Expectation value and momentum for an infinite square well

In summary: Since the momentum operator contains a derivative, it cannot be evaluated at a single point in time. So, it makes sense that it would be zero for this case.
  • #1
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Homework Statement

√[/B]
A particle in an infinite square well has the initial wave function:

[itex]Ψ(x, 0) = A x ( a - x )[/itex]

a) Normalize Ψ(x, 0)

b) Compute <x>, <p>, and <H> at t = 0. (Note: you cannot get <p> by differentiating <x> because you only know <x> at one instance of time)

Homework Equations

The Attempt at a Solution


For part a, I figured out A = [itex]\sqrt{30 / a^5}[/itex]

I'm sort of confused for part b. For <x>, I set up the integral like this:

[itex]\int_{0}^{a}x Ψ(x, 0)^2 dx[/itex]

And got [itex]\frac{a^6 A^2}{60}[/itex], but I'm not sure if I got it right.

For <p> and <H>, I don't know how to set up these integrals. How would I set up these integrals? I don't need you to solve it for me, I just wanted to know how to set them up.

Thank you for reading and helping.
 
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  • #2
How does the momentum operator look like in position space, i.e. ##p_x\psi(x) =\ldots##?
 
  • #3
Matt Q. said:
For <x>, I set up the integral like this:

[itex]\int_{0}^{a}x Ψ(x, 0)^2 dx[/itex]

And got [itex]\frac{a^6 A^2}{60}[/itex], but I'm not sure if I got it right.
If you plug in your result for ##A##, you get ##\langle x \rangle = \frac a2##. Given the symmetry of ##\Psi(x,0)##, does the answer seem reasonable?
 
  • #4
blue_leaf77 said:
How does the momentum operator look like in position space, i.e. ##p_x\psi(x) =\ldots##?

I figured it out and got 0 for <p>. That seems reasonable right?

vela said:
If you plug in your result for ##A##, you get ##\langle x \rangle = \frac a2##. Given the symmetry of ##\Psi(x,0)##, does the answer seem reasonable?

Ah I didn't think of that. It makes sense.
 
  • #5
I also figured out <H> which turns out to be ##\frac{5 h^2}{m a^2}## which seems pretty reasonable too right?
 
  • #6
Matt Q. said:
I figured it out and got 0 for <p>. That seems reasonable right?
Yes.
 

1. What is the expectation value for the momentum of a particle in an infinite square well?

The expectation value for the momentum of a particle in an infinite square well is zero. This means that, on average, the particle has no motion or velocity in the system.

2. How is the expectation value for the momentum of a particle calculated in an infinite square well?

The expectation value for the momentum of a particle in an infinite square well can be calculated using the formula pavg = ∫0L ψ*p ψ dx, where p represents the momentum operator and ψ is the wave function for the particle.

3. What is the significance of the expectation value for the momentum in an infinite square well?

The expectation value for the momentum in an infinite square well helps to understand the average behavior and motion of a particle in the system. It also provides information about the spread or uncertainty in the particle's momentum.

4. How does the expectation value for the momentum change for different energy levels in an infinite square well?

The expectation value for the momentum in an infinite square well increases as the energy level of the particle increases. This is because higher energy levels correspond to shorter wavelengths and faster velocities for the particle.

5. Can the expectation value for the momentum be negative in an infinite square well?

No, the expectation value for the momentum cannot be negative in an infinite square well. This is because the momentum operator and wave function values must be positive in order for the integral to be a real number.

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