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8.5: Dyson Series

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    08:36, 16 Nov 2014
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Let us now try to find approximate solutions of Equation (749) for a general system. It is convenient to work in terms of the time evolution operator, $ U(t_0, t)$ , which is defined

 

$\displaystyle \vert A, t_0, t\rangle = U(t_0, t) \,\vert A\rangle.$ (777)

 

 

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

 

$\displaystyle {\rm i}\, \hbar\, \frac{\partial U(t_0, t)}{\partial t} = (H_0 + H_1)\, U(t_0, t),$ (778)

 

 

subject to the initial condition

 

$\displaystyle U(t_0, t_0 ) = 1.$ (779)

 

 

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

 

$\displaystyle U(t_0, t) = \exp[-{\rm i} \, H_0\,(t-t_0)/\hbar].$ (780)

 

 

Let us switch on the perturbation and look for a solution of the form

 

$\displaystyle U(t_0, t) = \exp[ -{\rm i} \, H_0\,(t-t_0)/\hbar]\, U_I(t_0, t).$ (781)

 

 

It is readily demonstrated that $ U_I$ satisfies the differential equation

 

$\displaystyle {\rm i}\, \hbar\, \frac{\partial U_I(t_0, t)}{\partial t} = H_I(t_0, t)\, U_I(t_0, t),$ (782)

 

 

where

 

$\displaystyle H_I(t_0,t) = \exp[ +{\rm i} \, H_0\,(t-t_0)/\hbar] \, H_1\, \exp[ -{\rm i} \, H_0\,(t-t_0)/\hbar],$ (783)

 

 

subject to the initial condition

 

$\displaystyle U_I(t_0, t_0) = 1.$ (784)

 

 

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 Equation (744),

 

$\displaystyle \vert i, t_0, t\rangle = \sum_m c_m(t) \exp[ -{\rm i} \, E_m\,(t-t_0)/\hbar]\, \vert m\rangle.$ (785)

 

 

However, we also have

 

$\displaystyle \vert i, t_0, t\rangle = \exp[-{\rm i} \, H_0\,(t-t_0)/\hbar]\, U_I(t_0, t)\, \vert i\rangle.$ (786)

 

 

It follows that

 

$\displaystyle c_n(t) = \langle n\vert\, U_I(t_0, t)\, \vert i\rangle,$ (787)

 

 

where use has been made of $ \langle n\vert m \rangle = \delta_{n\,m}$ . 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

 

$\displaystyle P_{i\rightarrow n} (t_0, t) = \vert\langle n\vert\, U_I(t_0, t)\, \vert i\rangle\vert^{\,2}.$ (788)

 

 

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

Note that the differential equation (782), plus the initial condition (784), are equivalent to the following integral equation,

 

$\displaystyle U_I(t_0, t) = 1 - \frac{\rm i}{\hbar} \int_{t_0}^t dt' \,H_I(t_0, t')\, U_I(t_0, t') .$ (789)

 

 

We can obtain an approximate solution to this equation by iteration:

 

$\displaystyle U_I(t_0, t)$ $\displaystyle \simeq 1 - \frac{\rm i}{\hbar} \int_{t_0}^t H_I(t_0, t') \left[ 1 - \frac{\rm i}{\hbar} \int_{t_0}^{t'} dt'\,H_I(t_0, t'')\, U_I(t_0, t'')\right]$    
  $\displaystyle \simeq 1 - \frac{\rm i}{\hbar} \int_{t_0}^t H_I(t_0, t')\,dt' + \... ...\int_{t_0}^t dt' \int_{t_0}^{t'} dt''\, H_I(t_0, t' )\,H_I(t_0, t'' ) + \cdots.$ (790)

 

 

This expansion is known as the Dyson series. Let

 

$\displaystyle c_n = c_n^{(0)} + c_n^{(1)} + c_n^{(2)} + \cdots,$ (791)

 

 

where the superscript $ ^{(1)}$ refers to a first-order term in the expansion, etc. It follows from Equations (787) and (790) that

 

$\displaystyle c_n^{(0)}(t)$ $\displaystyle = \delta_{i\,n},$ (792)
$\displaystyle c_n^{(1)}(t)$ $\displaystyle = -\frac{\rm i}{\hbar} \int_{t_0}^t dt'\,\langle n \vert\,H_I(t_0, t')\,\vert i\rangle,$ (793)
$\displaystyle c_n^{(2)}(t)$ $\displaystyle = \left(\frac{-{\rm i}}{\hbar}\right)^2 \int_{t_0}^t dt' \int_{t_0}^{t'}dt''\, \langle n\vert\, H_I(t_0, t' )\,H_I(t_0, t'' )\,\vert i\rangle.$ (794)

 

 

These expressions simplify to

 

$\displaystyle c_n^{(0)}(t)$ $\displaystyle = \delta_{in},$ (795)
$\displaystyle c_n^{(1)}(t)$ $\displaystyle = -\frac{\rm i}{\hbar} \int_{t_0}^t dt'\, \exp[\,{\rm i} \,\omega_{ni}\, (t'-t_0)]\, H_{ni}(t') ,$ (796)
$\displaystyle c_n^{(2)}(t)$ $\displaystyle = \left(\frac{-{\rm i}}{\hbar}\right)^2 \sum_m \int_{t_0}^t dt'\i... ...t'-t_0)]\, H_{nm}(t') \, \exp[\,{\rm i} \,\omega_{mi}\,(t''-t_0)]\,H_{mi}(t''),$ (797)

 

 

where

 

$\displaystyle \omega_{nm} = \frac{E_n -E_m}{\hbar},$ (798)

 

 

and

 

$\displaystyle H_{nm} (t) = \langle n\vert\, H_1(t)\, \vert m\rangle.$ (799)

 

 

The transition probability between states $ i$ and $ n$ is simply

 

$\displaystyle P_{i\rightarrow n} (t_0, t) = \vert c_n^{(0)} + c_n^{(1)} + c_n^{(2)} +\cdots\vert^{\,2}.$ (800)

 

 

According to the above analysis, there is no chance of a transition between states $ \vert i\rangle$ and $ \vert n\rangle$ (where $ 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.

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