qft-lsn-04008
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I Session Overview
II Recall formulae needed
III Solution
Scattering-external-source-Schrodinger-field
V Useful Tips
It will be seen that squaring gives
\[(2\pi) \delta(\omega_{fi}) \stackrel{\text{square}}{\longrightarrow}(2\pi)^2 \delta(0) \delta(\omega_{fi})\] and computing transition probability per unit volume amounts to dropping the factor \((2\pi) \delta(0)\): \[\boxed{(2\pi) \delta(\omega_{fi}) \stackrel{\underbrace{\text{square}}}{\longrightarrow} (2\pi)^2 \delta(\omega_{fi}) \delta(\omega_{fi})= (2\pi)^2 \delta(0) \delta(\omega_{fi}) \stackrel{\underbrace{\text{per unit time}}}{\longrightarrow}(2\pi) \delta(\omega_{fi})}\]
This replacement can be intuitively understood as follows
\begin{eqnarray} (2\pi) \delta(\omega) &=& \int _{-\infty}^{\infty} e^{i\omega t} \,\, dt\\ &=& \lim_{T\to\infty} \int_{T/2}^{T/2} e^{i\omega t} \,\, dt\\ (2\pi) \delta(0) &=& \lim_{T\to\infty} \int_{T/2}^{T/2} e^{i\omega t} \,\, dt\big|_{\omega=0}\\ \int_{T/2}^{T/2} e^{i\omega t} \,\, dt\big|_{\omega=0}&=& \int_{T/2}^{T/2} \,dt\\ &=& T. \end{eqnarray}
Hence taking per unit time amounts to replacement
\[ (2\pi)\delta(\omega)\big|_{\omega=0} \to 1\] }
The above sequence of statement cannot be
justified, a complete argument has however been given earlier.