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Introductory quantum mechanics (particle in a box)

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data

    The (normalized) wavefunction of the particle is defined as:

    [tex]\psi_n = \sqrt{\frac{2}{L}} \sin \left ({\frac{\pi n}{L} x} \right )[/tex] for [tex]0 \leqslant x \leqslant L[/tex] and [tex]\psi_n = 0[/tex] otherwise, where L is the length of the one-dimensional box and n is the particle's energy level (so n is a nonzero positive integer).

    1. Find the expectation values for the particle's position.
    2. Find the expectation values for the particle's momentum.

    2. Relevant equations

    Quantum operators on the wavefunction.
    Calculus.

    3. The attempt at a solution

    1. We integrate over the wavefunction squared using the position operator (which is just x) to obtain the expectation values for the particle's position:
    \begin{align*}\int_0^L | \psi_n |^2 x ~~ dx &= \int_0^L x \frac{2}{L} \sin^2 \left ({\frac{\pi n}{L} x} \right ) ~~ dx \\
    &= \frac{2}{L} \int_0^L \frac{x}{2} \left (1 - \cos \left ({\frac{2 \pi n}{L} x} \right ) \right ) ~~ dx = \frac{1}{L} \int_0^L x - x\cos \left ({\frac{2 \pi n}{L} x} \right ) ~~ dx \\
    &= \frac{1}{L} \left [ \frac{x^2}{2} \right ]_{0}^{L} - \frac{1}{L} \left [ x\sin \left ({\frac{2 \pi n}{L} x} \right ) + \cos \left ({\frac{2 \pi n}{L} x} \right ) \right ]_0^L ~~ dx = \frac{L}{2}\end{align*} So the expectation value for position is always the middle of the box for any energy level.

    2. We use the momentum operator which is -ih d/dx, so:
    [tex]\frac{d}{dx} = \sqrt{2} \pi \left ( \frac{1}{L} \right )^\frac{3}{2} n \cos{ \left ( \frac{\pi n}{L} x \right ) }[/tex]
    \begin{align*}-ih \int_0^L | \psi_n |^2 \frac{d}{dx} ~~ dx &= -ih \int_0^L \frac{2}{L} \sin^2 \left ({\frac{\pi n}{L} x} \right ) \sqrt{2} \pi \left ( \frac{1}{L} \right )^\frac{3}{2} n \cos{ \left ( \frac{\pi n}{L} x \right ) } ~~ dx \\
    &= -ih \frac{2}{L} \sqrt{2} \pi \left ( \frac{1}{L} \right )^\frac{3}{2} n \int_0^L \sin^2 \left ({\frac{\pi n}{L} x} \right ) \cos{ \left ( \frac{\pi n}{L} x \right ) } ~~ dx \\
    &= c \int_0^L \sin^2 \left ({\frac{\pi n}{L} x} \right ) \cos{ \left ( \frac{\pi n}{L} x \right ) } ~~ dx
    \end{align*} where [itex]c = -ih \frac{2}{L} \sqrt{2} \pi \left ( \frac{1}{L} \right )^\frac{3}{2} n[/itex]

    Using substitution, we obtain:
    [tex]c \left [ \frac{\sin^3{ \frac{\pi n}{L} x }}{3 \frac{\pi n}{L}} \right ]_0^L = 0[/tex]So the momentum is zero for any energy level.

    ---

    Is the reasoning and maths correct? And does anyone have any tips to simplify the integrals, they are incredibly messy I find - would it help to just always bunch up constants into a single letter like I did for 2) and are there dangers in doing this?
     
    Last edited by a moderator: Jun 6, 2012
  2. jcsd
  3. Jun 6, 2012 #2

    vela

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    The expectation value of an operator ##\hat{A}## is given by
    $$\langle \hat{A} \rangle = \int \psi^* \hat{A} \psi \, dx.$$ Your calculation for <x> turns out to be correct, but the one for <p> isn't. You should have
    $$\langle \hat{p} \rangle = \int \psi^* \hat{p} \psi \, dx = \int \psi^* \left(\frac{\hbar}{i}\frac{d}{dx}\right) \psi \, dx$$
     
  4. Jun 6, 2012 #3
    Ah, I see. So if I replace my -ih with h / i it will be correct? It should still come to zero anyway because it's a constant difference, right?

    Is there a reason operators are placed in between the conjugate of the wavefunction and the wavefunction, rather than just multiplying it with the wavefunction's modulus squared? Just wondering.
     
  5. Jun 6, 2012 #4

    vela

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    h/i = -ih

    Yes, that was my point. Your method is wrong. The operator acts on ψ, and then you evaluate the resulting integral.
     
  6. Jun 6, 2012 #5
    Right, that makes sense.

    If I get this right, the difference is in the derivative for momentum, so I have to take the derivative of ψ, multiply it with -ih, then multiply that with ψ* and then integrate?

    But then why is position correct, shouldn't the operator x do nothing to ψ, and so I end up with |ψ|^2 and not x|ψ|^2 ?

    I'm very confused now. I think I understand the concept but it doesn't seem to be consistently applied for momentum and position.
     
  7. Jun 6, 2012 #6

    vela

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    The operator ##\hat{x}## simply ends up multiplying ψ by x, that is, ##\hat{x}\psi \to x\psi##, so you end up with
    $$\int \psi^* x \psi \,dx$$ which is equal to the integral you got.
     
  8. Jun 6, 2012 #7
    Right, that makes much more sense. Thanks a lot Vela! I think I can solve the problem correctly now :)
     
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