Product of complex conjugate functions with infinite sums

[Adolfo Scheidt]

Hello there. I'm here to request help with mathematics in respect to a problem of quantum physics. Consider the following function $$ f(\theta) = \sum_{l=0}^{\infty}(2l+1)a_l P_l(cos\theta) , $$ where $f(\theta)$ is a complex function $P_l(cos\theta)$ is the l-th Legendre polynomial and $a_l$ is a the complex term (known as partial wave amplitude). Then, the author calculates the product of $f(\theta)$ with its complex conjugate, and shows the result: $$ f(\theta)f^{*}(\theta) = |f(\theta)|^2 = \sum_{l}\sum_{l'}(2l+1)(2l'+1)(a_l)^{*}(a_{l'})P_l(cos\theta)P_{l'}(cos\theta). $$ My problem here is to understand how he obtained the second equation. I'm not familiar with operations with sums and real analysis, and I'm stuck with it. Any help will be very appreciated.

 

[mfb]

Which part is unclear? The second equation directly follows from the first. The sums work like with real numbers. The complex conjugation is only relevant for the complex amplitude, where you also find it in the second equation.

 

[Adolfo Scheidt]

Which part is unclear? The second equation directly follows from the first. The sums work like with real numbers. The complex conjugation is only relevant for the complex amplitude, where you also find it in the second equation.

Good pointed; more precisely, I do not understand why he chooses a different index for each sum.

 

[mfb]

Let's make an example, with a sum over just three terms: $(x_1 + x_2 + x_3) \cdot (y_1 + y_2 + y_3)$. This will lead to 9 terms: $x_i y_j$ for i=1,2,3, and j=1,2,3 separately. As formula, $(\sum_{i=1}^{3}x_i) (\sum_{i=1}^{3} y_i) = (x_1 + x_2 + x_3) \cdot (y_1 + y_2 + y_3) = (x_1 y_1 + x_1 y_2 + x_1 y_3 + x_2 y_1 + x_2 y_2 + x_2 y_3 + x_3 y_1 + x_3 y_2 + x_3 y_3) = \sum_{i=1}^{3} \sum_{j=1}^{3} x_i y_j$. You need two different indices to have to cross-terms (like $x_1 y_3$) included.

 

[Adolfo Scheidt]

Let's make an example, with a sum over just three terms: $(x_1 + x_2 + x_3) \cdot (y_1 + y_2 + y_3)$. This will lead to 9 terms: $x_i y_j$ for i=1,2,3, and j=1,2,3 separately. As formula, $(\sum_{i=1}^{3}x_i) (\sum_{i=1}^{3} y_i) = (x_1 + x_2 + x_3) \cdot (y_1 + y_2 + y_3) = (x_1 y_1 + x_1 y_2 + x_1 y_3 + x_2 y_1 + x_2 y_2 + x_2 y_3 + x_3 y_1 + x_3 y_2 + x_3 y_3) = \sum_{i=1}^{3} \sum_{j=1}^{3} x_i y_j$. You need two different indices to have to cross-terms (like $x_1 y_3$) included.

Oh, it's very clear now! Thank you so much :)