Recent content by silverthorne

Homework Statement This isn't really a textbook or a homework problem...rather it is a question I have in trying to write up my lab report. I am doing an experiment about the photoelectric effect in trying to determine Planck's constant h experimentally...and basically I am shining...

It's ok! So I guess the answer to part a is as above... Anyway, I very much appreciate your help!

Um...well, I am stuck then if we can't just tack on the exponential factor. :confused: How should I go about in doing this problem?

Ah! Thank you so much!

Yep, you're right. The previous C_{n}s I got were just the value of the integral...I forgot to multiply everything by the constant \frac{\sqrt{8}}{a}. Thanks for the heads up. :)

Hmm...I don't think we need to modify the energy eigenfunction because our question is still living from x=0 to x=a. OK, I am going to redo my calculations for the coefficients. I know part c is dependent on the number of terms in part b. But the thing is I couldn't figure out what does...

Ah...thanks! Yes, I forgot the fact that the reason why only the sines are the eigenfunctions is because of the initial conditions! So yeah, the cosine here works just fine. So I guess it's just \Psi(x,t) = \frac{1}{\sqrt{2}} \sqrt{\frac{2}{a}}cos\left \frac{\pi x}{a} \right \exp{\left(...

Hi guys, this assignment is driving me nuts! Thank you very much for the help! Homework Statement Consider the infinite square well described by V=0, -a/2<x<a/x, and V=infinity otherwise. At t=0, the system is given by the equation \Psi(x,0) = C_{1} \Psi_{1}(x) + C_{2} \Psi_{2}(x)...

Thanks for the advice! Yes, I reworked the problem (changing the limits of integration) and I got this: c_{n}=\int_{0}^{a/2} \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right \frac{2}{\sqrt{a}}sin\left \frac{2 \pi x}{a} \right dx and the c_{n}s came out to be c_{1}=\frac{2a}{3\pi}...

OK, I have done the integral for c_{n}, and here's what I got: C_{n}=0 for every n except 2. So...I guess the answer to part (a) of the question is \psi (x,0) = \sqrt{2} sin\left \frac{2 \pi x}{a} \right . There's only one term here...? It looks a bit suspicious...because it kind of...

Ah! I have here in my notes that..."by making use of the orthonormality of the solutions, the initial condition equation then gives us the mechanism for finding the coefficients of the series: c_{n}=\int_{0}^{a} \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right \psi(x,0)dx Is this the...

Yes, the latex didn't come out right :confused: It should be ok now...I think.

I THINK I'm in the same class as Lisa, actually...I got the same question on my assignment! :-p Anyway, I know the eigenfunctions for an infinite square well are \psi_{n}(x)= \sqrt{\frac{2}{a}}sin\left \frac{n \pi x}{a} \right and the corresponding eigenenergies are...

So does that mean plugging in \psi (x,0) = \frac{2}{\sqrt{a}}sin\left \frac{2 \pi x}{a} \right to the time independent Schrodnger Equation and solve?