# Find ψ(x, t) and P(En) at t > 0 infinite well

• rkdaniels

#### rkdaniels

Homework Statement
I am attempting to find a solution to the following problem
Relevant Equations
See below
I am currently stuck trying to work this out. I have an infinite potential with walls at x=0 and x=a, with the initial state:

$$\psi(x,0) = A_2(exp(i\pi(x-a)/a)-1)$$

I am trying to find psi(x,t). I know that

$$A_2(exp(i\pi(x-a)/a)-1) = A_2(-exp(i\pi/a)-1)$$

And this enables me to find the normalization constant:

$$A_2 = \frac{1}{\sqrt{a}}$$

I also know that I can expand psi(x,0) as:

$$\psi=\sum c_n\phi_n$$

where

$$c_n = \langle\phi_n|\psi\rangle$$

and

$$\phi_n = \sqrt{\frac{2}{a}}\sin\left( \frac{n\pi x}{a} \right)$$

I get stuck trying to find the coefficients here. When I try to integrate, I end up with something quite horrific and I'm sure it should be relatively simple.

Can you show your work where you are struck exactly?

Sure, so ignoring the normalization constants for now, I think I am supposed to do the following in order to find the coefficients:

$$\int_0^a\sin\frac{n\pi x}{a}\left( e^{\frac{i\pi x}{a}}+1 \right)dx$$

I can express the exponential as sines and cosines and I have to integrate:

$$\int_0^a\sin\left( \frac{n\pi x}{a}\right) \left(i \sin\left( \frac{\pi x}{a}\right) + \cos\left( \frac{\pi x}{a}\right) +1\right)dx$$

I can expand this out to give:

$$\int_0^a\sin\left( \frac{n\pi x}{a}\right) i \sin\left( \frac{\pi x}{a} \right) + \sin\left( \frac{n\pi x}{a}\right) \cos\left( \frac{\pi x}{a}\right) + \sin\left( \frac{n\pi x}{a}\right) dx$$

And using two identities:

$$\int_0^a \frac{i}{2}\cos\left( \frac{(n-1)\pi x}{a}\right) + \frac{i}{2} \cos\left( \frac{(n+1)\pi x}{a} \right) + \frac{1}{2}\sin\left( \frac{(n+1)\pi x}{a}\right) + \frac{1}{2}\sin\left( \frac{(n-1)\pi x}{a}\right) + \sin\left( \frac{n\pi x}{a}\right) dx$$

When I integrate, the imaginary parts will turn to sines and evaluate to zero for x = 0 and x = a, leaving:

$$\int_0^a \frac{1}{2}\sin\left( \frac{(n+1)\pi x}{a}\right) + \frac{1}{2}\sin\left( \frac{(n-1)\pi x}{a}\right) + \sin\left( \frac{n\pi x}{a}\right) dx$$

Is this right?

Performing the integrating gives:

$$\frac{a}{2\pi(n+1)}(\cos(n\pi+\pi)-1) + \frac{a}{2\pi(n-1)}(\cos(n\pi-\pi)-1) + \frac{a}{n\pi}(\cos(n\pi)-1)$$

And I get a $c_n$ of:

$$c_n = \frac{a-4an^2}{\pi n(n+1)(n-1)}\left( (-1)^n + 1 \right)$$

And so,

$$\psi(x,t) = \sum \frac{a-4an^2}{\pi n(n+1)(n-1)}\left( (-1)^n + 1 \right) \phi_n$$

But this just seems overly complicated...

Last edited:
So,why do you think it horrific? Doing integration we get for n=1,c=0 and the non-zero contribution from other terms. You found normalization. And now:
$$\psi(x,t)= \psi(x,0) e^{-iE_{n}t/h}$$
Where ##\psi(x,0) ## is expanded using the stationary states.

But when n is 1 c is undefined. Does this matter?

But when n is 1 c is undefined. Does this matter?
Substitute n=1 in the very first integral. You will get value of ##c_{1}## from there. As for equation,if you have done write integration(verify using softwares like Wolfram as I have not done your integrals but I think that they should be of these forms due to Sin terms) then,I don't see that equation is complicated.

Right but I have a formula for c and when n is 1, c is undefined, so perhaps my formula is wrong?

Right but I have a formula for c and when n is 1, c is undefined, so perhaps my formula is wrong?
In that case take c which is defined. It is usual thing in QM

Right but I have a formula for c and when n is 1, c is undefined, so perhaps my formula is wrong?
Your formula isn't valid for ##n=1## because when you integrated, you implicitly assumed ##n \ne 1## in evaluating the terms involving ##n-1##. For example,
$$\int_0^a \cos\left[\frac{(n-1)\pi x}a\right]\,dx = \int_0^a \cos 0\,dx = a$$ but you said it would be 0 because integrating would turn the cosine into a sine. That only happens when ##n \ne 1##.

So to find ##c_1##, set ##n=1##, simplify the integrand, and then integrate.