Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gaussian Integrals for Quantum States of well Defined Momentum

  1. Oct 18, 2013 #1
    Consider the Gaussian Integral (eqn 2.64).. is anyone able to explain how the constant of normalization is rationalised?
     

    Attached Files:

    Last edited: Oct 18, 2013
  2. jcsd
  3. Oct 18, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If by 'rationalized' you mean 'normalized' so that the integral of the pdf from -inf to + inf = 1, I would think in the usual fashion.
     
  4. Oct 18, 2013 #3
    Hey SteamKing, thanks for stopping by to help. I'm not referring to the normalisation process here, it's the constant on the exponential i'm having trouble with.. The equation shows a rationalisation of the numerator to give (2∏(hbar)^2 / 4σ^2)^-1/4.

    From what i understand if one wants to move a quantity from the numerator under the fourth principal root sign in the denominator, i must change the sign of the exponent and square that quantity twice but it doesn't work out. Perhaps i'm doing something wrong here?
     
    Last edited: Oct 18, 2013
  5. Oct 18, 2013 #4

    jtbell

    User Avatar

    Staff: Mentor

    It comes from normalizing ##\langle p | \psi \rangle##, that is, requiring that ##| \langle p | \psi \rangle |^2 = 1##.
     
  6. Oct 18, 2013 #5
    jtbell i apologise for any confusion here but I'm not referring to the normalisation process, it's the constant on the exponential. The equation shows a rationalisation of the numerator to give 1 /(2∏(hbar)^2 / 4σ^2)^1/4.

    From what i understand if one wants to move a quantity from the numerator under the fourth principal root sign in the denominator, i must change the sign of the exponent and square that quantity twice but it doesn't work out. Perhaps i'm doing something wrong here?
     
    Last edited: Oct 18, 2013
  7. Oct 18, 2013 #6
    [itex] h = 2\pi \hbar[/itex]

    [itex] \sqrt[4]{\frac{16\sigma^4\pi^2}{h^2 (2\pi \sigma^2)}} = \sqrt[4]{\frac{16\pi^2\sigma^2}{2\pi h^2}} = \sqrt[4]{\frac{16\sigma^2\pi^2}{2\pi\left(2\pi \hbar\right)^2}}[/itex]
    [itex] = \sqrt[4]{\frac{16\sigma^2}{2\pi 4\hbar^2}} = \sqrt[4]{\frac{4\sigma^2}{2\pi \hbar^2}} = \frac{1}{\sqrt[4]{2\pi \left(\hbar^2/(4\sigma^2) \right)}}[/itex]

    The momentum distribution does not describe a definite (well defined) momentum but rather corresponds to the case that both position and momentum are normal distributions, and this equation shows the relation between their standard deviations.
     
    Last edited: Oct 18, 2013
  8. Oct 18, 2013 #7

    jtbell

    User Avatar

    Staff: Mentor

    If you show us explicitly what you thought it should work out to be, in stepwise fashion, someone can probably tell you what your mistake is.
     
  9. Oct 18, 2013 #8
    Equation 2.64 implies that the normalisation constants are equivalent. that is..

    (2σ √∏) / √h(2∏σ^2)^1/4 = (2∏(hbar)^2 / 4σ^2)^-1/4.

    here's the algebra i did..


    (2σ √∏) / √h(2∏σ^2)^1/4 = √∏ / (2∏/16σ^2)^1/4 ..

    ..=√h (2∏/16σ^2 ^∏^2)^-1/4 = (2∏(hbar)^2 / 16σ^2)^-1/4.
     
  10. Oct 18, 2013 #9
    [itex]\frac{2\sigma \pi^{1/2}}{h^{1/2}(2\pi \sigma^2)^{1/4}}=\left [\frac{16\sigma^4\pi^2}{2\pi h^2\sigma^2}\right ]^{1/4}=\left [\frac{16\sigma^2\pi^2}{2\pi (2\pi\hbar)^2}\right ]^{1/4}=\left [\frac{16\sigma^2\pi^2}{8\pi^3 \hbar^2}\right ]^{1/4}=\left[\frac{4\sigma^2}{2\pi \hbar^2}\right ]^{1/4}[/itex]
     
  11. Oct 18, 2013 #10
    Hey that's great! i never thought of transforming it that way.. i guess what i was attempting to do at first before was nonsensical although i'm not sure why, it seems they should have revealed the same? i'm sure i'm on the right track with this now. Thanks for going to the time to show me where i was going wrong..
     
  12. Oct 18, 2013 #11
    i can see where i going wrong with this now.. thanks Dauto!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Gaussian Integrals for Quantum States of well Defined Momentum
Loading...