|
GR9677 #76
|
|
|
Problem
|
|
|
This problem is still being typed. |
Quantum Mechanics }Uncertainty
One can make a good stab at this problem by applying the uncertainty principle.
I. If the average momentum of the packet is 0, then one violates the uncertainty principle. See IV.
II. Maybe.
III. Maybe.
IV. True, recall the Gaussian uncertainty principle  .
Since I is false, choices (A), (C), and (E) are out. Choices (B) and (D) remain. Take the conservative approach and choose (B).
|
|
|
Alternate Solutions |
pablojm
2006-10-28 16:03:06 | I think buddy.epson's argument is even better than the original one, since can be 0 and not violate the uncertainty principle. This is because the uncertainty principle relates standard deviations, and the standard deviation of p is sqrt(-^2). Thus, you could have different from 0, so the average momentum is 0 and the uncertainty principle holds. However, the problem says that the statements should be correct for ALL wave packets, and if we have a wavepacket which starts motion to the right, its average momentum will continue to be to the right for all time because it's in free space. |  |
|
|
Comments |
sullx
2009-11-03 19:29:42 | For the visual learners:
http://rugth30.phys.rug.nl/quantummechanics/potential.htm
I watched these videos a few days before I took 9677 and I answered this problem correctly because of it.
|  | wangjj0120
2008-09-24 06:24:43 | I don't understand~the wave packet is traveling through "free space", it's a non-dispersive medium. Waves with different  and  propagate with  , from this point the wave packet should maintain its shape. What wrong with this point of view? | | jburkart
2007-11-02 18:04:15 | I thought along the lines of: [II] and [III] can't both be true, since if it's getting wider, due to normalization, its amplitude must be decreasing, so I could eliminate choices (D) and (E). [IV] I knew was true, from the uncertainty principle, so I eliminated (A). Lastly, [I] seemed highly suspect, and [II] and [III] both seemed about equally probable, so I went with the higher probability choice of (B).
99percent
2008-11-06 08:03:38 |
This is correct...
A gaussian wavepacket always expands in free space.. Thus, II is definitely true..!!
Now since II is correct, III has to be false..!!
Bingo..!!
For a better and more detailed explanation check this journal at IOP...
http://www.iop.org/EJ/abstract/0143-0807/18/3/022
" target="_blank">http://www.iop.org/EJ/abstract/0143-0807/18/3/022">http://www.iop.org/EJ/abstract/0143-0807/18/3/022
|
jmason86
2009-10-01 19:52:38 |
Also, I may seems suspect seen momentum is is a description of something velocity, and this is a "wave packet that TRAVELS through free space". So the average momentum shouldn't be zero.
|
| | dumbguy
2007-10-17 14:45:10 | I have never heard of a wave that has a constant amplitude. Doesn't as time goes on, the amplitude fluxuate back and forth? So isn't this enough reason to rule III as false? let me know
sawtooth
2007-10-29 09:30:13 |
By amplitude, usually we mean the maximum of the disturbance (some call this the peak amplitude). As I know it, amplitude is the coefficient  e.g. in the following wavefunction: =Asin(\omega t - k x)) . When doesn;t depent on  we call it constant. Ofcourse, the displacement (or whatever the oscillating "thing" is) fluxuates.
At least this is the case in this exercise-and usually it's used this way in quantum mechanics.
|
| | shoyer
2007-08-05 22:40:59 | I have a reason for the guess in the last stage of the official answer.rnrnOnce eliminate I and have decided that IV is true, you have two options: B or D. (II and IV or II, III and V).rnrnIf II must be true, then the wave spreads out. In that case, then amplitude remaining the same over time would mean that the total probability of measuring the wave packet integrated over all space would increase beyond 1. Since this can't be true, eliminate answer D. | | pablojm
2006-10-28 16:03:06 | I think buddy.epson's argument is even better than the original one, since can be 0 and not violate the uncertainty principle. This is because the uncertainty principle relates standard deviations, and the standard deviation of p is sqrt(-^2). Thus, you could have different from 0, so the average momentum is 0 and the uncertainty principle holds. However, the problem says that the statements should be correct for ALL wave packets, and if we have a wavepacket which starts motion to the right, its average momentum will continue to be to the right for all time because it's in free space. | | buddy.epson
2006-10-14 11:05:28 | I eliminated statement I by reasoning that a bound particle (e.g. in a 1-D well) could have time-avg'd momentum of zero as it bounces back and forth in the well, but a free particle would travel in one direction and hence have a nonzero avg. momentum.
After reading yosun's explanation above, I am curious if my reasoning is flawed. I encourage discussion, as my quantum mech. knowledge is limited. | | andreas
2005-11-10 03:46:31 | daschiach: True, and this can even be shown by IV. If anybody wants more info on this, check on the difference between Group and Phase velocity. For example:
http://en.wikipedia.org/wiki/Phase_velocity />
http://en.wikipedia.org/wiki/Group_velocity | | daschaich
2005-11-08 23:22:33 | As for II, we know that a Gaussian wave packet is the superposition of waves of several different momenta, which will travel at different speeds as time goes on. So the wave packet will spread out as time passes.
This shows that III is false, since if the wave packet is spreading out, its peak amplitude has to decrease to keep things properly normalized.
Townsend has a nice discussion on pgs. 160-164 of his "A Modern Approach to Quantum Mechanics" book. | |
|
| Post A Comment! |
|
|
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
|
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
|
