GR 8677927796770177 | # Login | Register

GR9277 #1
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\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$\Rightarrow$}Momentum Operator

$\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}\psi \Rightarrow p \psi = \hbar k \psi
$

Alternate Solutions
 casseverhart132019-07-04 10:57:43 Thanks for your post! electrical contactor\r\nReply to this comment tera2006-08-21 02:14:34 You can get the result by dimentional analysisReply to this comment
Comments
casseverhart13
2019-07-04 10:57:43
Thanks for your post! electrical contactor\r\n
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)
 zeper2013-03-09 09:11:39 no there is no mistake...1/i means -i...which is the same notation you mensioned.
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
 roofsing2010-06-21 23:37:57
grep
2006-10-03 12:37:48
Oops, I meant it also leaves choice D
 BerkeleyEric2010-05-31 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.
grep
2006-08-28 13:07:27
Hmm, doesn't dimensional analysis leave C as a possible choice? (or A for that matter)
 gina4eva2011-11-10 18:10:14 but isn't c the answer?
tera
2006-08-21 02:14:34
You can get the result by dimentional analysis
 alemsalem2010-09-24 08:30:20 the units for choice D also work but we dont know that the particle is moving with speed of light.
 Quark2011-10-25 12:04:21 More accurately, the particle can't be moving at the speed of light.
 myscifilullaby12012-03-08 17:32:19 indeed! dimensional analysis gives two possibilities, A & C. When you use the momentum operator, you get C as the final answer.
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.

Post A Comment!
You are replying to:
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.

 Username: Password:
Click here to register.
This comment is best classified as a: (mouseover)
Mouseover the respective type above for an explanation of each type.

## Bare Basic LaTeX Rosetta Stone

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...