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Expectation value and momentum for an infinite square well

  • #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.
 

Answers and Replies

  • #2
blue_leaf77
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How does the momentum operator look like in position space, i.e. ##p_x\psi(x) =\ldots##?
 
  • #3
vela
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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
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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?

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
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I also figured out <H> which turns out to be ##\frac{5 h^2}{m a^2}## which seems pretty reasonable too right?
 
  • #6
blue_leaf77
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I figured it out and got 0 for <p>. That seems reasonable right?
Yes.
 

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