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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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2 Answers 2

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 '13 at 20:57
    
I am using bra-ket notation. en.wikipedia.org/wiki/Bra%E2%80%93ket_notation –  Mark Eichenlaub Feb 28 '13 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 '13 at 21:05
    
I used $\psi$ because that is standard notation for a wavefunction –  Mark Eichenlaub Feb 28 '13 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 '13 at 18:04

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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