At our QM intro our professor said that we derive(adsbygoogle = window.adsbygoogle || []).push({}); uncertainty principleusing the integral of plane waves ##\psi = \psi_0(k) e^{i(kx - \omega t)}## over wave numbers ##k##. We do it at ##t=0## hence ##\psi = \psi_0(k) e^{ikx}##

[itex]

\psi = \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k

[/itex]

where ##\psi_0(k)## is a##k##-dependent normalisation factor(please correct me if I am wrong). This dependency was said to be a Gaussian function

[itex]

\psi_0(k)= \psi_0 e^{i(k-k_0)^2/4\sigma k^2}

[/itex]

where ##\psi_0## is anordinary normalisation factor(please correct me if I am wrong).

QUESTION 1:Why do we choose ##\psi_0(k)## as a gauss function? Why is this function so appropriate in this case?

QUESTION 2:I don't know how did our professor get a gauss function with an imagnary number ##i## in it. His gauss is nothing like the one on Wikipedia which is

[itex]

f(x) = a e^{-(x-b)^2/2c^2}

[/itex]

QUESTION 3:We used the first integral i wrote down to calculate the Heisenberg's uncertainty principle like shown below, but it seems to me that most of the steps are missing and this is the reason i don't understand this. Could anyone explain to me step by step how to do this.

[itex]

\begin{split}

\psi &= \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k\\

\psi &= \int\limits_{-\infty}^{+\infty} \psi_0 e^{i(k-k_0)^2/4\sigma k^2} \cdot e^{ikx} \, \textrm{d} k\\

\psi &= \psi_0 2 \sqrt{\pi} e^{ik_0x} e^{-x/2 \sigma k^2}

\end{split}

[/itex]

I think this is connected to a Gaussian integral, but it doesn't look quite like it to me. Well in the end our professor just says that out of the above it follows that

[itex]

\boxed{\delta x \delta k = \frac{1}{2}}

[/itex]

I don't understand this neither. It was way too fast or me.

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# Gaussian function to derive Heisenberg's uncertainty principle

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