Integrals in the derivation of the Optical Theorem

[maverick280857]

Hi everyone,

I'm trying to work out all the intricate details of the limits involved in computing the integrals leading to the Optical Theorem in nonrelativistic Quantum Mechanics. In doing so, I have arrived at the following integrals which are to be evaluated in the $r \rightarrow \infty$ regime:

$$r^{2}\int_{0}^{2\pi}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta)d\phi = I_{1}$$
$$r^{2}\int_{0}^{2\pi}\int_{-1}^{+1}f^{*}(\theta)e^{-ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)+\frac{1}{r^2}\right)d(\cos\theta)d\phi = I_{2}$$

Modulo the $\phi$ integral, the expression for $I_{1}$ is

$$r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta) = r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)\right)d(\cos\theta) - \int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}d(\cos\theta)$$

I am stuck at these integrals. I know what the "final" term in the optical theorem is, but without using the asymptotic solution, I don't know the right direction to proceed in. Specifically, if I consider $f(\theta)(1+\cos\theta)$ as the "first" function in the first integral, then integrating by parts I get

$$\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}(1+\cos\theta)d(\cos\theta) = \left[f(\theta)(1+\cos\theta)\frac{e^{ikr(1-\cos\theta)}}{-ikr}\right]_{-1}^{+1} - \int_{-1}^{+1}\frac{e^{ikr(1-\cos\theta)}}{-ikr}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)} + f(\theta)\right)d(\cos\theta)$$

Therefore,

$$r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)\right)d(\cos\theta) = ikr\left[\mbox{b.d.}\right] + \int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)} + f(\theta)\right)d(\cos\theta)$$

where b.d. denotes the boundary term in the second last equation. Put together, these equations give

$$r^{2}\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}\left(\frac{ik}{r}(1+\cos\theta)-\frac{1}{r^2}\right)d(\cos\theta) = ikr\left[\mbox{b.d.}\right] + \int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)}\right)d(\cos\theta)$$

Clearly the boundary term tends to zero as $r \rightarrow \infty$, so my original question reduces to the following question:

How does one evaluate the following integrals

$$\int_{-1}^{+1}e^{ikr(1-\cos\theta)}\left((1+\cos\theta)\frac{df(\theta)}{d(\cos\theta)}\right)d(\cos\theta)$$
$$\int_{-1}^{+1}f(\theta)e^{ikr(1-\cos\theta)}d(\cos\theta)$$

in the $r \rightarrow \infty$ regime? I want to get some more insight into the physics as $r \rightarrow \infty$ so I'm hoping that an explicit evaluation of all these terms will be helpful.

Thanks.

 

[maverick280857]

Someone please help...

 

[Count Iblis]

Doesn't the saddle point method work? If r is very large, then the contribution to the integral will come from a narrow range of theta near the point where the aregument of the exponential is extremal.

So, you expand i k r (1-cos(theta)) around its extremal values and then, for large r, you end up with Gaussian integrals.

 

[maverick280857]

If r is very large, then the contribution to the integral will come from a narrow range of theta near the point where the aregument of the exponential is extremal.

That seems intuitive. But I don't know how to rigorously justify it. Is this the saddle point method?

(I can plug in the asymptotic form of the spherical bessel functions in

$$e^{ikr\cos\theta} = \sum_{l=0}^{\infty}(2l+1)i^{l}j_{l}(kr)P_{l}(\cos\theta)$$

and get the limit.)