Integrals in the derivation of the Optical Theorem

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 [itex]r \rightarrow \infty[/itex] regime:

[tex]
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}
[/tex]
[tex]
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}
[/tex]

Modulo the [itex]\phi[/itex] integral, the expression for [itex]I_{1}[/itex] is

[tex]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)[/tex]

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 [itex]f(\theta)(1+\cos\theta)[/itex] as the "first" function in the first integral, then integrating by parts I get

[tex]\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)[/tex]

Therefore,

[tex]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)[/tex]

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

[tex]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)[/tex]

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

How does one evaluate the following integrals

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

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

Thanks.
 
Last edited:
Someone please help...
 
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.
 
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

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

and get the limit.)
 

Want to reply to this thread?

"Integrals in the derivation of the Optical Theorem" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top