0)} + c_n^{(1)} + c_n^{(2)} +\cdots\vert^2.
\end{displaymath} (791)

According to the above analysis, there is no chance of a transition between states $\vert i\rangle$ and $\vert n\rangle$ ($i\neq n$) to zeroth-order (i.e., in the absence of the perturbation). To first-order, the transition probability is proportional to the time integral of the matrix element $\langle n\vert H_1\vert i\rangle$, weighted by some oscillatory phase-factor. Thus, if the matrix element is zero, then there is no chance of a first-order transition between states $\vert i\rangle$ and $\vert n\rangle$. However, to second-order, a transition between states $\vert i\rangle$ and $\vert n\rangle$ is possible even when the matrix element $\langle n\vert H_1\vert i\rangle$ is zero.


next up previous
Next: Constant perturbations Up: Approximation methods Previous: Spin magnetic resonance
Richard Fitzpatrick 2006-02-16