Heisenberg uncertainty principle derivation - unexplained factor of $4 \sigma_k^2$ in Gaussian

I did a Fourier transform of a gaussian function $\scriptsize \mathcal{G}(k) = A \exp\left[-\frac{(k-k_0)^2}{2 {\sigma_k}^2}\right]$

$$\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}$$

It can be seen that Fourier transform equals $\scriptsize \mathcal{F} (x)= \sqrt{2\pi} {\sigma_k} A e^{ik_0x} \exp \left[ ({{\sigma_k}}^2 x^2) / 2\right]$. It is said on Wikipedia that the Gauss will be normalized only if $\scriptsize A=1 /(\sqrt{2 \pi} \sigma_k)$. I used this and got a result which corresponds with a result on Wikipedia - Fourier transform and characteristic function: $$\mathcal{F} (x)= e^{ik_0x} e^{\frac{{{\sigma_k}}^2 x^2 }{2}}\\$$ If i use a centralized Gauss whose mean value is $k_0=0$ i get: $$\mathcal{F} (x)= e^{\frac{{{\sigma_k}}^2 x^2 }{2}}\\$$ Which can be written as a: $$\mathcal{F} (x)= e^{\frac{x^2 }{2 \left(1/\sigma_k \right)^2}}\\$$

And i can see that $1/\sigma_k = \sigma_x$ BUT from this it follows that i get the Heisenberg uncertainty principle like this: $$\begin{split} \sigma_k \sigma_x &= 1\\ \Delta k \Delta x &= 1\\ \Delta p / \hbar \, \Delta x &= 1\\ \Delta p \Delta x &= \hbar\\ \end{split}$$

And this is a wrong result because i should get $\hbar/2$ in place of $\hbar$.

Question: On our university professor derived this in a simmilar way but in the beginning in Gaussian he used $4{\sigma_k}^2$ instead of $2 {\sigma_k}^2$. This contributed to the right result $\hbar/2$ in the end. But i want to know why do we use factor $4$ instead of $2$?

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1
Just asking where you went wrong in your work isn't really appropriate for this site. I'd suggest editing this to focus more on your first question, namely why you don't get the HUP from your calculation. You don't need to show your complete work, but if outline what you did in just a little more detail (right here in the question, don't make people follow a link) it would be quite fine. – David Zaslavsky♦ Feb 28 at 3:00

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I totaly rewrote the question and set some explicit questions which i cannot explain to myself. Today i was at university and my professor allso doesn't know why this is the case - i mean he couldn't explain to me where does facor 4 come from (please reread the question). – 71GA Feb 28 at 16:58

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That's indeed a much better question, thanks! – David Zaslavsky♦ Feb 28 at 17:24

2 Answers

You're treating this like a probability distribution instead of a wavefunction. Instead of assuming $2 \sigma_k^2$ vs. $4 \sigma_k^2$, I suggest setting the denominator equal to some constant and then finding the true variances in both position and wavenumber space directly--i.e. through the relation

$$\sigma_a^2 = \langle \psi | (\hat a - \bar a)^2 | \psi \rangle$$

for any observable $\hat a$ with expectation $\langle \hat a \rangle = \bar a$. Break this into two integrals and see what you get for $\hat a = \hat k$ and $\hat a = \hat x$.

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That's a long stream of equations to unravel, but it looks like you're trying to equate the standard deviation of the wavefunction in position space with $\Delta x$, which is not right. $\Delta x$ is defined by

$$(\Delta x)^2 = \langle \psi | X^2 | \psi \rangle - (\langle \psi | X|\psi \rangle)^2$$

You have to use that form to calculate the uncertainty.

For example, if your wavefunction is $\psi(x) = A e^{-x^2/\sigma^2}$, then

$$(\Delta x)^2 = A^2 \int e^{-x^2/\sigma^2} x^2 e^{-x^2/\sigma^2}$$

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,.Does this mean that my big integral is correct and i scew this up in my last steps? And what is that $X$ there. This notation $<\Psi|X|\Psi>$ is unknown to me. Is this maybee a average? Why is there $\Psi$ in notation? – 71GA Feb 28 at 20:57
– Mark Eichenlaub Feb 28 at 21:01
It is a bit long to explain in detail; your quantum mechanics textbook should explain it, though. I added a little detail on how to do this with a wavefunction. The answer to your question is essentially that the things you were call $\Delta x$ is not the $\Delta x$ in the uncertainty relationship. – Mark Eichenlaub Feb 28 at 21:05
I used $\psi$ because that is standard notation for a wavefunction – Mark Eichenlaub Feb 28 at 21:06
How did you know (i mean by what definition - please point me to this definition where i can read more) that: $(\Delta x)^2 = \int A e^{-x^2/\sigma^2} x^2 Ae^{-x^2/\sigma^2}$. So you think that my result $\sigma_k \sigma_x = 1$ is correct? Please confirm this. – 71GA Mar 1 at 18:04
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