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

[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.

 

[Abhishek11235]

Can you show your work where you are struck exactly?

 

[rkdaniels]

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...

 

[Abhishek11235]

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.

 

[rkdaniels]

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

 

[Abhishek11235]

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.

 

[rkdaniels]

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

 

[Abhishek11235]

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

 

[vela]

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.