Quantum Scattering: Do I Need to Lookup Orthogonal Functions?

[NewtonApple]

I need a little more explanation about the solution discussed in the following thread.

https://www.physicsforums.com/showthread.php?t=410830

Do I've to lookup for orthogonal functions?

PS. required thread is closed so I'm posting here.

 

[NewtonApple]

Can someone clarify how did they get equation 15.5.7?

 

[dauto]

They just compared the terms in the expansion term one by one. If the two expansions are equal to each other then each coefficient of the expansion is equal to each other. It's possible to prove that by use of the orthogonality of the functions as you guessed.

 

[NewtonApple]

ok after substituting values in Equation [15.5.6]

we get

$$\sum A_{l}\frac{e^{\iota(kr-\pi/2l+\delta_{l})}-e^{-\iota(kr-\pi/2l+\delta_{l})}}{2\iota kr}P_{l}(Cos\theta)= \sum[(2l+1)\iota^{l} \frac{e^{\iota(kr-\pi/2l+\delta_{l})}-e^{-\iota(kr-\pi/2l+\delta_{l})}}{2\iota kr}P_{l}(Cos\theta)
+ \frac{1}{r}\sum f_{l}P_{l}(Cos\theta)e^{\iota kr}]P_{l}(Cos\theta) $$

simplifying and making $P_{l}(Cos\theta) \rightarrow 1$

we have

$A_{l}\frac{e^{\iota{kr-\pi/2l+\delta_l}}}{2\iota kr} - A_{l}\frac{e^{-\iota kr-\pi/2l+\delta_l}}{2\iota kr} = (2l+1) i^{l}\frac{e^{\iota kr-\pi/2l+\delta_l}}{2\iota kr} - (2l+1) i^{l}\frac{e^{-\iota kr-\pi/2l+\delta_l}}{2\iota kr} + f_{l} \frac{e^{\iota kr}}{r}$

now what? please give me some hint.

 

[dauto]

"Now what" what? what are you trying to calculate? Also, you can't make P_l(cosθ) equal 1. That makes no sense. It is a function of θ.

 

[NewtonApple]

I'm trying to follow the steps in the given solution. I don't know how to use orthogonality condition here.