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Integrals with infinite well eigenfunctions

  1. Dec 24, 2016 #1
    1. The problem statement, all variables and given/known data
    This is problem 17 from Chapter 3 of Quantum Physics by S. Gasiorowicz
    "Consider the eigenfunctions for a box with sides at x = +/- a. Without working out the integral, prove that the expectation value of the quantity

    [tex] x^2 p^3 + 3 x p^3 x + p^3 x^2 [/tex]

    vanishes for all the eigenfunctions."

    2. Relevant equations
    The eigenfunctions for a symmetric box with sides at x = -a and x = a are

    [tex] u_n = A_n \sin\left(\frac{n \pi x}{a} \right)[/tex]
    or
    [tex] u_n = A_n \cos\left(\frac{n \pi x}{a} \right) [/tex]

    whether n is even or odd.

    [tex] p^3 = (- i \hbar \frac{d}{dx})^3 = i \hbar^3 \frac{d^3}{dx^3} [/tex]

    3. The attempt at a solution
    The integral is over a symmetric region so it should vanish if the integrand is an odd function of x.
    Starting with the odd eigenfuctions of the infinite well

    [tex] x^2 p^3 sin(x) = x^2 i \hbar^3 \frac{d^3}{dx^3} \sin(x) = - i \hbar^3 x^2 \cos(x) [/tex]
    which is an even function, so its integral from -a to a does not vanish

    [tex] 3 x p^3 x \sin(x) = 3 x i \hbar^3 \frac{d^3}{dx^3}(x \sin(x)) [/tex]

    x*sin(x) is a product of two odd functions so it would be an even function. The derivative of an even function is an odd function, so its second derivative should be an even function, and the third derivative will be an odd function again. But then this third derivative function is multiplied to the right by x, and the product of two odd functions is again an even function, So the integral does not vanish.

    [tex] p^3 x^2 \sin(x) i \hbar^3 \frac{d^3}{dx^3}(x^2 \sin(x)) [/tex]

    x^2*sin(x) is the product of an even function with an odd function, so it's an odd function. The first derivative is an even function, the second derivative is an odd function, and the third derivative will once again be even so the integral does not vanish.

    For the even eigenfunctions, that are something like A*cos(b*x), I can make the same argument, only replacing every "even" with "odd" and the expectation value would indeed vanish. But for the odd eigenfunctions that go like A*sin(b*x) this does not happen. How do I prove that the expectation value does vanish for eigenfunctions like sin(n*pi*x/a)?
     
  2. jcsd
  3. Dec 24, 2016 #2

    TSny

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    OK
    Did you leave out part of the integrand for finding the expectation value?
     
  4. Dec 24, 2016 #3

    Orodruin

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    How do you compute expectation values in quantum mechanics?
     
  5. Dec 25, 2016 #4
    Thank you for the replies.
    I see now that I forgot that the expectation value is calculated by integrating the square of the amplitude of the wavefunction, so if the wavefunction is an eigenfunction of the infinite well u_n (x) = A sin(pi*n*x/a), then the expectation value will be

    [tex] \int_{-a}^{a} dx |A|^2 \sin(\pi n x/a) Ô \sin(\pi n x/a) [/tex]

    Ô is the long operator with x^2,x and p^3, that produces an even function when acting on sin(x). But because I have to multiply it to the right by the eigenfunction (in fact by its complex conjugate, but because the eigenfunctions are real in this case, it makes no difference), then I have sin(x) times an even function which is an odd function. The integral of an odd function over a symmetric interval vanishes.

    The operator produces an odd function when it acts on the cos(x) eigenfunctions and so the integrand is cos(x) times an odd function of x, which is an odd function of x too. Then, its integral over a symmetric region is 0.

    Am I right?
    I was very much focused only on tracking the parity changes and forgot something more important than that.
    Again, thank you for helping me.
     
  6. Dec 25, 2016 #5

    Orodruin

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    Yes.
     
  7. Dec 25, 2016 #6
    Great! Thank you so much.
     
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