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Normalizing a wave function and finding probability density

  1. Apr 16, 2014 #1
    1. The problem statement, all variables and given/known data

    A state of a particle bounded by infinite potential walls at x=0 and x=L is described by a wave function [itex]\psi = 1\phi_1 + 2\phi_2 [/itex] where [itex]\phi_i[/itex] are the stationary states.
    a) Normalize the wave function.
    b) What is the probability to find the particle between x=L/4 and x=3L/4?
    c) Calculate the expectation value of the Hamiltonian operator [itex]\langle \hat{H} \rangle[/itex]

    2. Relevant equations

    I wasn't sure Schrödinger's equation was necessary here: [itex]i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + V(x)\psi(x,t)[/itex]
    but I did remember that the solution to any wavefunction -- at least the stationary state -- is going to be [itex]\psi(x) = A sin(\frac{n\pi x}{a})[/itex] where a is the length of your "box" for the particle. So in this case a=L.

    3. The attempt at a solution

    We have a situation where V(x) = 0 0 < x < L and V(x) = infinity outside of that. So the V(x) term for inside the well disappears (it's zero).

    The probability of the particle being at any point from 0 to L is 1. So I need to integrate the wave functions squared over that interval. By the superposition principle it is OK to just add them.

    [itex]\psi = 1\phi_1 + 2\phi_2 [/itex]
    [itex]\psi = (1\phi_1 + 2\phi_2)(1\phi_1^* + 2\phi_2^*)[/itex]

    multiply this out
    [itex]\psi = (1\phi_1^* \phi_1 + 2\phi_1 \phi_2^* + 2\phi_1^* \phi_2 + 4\phi_2^*\phi_2)[/itex]

    SInce the phi functions are eigenvalues, the ones on the diagonal of the matrix are the only ones not zero. So we get
    [itex]\psi = (1\phi_1^* \phi_1 + 4\phi_2^*\phi_2) = (1 + 4) [/itex]

    because the complex conjugate of a function multiplied by a function is 1.

    That makes the whole thing add up to five. and since the probability of finding the particle on the interval 0 to x is

    [tex]\int^L_0 |\psi|^2 dx = 1 \rightarrow \int^L_0 |5|^2 dx = 1 \rightarrow 25x = 1[/tex]

    so x = 1/5 for the whole interval, (since that is the square root of 1/25) so normalizing the wave function I should get

    [itex]\psi = \frac{1}{5}\phi_1 + \frac{2}{5}\phi_2 [/itex]

    and for the probability that the particle is at L/4 and 3/4 L

    (25L/4) and (75L/4)

    Now, if someone could tell me where I am getting lot and doing this completely wrong :-)

    Actually I know this is wrong, because the probabilities should add up to one, at least with the stationary states.

    After that I get even more confused. I think -- and I stress think -- I have some vague idea of how to get expectation values, but whenever someone mentions stuff like "Hamiltonian operator" I want to run away and hide. :-) More seriously, I am trying to grasp what exactly is meant by the notation [itex]\langle \hat{H} \rangle[/itex] - some of it is a notational question, but i get confused because I am never sure if they want the Hamiltonian like what you do in mechanics or something else. I feel like if someone could explain that I'd be a lot further along.

    thanks in advance.

    PS I put this up before in advanced HW, but perhaps it goes here. (Nobody answered, I might have put up the wrong tags) The line between what counts as advanced around here and what is basic seems rather fluid :-) But since I am clearly the dumbest guy in the class I'll start here. :-)
  2. jcsd
  3. Apr 16, 2014 #2
    You have the right idea, but you're making some silly mistakes.
    withe ψ = N (ψ1 + 2 ψ2) and integrate ∫|ψ|2 dx from zero to L and set the integral to 1 to find the normalization factor N. Tip: N is not equal to 5.
  4. Apr 16, 2014 #3
    I did that tho, didn't I? I got x=1/25. I'm not sure I understand what you are referring to.

    When I do the integration I get the ∫|ψ|2 dx = ∫|5|2 dx. Is that not correct? Because if you integrate that from 0 to L you would get 25* (x) from 0 to L, or 25L = 1. That would make L = 1/25, no?

    And in that case my N would be 1/25? Is that what you are saying?
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