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Problem
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\prob{1}
The wave function of a particle is $e^{i(kx-\omega t)}$ where x is distance, t is time, and k and $\omega$ are positive real numbers. The x-component of the momentum of the particle is

  1. 0
  2. $\hbar\omega$
  3. $\hbar k$
  4. $\frac{\hbar\omega}{c}$
  5. $\frac{\hbar k}{\omega}$

Quantum Mechanics}Momentum Operator


See below for user comments and alternate solutions! See below for user comments and alternate solutions!
Alternate Solutions
casseverhart13
2019-07-04 10:57:43
Thanks for your post! electrical contactor\r\nAlternate Solution - Unverified
tera
2006-08-21 02:14:34
You can get the result by dimentional analysisAlternate Solution - Unverified
Comments
casseverhart13
2019-07-04 10:57:43
Thanks for your post! electrical contactor\r\nAlternate Solution - Unverified
deepblue
2013-03-04 18:35:26
This website is great.

I was wondering if there was a possible miscalculation.

It appears using the momentum operator normally defined you would have

-(i)*(i)hk which leads to an imaginary number times an imaginary number which is -1 .

So you have - [(i)*(i)] = (-)(-1)hk. So you get answer (C) which is just hk.
=================================================
(My point is the author divided by (i) in the definition of the momentum operator instead of multipling the two imaginary numbers)
zeper
2013-03-09 09:11:39
no there is no mistake...1/i means -i...which is the same notation you mensioned.
NEC
grep
2006-10-30 13:28:58
You can also recognize that the wave function is just a standard transverse wave (like a light wave) for which de Broglie tells us h=lambda*p and the propagation number tells us k=2 pi/lambda
roofsing
2010-06-21 23:37:57

NEC
grep
2006-10-03 12:37:48
Oops, I meant it also leaves choice D
BerkeleyEric
2010-05-31 15:18:12
(D) actually gives units of kg * m/s^2, not kg * m^2/s as needed. (A) cannot be eliminated from dimensional analysis, but from physical intuition we know that a wave traveling in the x-direction should have momentum in the x-direction.
NEC
grep
2006-08-28 13:07:27
Hmm, doesn't dimensional analysis leave C as a possible choice? (or A for that matter)
gina4eva
2011-11-10 18:10:14
but isn't c the answer?
NEC
tera
2006-08-21 02:14:34
You can get the result by dimentional analysis
alemsalem
2010-09-24 08:30:20
the units for choice D also work but we dont know that the particle is moving with speed of light.
Quark
2011-10-25 12:04:21
More accurately, the particle can't be moving at the speed of light.
myscifilullaby1
2012-03-08 17:32:19
indeed! dimensional analysis gives two possibilities, A & C. When you use the momentum operator, you get C as the final answer.
Alternate Solution - Unverified
gottlob
2006-06-13 12:18:50
I am a physics teacher from Greece and am trying to download the gre.pdf forms of the gre tests. I have downloaded the GRE0177 but i can't get the others (GRE9677, GRE9277 etc. My e-mail is nchalk@phs.uoa.gr. Please, if any one can sent me, or give me information about, the tests i would be gratefull to him.NEC

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This website is great. I was wondering if there was a possible miscalculation. It appears using the momentum operator normally defined you would have -(i)*(i)hk which leads to an imaginary number times an imaginary number which is -1 . So you have - [(i)*(i)] = (-)(-1)hk. So you get answer (C) which is just hk. ================================================= (My point is the author divided by (i) in the definition of the momentum operator instead of multipling the two imaginary numbers)

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