# On my way to Heisenberg uncert. princ.

1. Mar 1, 2013

### 71GA

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:

$\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}$

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:
$\mathcal{F} (x)= e^{ik_0x} e^{\frac{{{\sigma_k}}^2 x^2 }{2}}\\$
I used a centralized Gauss whose mean value is $\scriptsize k_0=0$ and got:
$\mathcal{F} (x)= e^{\frac{{{\sigma_k}}^2 x^2 }{2}}\\$
Which can be rewritten as :
$\mathcal{F} (x)= e^{\frac{x^2 }{2 \left(1/\sigma_k \right)^2}}\\$
And i can see that:
$\begin{split} &~~1/\sigma_k = \sigma_x\\ &\boxed{\sigma_k \sigma_x = 1} \end{split}$

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. Mar 1, 2013

### 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