Why is Griffiths Treating the Summation Like This?

[bluepilotg-2_07]

TL;DR

From line one to two in the image, the summations go from $\int(\sum_{m=1}^\infty c_m\psi_m)* \, (\sum_{n=1}^\infty c_n\psi_n)dx$ to $\sum_{n=1}^\infty \sum_{m=1}^\infty c_m * c_n \int \psi_m * \psi_n dx$. Can someone explain why please.

I am self-studying quantum mechanics from Griffiths' textbook and some other sources. I have come across this derivation shown in the photo. I've taken all three major calculus courses for physics, linear algebra, ODE, PDE, Complex Analysis, etc.
However, I do still struggle with rules for summations. I do not understand why Griffiths goes from $\int(\sum_{m=1}^\infty c_m\psi_m)* \, (\sum_{n=1}^\infty c_n\psi_n)dx$ to $\sum_{n=1}^\infty \sum_{m=1}^\infty c_m * c_n \int \psi_m * \psi_n dx$. I was under the impression that summations do not "multiply" in this way. I have not found another explanation online so far. Could someone kindly explain. Thanks.

 

[fresh_42]

Just multiply the sums:
$$
\left( \sum_{n=1}^\infty a_n \right) \cdot \left( \sum_{m=1}^\infty b_m \right)= \sum_{k=2}^\infty \sum_{n+m=k}a_nb_m= \sum_{n=1}^\infty \sum_{m=1}^\infty a_nb_m
$$
by the distributive law. The real question is why you can swap the infinite summation with the integral.

 

[bluepilotg-2_07]

Hi, thanks for responding! Just after posting, I finally found a resource that explained what I wanted to know. It feels as if it should have been obvious now that I have read about it. In response to the question you posited, we should be able to swap the summations and integral because the integral is only dependent on x and functions of x while the summations are only dependent on n. Then we use the fact that we can break up any summations in the integrand into separate integrals. So if we view our double summation notation as what it really is: just a sum over a bunch of terms, then we can break all of those terms into separate integrals, and thus we get a summation over integrals.

 

[Office_Shredder]

Just multiply the sums:
$$
\left( \sum_{n=1}^\infty a_n \right) \cdot \left( \sum_{m=1}^\infty b_m \right)= \sum_{k=2}^\infty \sum_{n+m=k}a_nb_m= \sum_{n=1}^\infty \sum_{m=1}^\infty a_nb_m
$$
by the distributive law. The real question is why you can swap the infinite summation with the integral.

Well hold on, even the distribution is sketchy if you don't have absolute convergence. I would be pretty nervous distributing
$$\sum \frac{(-1)^n}{\sqrt{n}} \sum \frac{(-1)^n}{\sqrt{n}}$$

 

[fresh_42]

Well hold on, even the distribution is sketchy if you don't have absolute convergence. I would be pretty nervous distributing
$$\sum \frac{(-1)^n}{\sqrt{n}} \sum \frac{(-1)^n}{\sqrt{n}}$$

Yes. I thought about correcting it, but as we need absolute convergence for the integral exchange anyway, I was lazy.