- #1

maverick280857

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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:

[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.

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.

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