The complete question I am trying to answer is the following:
Show that the probability density of a linear harmonic oscillator in an arbitrary superposition state is periodic with period equal to the period of the oscillator.
I have: $$\Psi(x,t)=\sum\limits_n c_n \psi_n(x) \exp\left(-\frac{iE_nt}{\hbar}\right)$$
Therefore, the probability density is: $$P(x,t)=\Psi^*(x,t)\Psi(x,t)=\sum\limits_n c_n^2 \psi_n^2(x) $$
(The exponential term should cancel out because of the conjugation). So I get the probability density as being time-independent, which seems wrong, according to how the question is stated.
In the solution in the book, we have:$$P(x,t)=\Psi^*(x,t)\Psi(x,t)=\sum\limits_m \sum\limits_n c_m^* c_n\psi_m^*(x)\psi_n(x) \exp\left(\frac{i(E_m-E_n)t}{\hbar}\right)$$
I don't understand why, in the solution, we have the conjugate of the wavefunction have the subscript $m$, while the wavefunction itself has the subscript $n$. It changes the whole thing, when we introduces different subscripts. Moreover, the formula for probability density is $$P(x,t)=\Psi^*(x,t)\Psi(x,t)$$not $$P(x,t)=\Psi_m^*(x,t)\Psi_n(x,t)$$
What is going on here?
Best Answer
Your problem essentially amounts to multiplying two sums of numbers. I would also say this seems like more of a homework problem than a research level question, but since I'm new here and feel like answering my first question, I will help you out.
Let $A = (a_1 + a_2 + \dots)$ and $B = (b_1 + b_2 + \dots)$.
So the product is
$AB = a_1 (b_1 + b_2 + \dots) + a_2 (b_1 + b_2 + \dots) + \dots$.
Rearranging this into single-index and double-index terms,
$AB = (a_1 b_1 + a_2 b_2 + a_3 b_3 + \dots) + (a_1 b_2 + a_1 b_3 + \dots) + (a_2 b_1 + a_2 b_3 + \dots) + \dots$.
What you're doing when you have n = m is only allowing terms in the first group. Terms with identical indices. As you can see there are many more "cross-terms" that you also need to include. This is why when taking the product of two sums you need to use a dummy variable on one of the sums (ex. changing n to m). This way you get all the cross-terms as well as the direct terms.
This is pretty fundamental, so make sure you understand the reasoning. You're only going to encounter these types of things more and more often.