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First Order Perturbation Theory

This is achieved by doing the first iteration in
$\displaystyle \frac{d}{dt}\vert\Psi_I(t)\rangle$ $\displaystyle =$ $\displaystyle -i V_I(t) \vert\Psi_I(t)\rangle$  
$\displaystyle \leadsto \vert\Psi_I(t)\rangle$ $\displaystyle =$ $\displaystyle \Psi_I(t_0)\rangle -i \int_{t_0}^t dt' V_I(t') \vert\Psi_I(t')\rangle$  
  $\displaystyle =$ $\displaystyle \vert\Psi_I(t_0)\rangle -i \int_{t_0}^t dt' V_I(t') \vert\Psi_I(t_0)\rangle +O(V_I^2).$ (3.6)

We take \bgroup\color{col1}$ t_0=0$\egroup for simplicity and the initial state therefore is \bgroup\color{col1}$ \vert\Psi_I(0)\rangle =\vert\Psi(0)\rangle$\egroup,
$\displaystyle \vert\Psi_I(t)\rangle$ $\displaystyle =$ $\displaystyle \vert\Psi(0)\rangle -i \int_{0}^t dt' V_I(t') \vert\Psi(0)\rangle +O(V_I^2).$ (3.7)

This can be worked out in some more detail by assuming a basis \bgroup\color{col1}$ \vert n\rangle$\egroup of eigenstates of \bgroup\color{col1}$ H_0$\egroup,
$\displaystyle H_0\vert n\rangle$ $\displaystyle =$ $\displaystyle \varepsilon_n \vert n\rangle,\quad \hat{1}=\sum_{n'}\vert n'\rangle \langle n'\vert\leadsto$ (3.8)
$\displaystyle \langle n\vert\Psi_I(t)\rangle$ $\displaystyle =$ $\displaystyle \langle n\vert\Psi(0)\rangle
-i\sum_{n'} \int_{0}^t dt' \langle n\vert V_I(t')\vert n'\rangle \langle n' \vert\Psi(0)\rangle +O(V_I^2).$  

Let us assume the initial state \bgroup\color{col1}$ \vert\Psi(0)\rangle = \vert m\rangle$\egroup is an eigenstate of \bgroup\color{col1}$ H_0$\egroup, then
$\displaystyle \langle n\vert\Psi_I(t)\rangle$ $\displaystyle =$ $\displaystyle \delta_{nm} -i \int_{0}^t dt' \langle n\vert V_I(t')\vert m\rangle +O(V_I^2).$ (3.9)

The probability to find the system in state \bgroup\color{col1}$ \vert n\rangle$\egroup after time \bgroup\color{col1}$ t$\egroup is then a transition probability. Use \bgroup\color{col1}$ \vert\langle n\vert\Psi(t)\rangle\vert^2 = \vert\langle n\vert\Psi_I(t)\rangle\vert^2$\egroup (EXERCISE:CHECK!) to find within first order perturbation theory
$\displaystyle P_{m\to n}(t)$ $\displaystyle =$ $\displaystyle \vert\langle n\vert\Psi(t)\rangle\vert^2 = \left\vert\int_{0}^t dt' \langle n\vert V_I(t')\vert m\rangle\right\vert^2$   first order  
$\displaystyle \vert\Psi(0)\rangle$ $\displaystyle \equiv$ $\displaystyle \vert m\rangle \ne \vert n\rangle.$ (3.10)



Subsections
next up previous contents index
Next: Time-Independent Hamiltonian Up: Time-Dependent Perturbation Theory Previous: The Interaction Picture   Contents   Index
Tobias Brandes 2005-04-26