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I want to find out how big a Gaussian wave packet of a $C_{60}$ beam of 300 m/s is after 100 ms of propagation if the initial $1/e^2$ confinement was 12 pm?

I know for a Gaussian wave packet the spacial width grows with $$ \sigma_x(t)=\sqrt{\sigma_x^2+\frac{\hbar^2 t^2}{4m^2\sigma_x^2}}. $$

Can I now just say $\sigma_x(0)=\sigma_x=12\cdot 10^{-12}$ m ? Then knowing the mass $m$ of a $C_{60}$ molecule I calculate $\sigma_x(10^{-1}s)=0.367$ m. How can this have any meaning without taking in account the velocity of the beam? To my intuition the uncertainty in momentum and maximum of the wave function in momentum space should be the most important quantities here, yet I don't need them?

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To my intuition the uncertainty in momentum and maximum of the wave function in momentum space should be the most important quantities here, yet I don't need them?

Who says that you don't? The formula that you've used for $\sigma_x(t)$ requires a fully coherent initial gaussian state for which the uncertainty principle $$ \Delta x \,\Delta p\geq \hbar $$ is saturated, i.e. for which $$ \frac{\hbar}{\Delta x} =\Delta p. $$ Since the combination on the left appears in your $\sigma_x(t)$, so does $\Delta p$.

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  • $\begingroup$ If the initial state is not fully coherent, $$ \sigma_x(t)=\sqrt{\frac{1}{\sigma_k^2}+\frac{\hbar^2 t^2}{m^2\sigma_k^2}},$$ correct? Does that mean the velocity of the beam doesn't matter? $\endgroup$ – ty. Mar 18 '18 at 21:41
  • $\begingroup$ If the initial state is not fully coherent then that expression is incorrect. $\endgroup$ – Emilio Pisanty Mar 18 '18 at 22:27

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