Integrals in the derivation of the Optical Theorem

In summary, the conversation is about evaluating integrals in the r \rightarrow \infty regime in nonrelativistic Quantum Mechanics for the Optical Theorem. The integrals, denoted as I_1 and I_2, involve the functions f(\theta) and its derivative. The question arises about the evaluation of these integrals in the limit of r \rightarrow \infty, and the saddle point method is proposed as a possible approach. The use of the asymptotic form of spherical bessel functions is also mentioned as a way to evaluate the integrals.
  • #1
maverick280857
1,789
4
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:
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  • #2
Someone please help...
 
  • #3
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.
 
  • #4
Count Iblis said:
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.)
 

Related to Integrals in the derivation of the Optical Theorem

1. What is the Optical Theorem?

The Optical Theorem is a fundamental law in optics that relates the scattering of light by an object to its absorption. It states that the total cross section of an object, which measures the likelihood of light being scattered or absorbed by the object, is equal to 4 times the imaginary part of the forward scattering amplitude.

2. How are integrals used in the derivation of the Optical Theorem?

The derivation of the Optical Theorem involves complex mathematical calculations, including the use of integrals. Integrals are used to calculate the total cross section of an object by integrating the scattering amplitude over all possible scattering angles. This allows us to determine the total amount of light scattered or absorbed by the object.

3. What is the significance of the Optical Theorem in optics?

The Optical Theorem is a powerful tool in understanding the behavior of light interacting with matter. It allows us to relate the macroscopic properties of an object, such as its total cross section, to its microscopic properties, such as its scattering amplitude. This has important applications in fields such as spectroscopy, where we can use the Optical Theorem to study the composition and structure of materials.

4. How is the Optical Theorem related to the conservation of energy?

The Optical Theorem is closely linked to the principle of energy conservation. In optics, the total amount of energy in a system is conserved, meaning that the energy of the incident light must be equal to the energy of the scattered or absorbed light. The Optical Theorem helps us understand how this energy is distributed among different scattering angles, providing valuable insight into the scattering process.

5. Are there any limitations to the application of the Optical Theorem?

While the Optical Theorem is a powerful and widely applicable law in optics, there are some limitations to its use. It assumes that the interactions between light and matter are linear, meaning that the scattering and absorption processes are proportional to the intensity of the incident light. It also assumes that the object is much smaller than the wavelength of the incident light. In certain cases, these assumptions may not hold and the use of the Optical Theorem may not be appropriate.

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