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

In summary, the conversation involves a person trying to find the coefficients for a quantum mechanical system with an infinite potential. The person gets stuck trying to integrate and finds a formula for the coefficients that is not valid for n=1. They are advised to set n=1 and simplify the integrand to find the correct value for c1.
  • #1
rkdaniels
4
0
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.
 
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  • #2
Can you show your work where you are struck exactly?
 
  • #3
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:
  • #4
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.
 
  • #5
But when n is 1 c is undefined. Does this matter?
 
  • #6
rkdaniels said:
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.
 
  • #7
Right but I have a formula for c and when n is 1, c is undefined, so perhaps my formula is wrong?
 
  • #8
rkdaniels said:
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
 
  • #9
rkdaniels said:
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.
 

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

1. What is ψ(x, t) in the context of an infinite well?

ψ(x, t) is the wave function that describes the probability amplitude of finding a particle at a specific position (x) and time (t) within an infinite well. It is a complex-valued function that represents the quantum state of the particle.

2. How is ψ(x, t) related to the energy levels (En) in an infinite well?

The energy levels (En) in an infinite well are quantized, meaning they can only take on certain discrete values. The wave function ψ(x, t) describes the probability of finding the particle in a specific energy level (En) at a given time (t).

3. How do you calculate ψ(x, t) for an infinite well at t > 0?

The equation for ψ(x, t) in an infinite well is ψ(x, t) = √(2/L) * sin(nπx/L) * e^(-iEn t/ħ), where L is the length of the well, n is the energy level, and ħ is the reduced Planck's constant. To calculate ψ(x, t) at t > 0, simply plug in the values for x, t, L, n, and En into the equation.

4. What is P(En) in the context of an infinite well?

P(En) is the probability of measuring the particle to have a specific energy level (En) if a measurement is made on the system. It is given by the absolute square of the wave function ψ(x, t), meaning P(En) = |ψ(x, t)|^2.

5. How does P(En) change over time in an infinite well?

In an infinite well, the energy levels (En) and the corresponding probabilities (P(En)) do not change over time. This is because the system is in a stationary state, meaning it is not evolving or changing over time. The probabilities will only change if an external force or measurement is applied to the system.

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