Showing that the probability density of a linear harmonic oscillator is periodic

Question

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?

Answer 1

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.

Answer 2

You have raised the typical mistakes by the beginner of the quantum mechanics. Consider the most general total time dependent wave function is $$\Psi(x,t)=\sum\limits_n c_n \psi_n(x) \exp\left(-\frac{iE_nt}{\hbar}\right)$$

(a) Never write $\psi_n^*(x) \psi_n(x) = \psi_n^2(x) $

The correct equation is $\psi_n^*(x) \psi_n(x) = |\psi_n(x)|^2$. Since the $\psi_n(x)$ is in general complex number. For a general complex number $(a+bi)$, these two different equations will evaluated to $$(a+bi)^2=(a^2-b^2)+2abi$$ $$|a+bi|^2 =\sqrt{a^2+b^2}$$ respectively, where the first one is complex number and the second is always real.

(b) The general total time dependent probability density is $$P(x,t) = |\Psi(x,t)|^2 = \Psi^*(x,t) \Psi(x,t)$$ which gives the probability of any point in the position space and it is cleary more general than $\psi_m^*(x,t) \psi_n(x,t)$ which only includes particular eigenstate. Indeed, $P(x,t)$ defined is the most general probability with spatial component only. Since the completeness property of the wavefunction implies that all wave function can always be written as the Eq (1).

(c) For the explicit form of the total probability, first rewrite the complex conjugate of Eq (1) as $$\Psi^*(x,t)=\sum_m c_m^* \psi_m^*(x) \exp\left(\frac{iE_mt}{\hbar}\right)$$ The index is changed here for clarity. As explained above, the total probability density is $$P(x,t)=\Psi^*(x,t)\Psi(x,t)=\sum_m \sum_n c_m^* c_n\psi_m^*(x)\psi_n(x) \exp\left(\frac{i(E_m-E_n)t}{\hbar}\right)$$ The exponential term cannot be cancelled in general. Note that there are two summation over two indexes $m$ and $n$. If the wavefunction $\psi(x,t)$ have finite, say $N$, non-zero $c_n$, then there are total $N^2$ terms being summed in $P(x,t)$.

(d) The total probability over the whole space is $$\int P(x,t) dx = \sum_m \sum_n c_m^* c_n \exp\left(\frac{i(E_m-E_n)t}{\hbar}\right) \int \psi_m^*(x)\psi_n(x) dx$$ where $\int \psi_m^*(x)\psi_n(x) dx = \delta_{nm}$ is the Kronecker delta function due to the orthonormality of the wavefunction, so $$\int P(x,t) dx = \sum\limits_n c_n^* c_n = \sum_n |c_n|^2$$ In this case, all the term with $n \ne m$ are zero so only the "diagonal" term remains. For $n=m$, we have $\exp(i(E_n-E_n)t/\hbar) = 1$ and so the result. This integral gives the normalization criteria and its value must be equal to 1: $$\sum_n |c_n|^2=1$$ Also, once the normalization constant is found, the total probability is always 1 and independent of time which means the probability of wavefunction is conserved.