Physical meaning of a wave packet w/ respect to HUP&duality

[CentralSand]

I'm a QM noob/newb trying to understand the physical implication of a wave packet, in my mind it is something like this:

On the x-axis there is displacement (vibration), probability on the y. I Imagine stretching and compressing the wave packet. When I stretch it out, the amplitude must decrease, as the total area must stay the same? As it stretches further and further it becomes closer to a line (being equally present along the trajectory?). So this equivalent to a constant massive particle speeding up? In reverse as something decreases in momentum, it's wave like character decreases as the wave packet becomes more localized, so it is easier to tell were it actually is. Increase it to infinity (which is of course impossible) and its position becomes completely defined, whist the probability of it being detected outside of its position becomes undefined, so we have no idea where it is going (momentum).
i.e. a normal wave has no position, because it continues to infinity, but we know it will oscillate the same forever. Yet, a wave pulse (packet) has a more specified definition, but it's oscillation changes (it's wave character dissipates), so we don't know how much momentum it really has.

And of course, how does this relate to the momentum of a wave? if the wave packet localizes, won't the wavelength get shorter, increasing momentum, thus being contradictory? Does the de Broglie momentum apply to all matter, and override the simple: p=mv? And p tends to infinity, wavelength tends to zero, so I'm guessing that relates to the constituent waves as producing a wave packet that is more confined means a wider distribution of wavelengths?

 

[blue_leaf77]

if the wave packet localizes, won't the wavelength get shorter, increasing momentum, thus being contradictory?

Wavelength won't change as long as there is no potential. When the wave packet is localized, what gets smaller is variance in space and what gets bigger is variance in space. However that statement is only valid in static case. As you have known, wavepackets disperse as time evolves (variance in space gets wider) but the momentum distribution stays unchanged, again assuming no potential on the way.

 

[CentralSand]

what gets smaller is variance in space and what gets bigger is variance in space.

Umm, what?

My understanding of momentum is essentially “how much something is effected when an object hits”, i.e. quantity of motion. Yet I'm actually confused as to whether there is a separate wave function in position representation, to a wave function in momentum representation. Like, what quality of the wave packet effects certainty of momentum and which effects certainty of position? When a wave packet becomes localized that means its position is more certain right? So the momentum must decrease in certainty somehow, which is physically drawn from which change in the wave packet?

 

[bhobba]

My understanding of momentum is essentially “how much something is effected when an object hits”, i.e. quantity of motion.

That's not what momentum is. At the deepest level its the conserved quantity associated with spatial symmetry implied by Noethers theorem:
http://www.physics.ucla.edu/~cwp/articles/noether.asg/noether.html

For wave-packets the Wikipedia article is quite good:
http://en.wikipedia.org/wiki/Wave_packet

Thanks
Bill

 

[CentralSand]

After watching this video: and some ones on noethers theorem (geometric as opposed to algebraic understanding) I think my analogy was quite close, I think my question is ultimately interpretational.

 

[CentralSand]

Oh and this one as well:

 

[bhobba]

After watching this video

The language its using is misleading - its not what I would call wrong - some are downright wrong - but it is misleading.

You can measure position and momentum to any accuracy you like - perfectly even.

The Heisenberg uncertainly principle is a statistical statement about similarly prepared systems. Take a large number of similarly prepared systems and divide them into two lots. If you measure the position is the first lot you will get an exact answer each time - it may be different for each measurement - but it's an exact answer. The same if you measure the the momentum in the second lot you will get an exact answer each time. But if you compare the standard deviation of the answers in each lot you will find it obeys the Heisenberg Uncertainty relation.

Thanks
Bill