QS 17: Born Series

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We have seen that all information about scattering is contained in the T-matrix which is the matrix representation of $V\molp$.  In order to calculate the transition amplitude  we need the operator $\molp$ for calculating which there is the Lippmann-Schwinger equation
\begin{eqnarray*} \molp=1+\frac{1}{i\hbar}\int_{-\infty}^0U_0(t)^{-1}V\molp U_0(t) dt\end{eqnarray*}
or equivalently
\begin{eqnarray*} \bb{E\alpha}|\molp\kk{E'\alpha'}=\delta(E-E')\dd{\alpha\alpha'}
+\frac{\bb{E\alpha}|V\molp\kk{E'\alpha'}}{E'-E+i\epsilon} \end{eqnarray*}
If the interaction $V$ can be regarded as small in some sense then we can obtain $\molp$ by iteration : put $V=0$ on the right hand side of the above equation
\begin{eqnarray*} \bb{E\alpha}|\molp_0\kk{E'\alpha'}=\delta(E-E')\dd{\alpha\alpha'}\end{eqnarray*}
This is called the {\bf first Born approximation} because for this lowest order case the transition amplitude is of first order in interaction
\begin{eqnarray*} T_1(E\alpha,E'\alpha')=\bb{E\alpha}|V\molp_0\kk{E'\alpha'}=

Next put $\molp_0$ into the right hand side of the L-S equation to obtain the next correction $\molp_1$
\begin{eqnarray*} \bb{E\alpha}|\molp_1\kk{E'\alpha'}=\delta(E-E')\dd{\alpha\alpha'}
+\frac{\bb{E\alpha}|V\kk{E'\alpha'}}{E'-E+i\epsilon} \end{eqnarray*}
Then this can in turn be substituted for $\molp$ to give the second order transition amplitude (the second Born approximation) :
\begin{eqnarray*} T_2(E\alpha,E'\alpha')=\bb{E\alpha}|V\kk{E'\alpha'}
+\sum_{\alpha''}\int dE''\frac{\bb{E\alpha}|V\kk{E''\alpha''}
\bb{E''\alpha}|V\kk{E'\alpha'}}{E'-E''+i\epsilon} \end{eqnarray*}
This can be continued to obtain the {\bf Born series}, though normally first two approximations are good enough.

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