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

On my way to Heisenberg uncert. princ.

  1. Mar 1, 2013 #1
    I did a Fourier transform of a gaussian function ##\scriptsize \mathcal{G}(k) = A \exp\left[-\frac{(k-k_0)^2}{2 {\sigma_k}^2}\right]## and derived a result ##\sigma_k \sigma_x = 1## which is the same they get on this webpage - read paragraph right above the picture. Here is a procedure i used described in 2 steps:

    1 st) I did the Fourier transform of the Gaussian:

    [itex]
    \scriptsize
    \begin{split}
    \mathcal{F}(x) &= \int\limits_{-\infty}^{\infty} \mathcal{G}(k) e^{ikx} \, \textrm{d} k = \int\limits_{-\infty}^{\infty} A \exp \left[-\frac{(k-k_0)^2}{2 {\sigma_k}^2}\right] e^{ikx}\, \textrm{d} k = A \int\limits_{-\infty}^{\infty} \exp \left[-\frac{(k-k_0)^2}{2 {\sigma_k}^2} \right] e^{ikx}\, \textrm{d} k =\\
    &= A \int\limits_{-\infty}^{\infty} \exp \left[-\frac{m^2}{2 {\sigma_k}^2} \right] e^{i(m+k_0)x}\, \textrm{d} m = A \int\limits_{-\infty}^{\infty} \exp \left[-\frac{m^2}{2 {\sigma_k}^2} \right] e^{imx} e^{ik_0x}\, \textrm{d} m =\\
    &= A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-\frac{m^2}{2 {\sigma_k}^2} \right] e^{imx}\, \textrm{d} m = A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-u^2 \right] e^{iu \sqrt{2} {\sigma_k} x} \sqrt{2} {\sigma_k} \textrm{d} u = \\
    &=\sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-u^2 \right] e^{iu \sqrt{2} {\sigma_k} x}\, \mathrm{d} u = \sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-u^2 + i u \sqrt{2} {\sigma_k} x \right]\, \mathrm{d} u =\\
    &= \sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-\left(u + \frac{i {\sigma_k} x}{\sqrt{2}} \right)^2 - \frac{i^2 {\sigma_k}^2 x^2 }{2}\right]\, \mathrm{d} u =\\
    &= \sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} \exp \left[-\left(u + \frac{i {\sigma_k} x}{\sqrt{2}} \right)^2 + \frac{{\sigma_k}^2 x^2 }{2}\right]\, \mathrm{d} u = \\
    &= \sqrt{2} {\sigma_k} A e^{ik_0x} \int\limits_{-\infty}^{\infty} e^{-z^2} \exp \left[ \frac{{{\sigma_k}}^2 x^2 }{2} \right]\, \mathrm{d} z = \sqrt{2} {\sigma_k} A e^{ik_0x} \exp \left[ \frac{{{\sigma_k}}^2 x^2 }{2} \right] \underbrace{\int\limits_{-\infty}^{\infty} e^{-z^2} \, \mathrm{d} z}_{\text{Gauss integral}}=\\
    &= \sqrt{2} {\sigma_k} A e^{ik_0x} \exp \left[ \frac{{{\sigma_k}}^2 x^2 }{2} \right] \sqrt{\pi}\\
    \mathcal{F} (x)&= \sqrt{2\pi} {\sigma_k} A e^{ik_0x} \exp \left[ \frac{{{\sigma_k}}^2 x^2 }{2} \right]\\
    \end{split}
    [/itex]

    2nd) I did some modifications to get the desired result:

    It is said on Wikipedia that the Gauss will be normalized only if ##\scriptsize A=1 /(\sqrt{2 \pi} \sigma_k)##. I used this on ##\mathcal{F}(x)## and got a result which corresponds with a result on Wikipedia - read chapter "Fourier transform and characteristic function", so i think it must be ok but please confirm:
    [itex]
    \mathcal{F} (x)= e^{ik_0x} e^{\frac{{{\sigma_k}}^2 x^2 }{2}}\\
    [/itex]
    I used a centralized Gauss whose mean value is ##\scriptsize k_0=0## and got:
    [itex]
    \mathcal{F} (x)= e^{\frac{{{\sigma_k}}^2 x^2 }{2}}\\
    [/itex]
    Which can be rewritten as :
    [itex]
    \mathcal{F} (x)= e^{\frac{x^2 }{2 \left(1/\sigma_k \right)^2}}\\
    [/itex]
    And i can see that:
    [itex]
    \begin{split}
    &~~1/\sigma_k = \sigma_x\\
    &\boxed{\sigma_k \sigma_x = 1}
    \end{split}
    [/itex]

    Could you please tell me what is this that i just derived and tell me how can i continue to get (derive) Heisenberg uncertainty principle ##\scriptsize \Delta x \Delta p = \frac{\hbar}{2}##? I am kind of newbie with Dirac notation so take it easy on me please.
     
  2. jcsd
  3. Mar 1, 2013 #2

    jtbell

    User Avatar

    Staff: Mentor

    What you have is simply a different notation for ##\Delta k \Delta x = 1##, although it should actually be 1/2, not 1. I haven't looked at your math in detail, but I suspect that either (a) you have misplaced a 2 somewhere, or (b) you are defining σx in terms of the probability amplitude ψ and not the probability distribution |ψ|2, or similarly for σk.


    k is the wavenumber, equal to ##2\pi/\lambda##. And of course ##\lambda = h/p## (de Broglie).
     
    Last edited: Mar 1, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: On my way to Heisenberg uncert. princ.
  1. Heisenberg Picture (Replies: 3)

  2. Heisenberg Equation (Replies: 5)

Loading...