Stationary States and Spreading of Wave Function

[boderam]

What I know: In stationary states the time dependence is factored out so it is of the form phi(q) * e^(-i omega t), thus in its appearance there is no wave function spread. However I recall from texts that wave packet spread is considered a universal phenomena in quantum mechanics, so I am looking to resolve this contradiction.

 

[jtbell]

Wave packets are not stationary states.

 

[K^2]

No. It's just that a packet consists of stationary states with different frequencies and wave vectors.

Note that how quickly packet is supposed to spread depends on uncertainty in momentum, which is zero for the wave function you wrote. That, of course, is compensated by infinite uncertainty in position. If you try to compute <q²>-<q>², the first integral will diverge.

To construct a localized packet, you have to use different frequencies, so expectation of momentum will have uncertainty to it, and that will cause packet spread, unless the packet happens to travel at the speed of light (m=0). But that's relativistic QM already.

 

[boderam]

Note that how quickly packet is supposed to spread depends on uncertainty in momentum, which is zero for the wave function you wrote. That, of course, is compensated by infinite uncertainty in position. If you try to compute <q²>-<q>², the first integral will diverge.

Does that mean stationary states do not obey the Uncertainty Principle in the usual sense? We would need a sort of limiting process that would this work. I am having a hard time understanding this. I imagine a function like a gaussian being the phi(q) and then it multiplied by the phase factor, so I don't see how the uncertainty in position is infinite.