Free particle wave packet example

[@YourAverageHater]

1. Equal contribution from both the right going and left going waves, so i guess the wave is symmetric on both sides (maybe stationary?) 2. Probably not, if the square modulus of psi spreads out then the uncertainty in position increases, which means the uncertainty in momentum decreases?

[@ChanawerebiChanawerebi]

if wave function spreads out, does not it mean that wave length spreads out? and then by de Broglie formula, momentum spreads out too

[@albertliu2599]

@@ChanawerebiChanawerebi I think the thing is, when it spreading out, the function looks more and more like a plane wave (cos in this case). So We have more and more confidence about wavelength. In other words. The uncertainty of wavelength decreases, and so does the momentum.

[@ChanawerebiChanawerebi]

@@albertliu2599 Oh, okay! Thank You!

[@albertliu2599]

@@ChanawerebiChanawerebi You are welcome! Have a good time with QM. 😀

[@Dekoherence-ii8pw]

5:20 Shouldn't there be a square root on that 1/2pi?

[@betadistribution6534]

What about initial conditions which have momentum? How would you go about constructing the wavefunction for a free particle which has a net group velocity?

[@JonSedlak-xe5ef]

so from the wave animation the wavelet flattens out and there is no more wavelet and the probability animation suggest that the particle can be anywhere. Is that right?

[@the-fantabulous-g]

1:13 how did we get probability density function? He asked for us to recall but I don't remember much right now

[@mayanktiwari591]

It's simple just square the wavefuntions and we can already see them as parabolas.

[@dashywashy34]

Does anyone know the syntax for the wave function in the last example? I try to type it into sage, but I can't figure it out exactly.

[@aidankelly8013]

For your inverse fourier transform is it not 1/sqrt(2pi)?

[@timetraveller1237]

notice how the phi(k) actually has a 1/sqrt(2pi) but he just did not write it down i believe and the inverse transform actually has a 1/sqrt(2pi) also so they multiplied together give you 1/2pi

[@kontiimanalatit8987]

In signal analysis in engineering, we use 1/2pi before integral for inverse and 1 before integral for fourier transform. It depends on how you derive it I guess, but in my engineering studies, we use it like this

[@davidhand9721]

So if a wavefunction of a free particle is a superposition of several wavelengths (e.g. an envelope and a higher frequency inner wave) then the particle is observed, does it collapse into just one of these waves? A random one? Wouldn't we then get wacky, inconsistent results for e.g. velocity and previous position because the wavelength and velocity will be different from the whole ensemble? Does the observed wave suddenly eat the energy in the others and appear as if it alone had that energy?

[@betadistribution6534]

Wavefunction collapse depends on the basis used for measurement. I'm not sure how that works for free particles, since the basis is uncountably infinite and none of them is normalizable. But if you think about it, collapsing to a single momentum would imply \Delta p = 0, which would violate the uncertainty principle unless \Delta x = \infty. That doesn't seem like a physically realizable experiment.

[@davidhand9721]

@@betadistribution6534 but don't we make these measurements all the time? Like every time one electron scatters off another, or runs into a sensor?

[@yoyo-pi4mm]

Can somebody suggest another course. This one is so confusing. He doesn't explain things clearly

[@vintagejock3951]

Try MIT open courseware