The Delta Function of Energy Conservation

For harmonic perturbations, we have derived a probability to be in the final state $\phi_n$ proportional to the following.

$$\bgroup\color{black} P_n\propto \left[{4\sin^2\left((\omega_{ni}+\omega) t/2\right)\over (\omega_{ni}+\omega)^2}\right] \egroup$$

For simplicity of analysis lets consider the characteristics of the function

$$\bgroup\color{black} g(\Delta\equiv\omega_{ni}+\omega)=\left[... ...equiv {4\sin^2\left(\Delta t/2\right)\over \Delta^2t^2} \egroup$$

for values of $t»{1\over\Delta}$. (Note that we have divided our function to be investigated by $t^2$. For $\Delta=0$, $g(\Delta)=1$ while for all other values for $\Delta$, $g(\Delta)$ approaches zero for large $t$. This is clearly some form of a delta function.

To find out exactly what delta function it is, we need to integrate over $\Delta$.

$$\begin{eqnarray*} \int\limits_{-\infty}^\infty d\Delta\; f(\Delta) g(\Delta) &=&... ...over t}\int\limits_{-\infty}^\infty dy\;{\sin^2(y)\over y^2} \\ \end{eqnarray*}$$

We have made the substitution that $y={\Delta t\over 2}$. The definite integral over $y$ just gives $\pi$ (consult your table of integrals), so the result is simple.

$$\begin{eqnarray*} \int\limits_{-\infty}^\infty d\Delta\; f(\Delta) g(\Delta) &=&... ...mega_{ni}+\omega)^2}\right]&=&2\pi t\:\delta(\omega_{ni}+\omega) \end{eqnarray*}$$

Q.E.D.