# Analysis of time-dependent SE (infinite square potential)

1. Dec 28, 2009

### scorpion990

Hey all :)

I'm a second-semester junior who is studying chemistry and mathematics. I've self-studied basic quantum chemistry before, and I picked up Griffith's Introduction to Quantum Mechanics a few days ago to get a broader/more rigorous taste of the subject. I've been going through the reading/problems, and I'm really confused about something...

For those who want to follow along in the book, I'm referring to question 2.8 in the first edition.

Problem: Given $$\Psi(x,0)=Ax(a-x)$$ for a particle in a 1-dimensional box $$(0<x<a)$$, find the time-dependent wave function for any given time.

Solution:
$$\Psi(x,t)=\sqrt\frac{2}{a}\sum^{\infty}_{n=1}c_n e^{\frac{-in^2 \pi^2 \hbar t}{2ma^2}}} sin\frac{n\pi x}{a}$$

I found that:
$$c_n = \frac{\sqrt{60}}{a^3}\int^{a}_{0}x(a-x)sin\frac{n\pi x}{a}dx = \frac{8\sqrt{15}}{n^3\pi^3}$$
... if n is odd. Its 0 if n is even.

Thus, the wavefunction is equal to:
$$\Psi(x,t)=\frac{4\sqrt{30}}{\pi^3n^3\sqrt{a}}\sum^{\infty}_{n=1}(1-(-1)^n)e^{\frac{-in^2 \pi^2 \hbar t}{2ma^2}}} sin\frac{n\pi x}{a}$$

I know that I could rewrite that sum over just the odd integers. But it doesn't matter that much for now.

My questions are:

1. How can a particle in a box actually have such an initial wavefunction? I was under the impression that a particle in a box MUST have a wavefunction described by the simple sinusodidal waves that you get by solving the time-independent Schrodinger equation!

2. If a particle in a box can have any initial wavefunction, doesn't that mean that it can also have any initial energy? Where does the concept of quantization of energy come into play in this analysis?

3. Does the particle have a definite energy? How would I verify that its constant, as it should be?

Sorry for the confusion. Quantum chemistry books hardly ever cover time-dependence in quantum mechanics! I'm going to analyze my solution further. However, it's 4:40 AM and I should go to bed :)

2. Dec 28, 2009

### scorpion990

Hey... Cool.
I just noticed that my solution and the orthonormality of the 1D particle in a box wavefunctions can be used to prove that:

$$\sum^{\infty}_{j=0}\frac{1}{(2j+1)^6}=\frac{\pi^6}{960}$$

I have no idea what that has to do with quantum mechanics, but it sure is beautiful!

3. Dec 28, 2009

### Staff: Mentor

No. When $\Psi$ is a superposition of energy eigenfunctions:

$$\Psi = \sum {c_k \Psi_k}$$

then if you measure the energy of the particle, you always get one of the discrete energy eigenvalues $E_k$, at random, with probablilties determined by the coefficients $c_k$. Specifically, if $\Psi$ is normalized, then the probability of getting energy $E_k$ is $|c_k|^2 = c_k^* c_k$.

4. Dec 28, 2009

### scorpion990

So is the particle's energy changing with time? Or is it constant but we are not able to know it absolutely due to the uncertainty principle?

5. Dec 28, 2009

### Staff: Mentor

There is no generally-agreed-upon answer to questions about what a QM particle is "really doing" before we measure its energy (or whatever other quantity). The mathematical formalism that you're learning simply doesn't address questions like these.

Most physicists, as a practical matter, probably would say that the exact value of the energy (and other quantities) is simply "undefined" before being measured, unless the particle is in an eigenstate of that quantity (e.g. in a specific "energy level" of the infinite square well). People have come up with various possible descriptions of what is "really" happening. We call these descriptions interpretations of QM. Unfortunately, there is (as yet) no way to distinguish among these interpretations experimentally, even in principle (so far as I know). It basically comes down to personal preferences.

Last edited: Dec 28, 2009
6. Dec 29, 2009

### scorpion990

Hmmm... I've read up on indeterminacy, interpretations, and wave-function collapse. I think what gets me is the fact that the potential energy of the particle is necessarily 0. It just seems strange to me that its energy wouldn't be known exactly, and that repeated measurements would give me different results.

See... I'm a chemist who hasn't really touched the time-dependent Schrodinger equation until now.

Hmmm... Thanks for the responses so far! I'd love to discuss this matter further if anybody wishes. Any comments, questions, answers would be appreciated!

7. Dec 29, 2009

### RedX

This is not really quantum mechanics, or physics, for that matter. If you take a look at how the Riemann zeta function is calculated in terms of the Bernoulli numbers, then it uses a Fourier series which is a discrete sum, but the coefficients of this discrete Fourier series is calculated by an integral (the integral of the function). Going back and forth between discrete Fourier n-space and continuous x-space is what allows you to evaluate these things. For some reason integrals are easier than sums.

Otherwise you have to be conventional:

$$\zeta(6)=\frac{\pi^6}{945}=\frac{1}{1^6}+\frac{1}{2^6}+\frac{1}{3^6}+...= (\frac{1}{1^6}+\frac{1}{3^6}+\frac{1}{5^6}) +\frac{1}{2^6}(\frac{1}{1^6}+\frac{1}{3^6}+\frac{1}{5^6}) +\frac{1}{4^{6}}(\frac{1}{1^6}+\frac{1}{3^6}+\frac{1}{5^6}) +\frac{1}{8^{6}}(\frac{1}{1^6}+\frac{1}{3^6}+\frac{1}{5^6})= (\frac{1}{1^6}+\frac{1}{3^6}+\frac{1}{5^6}+...)(\frac{1}{1-\frac{1}{2^6}})$$

8. Dec 29, 2009

### Staff: Mentor

When you measure a quantity, the system's wave function "collapses" into the eigenstate for whatever eigenvalue you get as the result of the measurement, and develops from there as time goes on. In your case, if you measure the energy of the particle in the box to be $E_5$, then immediately afterward $\Psi = \Psi_5$. If nothing else interacts with the particle thereafter, it remains in that energy eigenstate.