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Gaussian Wave packet

  1. Sep 2, 2014 #1
    1. The problem statement, all variables and given/known data

    Wave function of an electron:

    ##\frac{1}{N}∫_a^b[e^{ik_0x}(1+\frac{1}{2}e^{i\frac{0.1}{2}x}+\frac{1}{2}e^{-i\frac{0.1}{2}x})]dx##

    The integrand becomes zero both to the left and right of x = 0 . Let a be the first time it hits zero to the left and b the first time it hits zero to the right. What value of N is required to normalize this wave-function?

    What is the probability that the electron is between (−1, 1) ?

    What is the expected value of the electron location?

    What value of α minimizes the variance of the electron?


    2. Relevant equations



    3. The attempt at a solution

    I'm pretty lost on this question. I have that the zero points as +-2(π)/0.1. I'm not sure how to get N from that.

    From what I've read online (I have no notes describing this from class) I should take the integral for the squared function with the limits -1 to 1. I don't even know where to start on this integral. The first exponential resembles one I've seen (it was given) before that was δ(k)2π when solved with limits of +-infinity. From what I understand it should look something like:

    ##\frac{1}{N}∫_a^b[e^{ik_0x}(1+\frac{1}{2}e^{i\frac{0.1}{2}x}+\frac{1}{2}e^{-i\frac{0.1}{2}x})]^2dx=1##

    Surely there must be a way to simplify this right?

    For the last problem I was told I have to multiply the inside of the integral with x so that it looks like

    ##\frac{1}{N}∫_a^b[e^{ik_0x}(1+\frac{1}{2}e^{i\frac{0.1}{2}x}+\frac{1}{2}e^{-i\frac{0.1}{2}x})]^2x dx##

    But I don't know what the limits should be, still -1,1?

    It's the first homework of the class and I'm immediately overwhelmed and confused. Thanks for any help!
     
  2. jcsd
  3. Sep 2, 2014 #2

    nrqed

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    Is the question phrased exactly like this? It does not make any sense as stated.
    First, there should be no integration over x there if this is meant to define a wave function. And the overall factor should be ## 1/\sqrt{N}##.
    Second, the statement "The integrand becomes zero both to the left and right of x = 0" is confusing. I guess they mean that it is zero at some point.

    Then the question "What value of N is required to normalize this wave-function?" is also impossible to understand. Maybe they mean that you are supposed to set the wave function equal to zero to the left and right of the first zeros. If so, they really did not state clearly what they meant.

    If that's the case, then simply impose [itex] \int_{x_a}^{x_b} | \psi(x)|^2 dx = 1 [/itex]
    You found the values of ##x_a## and ##x_b##.
     
  4. Sep 2, 2014 #3
    Yeah, he did a really poor job of explaining clearly what he wanted. One of his habits so far. He originally had the first question as simply find N with bounds neg infinity to infinity. But changed it today when it was clear that this wasn't possible.

    So I got ## | \psi(x)|^2## now but the way in which I have to solve the probability between -1 and 1 is confusing.

    If I solve the integral with ## | \psi(x)|^2## from -1 to 1 I get 10.99. Clearly this isn't the probability. How do I extrapolate the probability from that?
     
  5. Sep 2, 2014 #4

    nrqed

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    Did you find the normalization constant that forces the wave function to be normalized between ## \pm 20 \pi ##? If this is correct, the integral ##\int_{-1}^1 | \psi(x)|^2 dx ## has to give a value less than 1.
     
  6. Sep 2, 2014 #5
    I keep forgetting about N. In order to find it do I take the initial integral and just use the ##\pm \frac{2}{0.1} \pi## limits? I plugged it into wolfram and it gave me something crazy back.
     
  7. Sep 2, 2014 #6

    nrqed

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    Yes, you use those limits. You know that ##|\psi(x)|^2 = \psi^*(x) \psi(x) ##, right? Then the integrand is the sum of constants and trig functions.
     
  8. Sep 2, 2014 #7
    I'm just trying to figure out the order/process of things. I have done the integral of ##\psi## and it's complex conjugate. So, I guess when I'm trying to do the integral to find ##\frac{1}{N}## am I suppose to do the initial integral with those limits then square it or take the integral of ##|\psi(x)|^2## set it = to 1 and then solve for N?
     
  9. Sep 2, 2014 #8

    nrqed

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    Yes. Again, the statement of the question is very confusing but assuming he wants you to set the wave function equal to zero outside of these first zeros, then you impose that ##\int dx |\psi(x)|^2 =1## between these limits (where ##\psi(x)## contains the normalization factor). Then you will be able to find the probability between -1 and +1 by integrating ##\int dx |\psi(x)|^2 ## between these two limits (and you will get something less than 1).
     
  10. Sep 3, 2014 #9
    Okay, so I got a high number from the integral 376.99 and found that the 2 cosine functions from my ##|\psi(x)|^2## basically added nothing (both came out to ~##10^{-15}##). So, I have ##N(376.99)=1##
     
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