I am trying to show the uncertainty relation of $\Delta x\Delta p = \frac{\hbar}{2}$ using a gaussian wave function and its Fourier transform. I have found correctly the uncertainty in position $\Delta x = \frac{\sigma}{\sqrt 2}$ and am trying to find the uncertainty of the wavenumber to get the uncertainty in momentum. I am getting the reciprocal of what it is supposed to be, can someone tell me what I am doing wrong? Here is my working for the uncertainty in the wave number, I am getting $\Delta k= \sqrt 2 \sigma$ but it should be $\Delta k = \frac{1}{\sqrt 2 \sigma}$ : 
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2 Answers
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In the expression $$ e^{-k^2 \sigma^2} $$ to find the standard deviation you can compare it to the function $$ e^{-x^2/2 a^2}. $$ In this function of $x$, the standard deviation is $a$. So the general statement about Gaussian functions is that they have the form $$ \exp( - x^2 / 2 \Delta x^2 ) $$ or in the case you are interested in, $$ e^{ - k^2 / 2 \Delta k^2 }. $$ Equating this to $e^{- k^2 \sigma^2}$ you get $$ \frac{1}{2 \Delta k^2} = \sigma^2 $$ So we have $$ \Delta k = \frac{1}{\sqrt{2} \sigma}. $$ Combining this with $\Delta x = \sigma / \sqrt{2}$ gives $$ \Delta x \Delta k = \frac{1}{2} $$
answered Nov 21, 2020 at 14:02
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$\begingroup$ So I compare the squared $\widetilde{ \psi }(k)$ to the standard gaussian in order to get the uncertainty in k? $\endgroup$– Σ baryonCommented Nov 21, 2020 at 14:12
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2$\begingroup$ Yes, because the standard deviation is an aspect of the probability distribution. $\endgroup$ Commented Nov 21, 2020 at 14:31
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When solving for $\Delta k$ you write $$\frac{(\Delta k)^{2}}{2}=\sigma^{2}$$ Where did this statement come from? Maybe start with $$\frac{(\Delta k)^{2}\sigma^2}{2}=1$$
