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Normalizing a wave function and calculating probability of position

  1. Apr 8, 2014 #1
    Forgive me if this goes in elementary physics, but I think since it's an upper level undergrad class
    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?

    2. Relevant equations

    OK so was looking at how to approach this. One thought was to start with Schrödinger's equation

    [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]

    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).

    And Energy, [itex]E = \frac{\hbar^2 k^2}{2m}[/itex] where k is a constant, in this case 1.

    That leaves us with a partial differential equation

    [itex]i \hbar \frac{\partial \psi}{\partial t} = - \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}[/itex]

    which, moving it around a little,

    [itex]i \hbar \frac{\partial \psi}{\partial t} - \frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} = 0 [/itex]

    and that's a 2nd order differential equation. The solutions are

    [itex]\psi(x,t) = [A \sin(kx) + B \cos(kx)] e^{-i\omega t}[/itex]

    k is a wavenumber and for this to work -- for it to be zero when x is greater than L or less than 0, and some value anywhere else, the cosine term has to go, and since the wavenumber is limited to nπ/L (that's the only way you get an integral number of waves) So I should end up with

    [itex]\psi(x,t) = [A \sin(k_nx)] e^{-i\omega_n t}[/itex]


    But I am not entirely sure why the cos term has to go; I feel like I am missing a step. Either way, the probability of the particle being at any point from 0 to L is 1. So I need to integrate the wave function squared over that interval:

    [itex]\int^L_0 |[A \sin(k_nx)] e^{-i\omega_n t}|^2 dx[/itex]

    but that's the thing, I feel like I have lost the plot with this.

    ANother way to approach it was to assume that the wave function given can be

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

    and then 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.

    Tht 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 = 1/25 for the whole interval, so normalizing the wave function I should get

    [itex]\psi = \frac{1}{25}\phi_1 + \frac{2}{25}\phi_2 [/itex]

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

    (25/4)^2 and (75/4)^2

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

    thanks in advance.
     
    Last edited: Apr 8, 2014
  2. jcsd
  3. Apr 8, 2014 #2
    In the original question you state that the wave function is "=11+22, where are the stationary states" are those boxes a mistake?
     
  4. Apr 8, 2014 #3
    try it now...
     
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