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Homework Help: Particle bound by quadratic potential

  1. Nov 5, 2008 #1
    1. The problem statement, all variables and given/known data

    A particle that can move in one dimension and that is in a stationary state, is bound by a potential V(x) = (1/2)kx^2. The wave function is [tex]\Psi[/tex](x,t) = [tex]\psi[/tex](x)exp(-iEt/[tex]\hbar[/tex])

    We look at a state in which [tex]\psi[/tex](x) = Aexp(-x^2/2a^2a^2), where a is a constant and A is the normalisation constant. Determine a so that [tex]\psi[/tex](x) is an energy state. What is the energ of the particle?

    3. The attempt at a solution

    I don't really know what to do here, but setting up the wave equation tends to be a good start:

    H[tex]\psi[/tex](x) = E[tex]\psi[/tex](x), where H = -([tex]\hbar[/tex]^2/2m)(d^2/dx^2) + kx^2/2

    How can I determine a so that [tex]\psi[/tex](x) is an energy state?
    Last edited: Nov 6, 2008
  2. jcsd
  3. Nov 5, 2008 #2
    Just calculate the left-hand and right-hand sides, and set them equal.
  4. Nov 5, 2008 #3
    That the wave function is an energy state simply means that it's described by the Schrödinger equation?
  5. Nov 5, 2008 #4
    Yes, it's a solution to the *time-independent* SE, to be precise.
  6. Nov 5, 2008 #5
    OK, but I still don't see how to find a when E is unknown. My expression after differentiating twice is:

    ([tex]\hbar[/tex]^2/2ma^2) (A - x^2/a^2) + kx^2/2 = E
  7. Nov 5, 2008 #6
    First, you didn't calculate it right.

    Second, the correct equation you'll get will have to be true for every x.
  8. Nov 6, 2008 #7
    Now I ended up with

    ([tex]\hbar[/tex]^2/2ma^2)(1 - x^2/a^2) + kx^2/2 = E

    What I should do here is to eliminate x so that the equation holds for all x-es? In that case, a must be ([tex]\hbar[/tex]^2/km)^0.25
  9. Nov 6, 2008 #8
    Your equation is now of the form (if you move E to the left-hand side)


    where C and D are constants (independent of x).
    If this is to be true for all x, then both C and D must be zero.
  10. Nov 6, 2008 #9
    I got it right now, didn't I?

    My next problem is to calculate the expectation value. That's not hard, I just wonder why the integral is written

    [tex]\int[/tex]dx [tex]\psi[/tex]*(x)x[tex]\psi[/tex](x)

    and not

    [tex]\int[/tex] [tex]\psi[/tex]*(x)x[tex]\psi[/tex](x) dx
    Last edited: Nov 6, 2008
  11. Nov 6, 2008 #10


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    Science Advisor
    Gold Member

    In this case it doesn't matter where you put the "dx", it is the same thing.

    (note that there are cases where it DOES matter; e.g. if dx is a vector)
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