Calculating the momentum of a debroglie wave from its wave function

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Discussion Overview

The discussion revolves around calculating the momentum of a de Broglie wave from its wave function, specifically the wave function psi = Asin(8.92e10 x). Participants explore the relationship between the wave function, its wavelength, and the corresponding momentum, while grappling with concepts from quantum mechanics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant seeks guidance on calculating momentum from the wave function, expressing uncertainty about extracting information from it.
  • Another participant points out that the number in front of x corresponds to the wave number k, leading to a discussion about using this wavelength for de Broglie wave calculations.
  • A participant asserts that the given wave function is not an eigenstate of the momentum operator and suggests that it represents a superposition of two opposite momenta, resulting in an average momentum of zero.
  • One participant questions the validity of the zero average momentum conclusion, expressing skepticism about the simplicity of that answer in the context of their assignment.
  • Another participant speculates that the exercise may be part of an introductory course where quantum concepts are not fully explained, suggesting that proceeding with the initial approach may satisfy the instructor despite its inaccuracies.
  • A participant raises a question about converting momentum into keV/c, discussing the physical meaning of this unit and how to derive it without unnecessary conversions.
  • One participant explains the dimensional analysis of energy and momentum, clarifying how dividing energy by the speed of light yields momentum units.
  • A participant shares their calculated momentum values and seeks validation on whether these values are reasonable.
  • Another participant inquires about the applicability of classical kinetic energy equations in the context of quantum mechanics, specifically regarding the de Broglie relation.
  • One participant specifies that the wave function pertains to an electron and questions the validity of using k = 2pi/wavelength for matter waves.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the wave function and its implications for momentum. There is no consensus on whether the average momentum should be considered zero or if the initial approach to calculating momentum is valid.

Contextual Notes

Participants highlight limitations in understanding quantum mechanics concepts, particularly regarding operators and eigenstates, which may affect their interpretations and calculations.

mitch_1211
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I want to calculate the momentum of the wave given by
psi = Asin(8.92e10 x) where x is in metres.

I know that momentum = planks/wavelength, but I'm unsure of how to get any information out of the wavefunction alone...

any guidance would be appreciated.

mitch
 
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Do you know what the wavelength of a sin function is?
 
Matterwave said:
Do you know what the wavelength of a sin function is?

Ahh so the number infront of the x corresponds to k in y = Asin(kx-wt) where k = 2pi/wavelength.

From here i can just use this wavelength as per normal calculations for debroglie waves?

mitch
 
mitch_1211 said:
I want to calculate the momentum of the wave given by
psi = Asin(8.92e10 x)
Your psi is not an eigenstate of the momentum operator. It is a superposition of two opposite momenta, with the average momentum equal to zero.
 
@Demystifier

So you are saying the 'correct' answer should be zero?
I don't doubt your reasoning, but for this set of problems I'm looking at i think that would be a little to easy, especially since it is asking for 3 decimal places. Thanks for your answer though! I'm finding it really hard to get my head around these 'matter' waves...
 
Mitch, I suspect that your exercise is a part of some introductory course to quantum mechanics, where concepts like operators, eigenstates and eigenvalues are not properly explained. Am I right? If that is so, then I'm sure your teacher will be satisfied if you proceed along the lines you indicated in post #3. Despite the fact that, strictly speaking, it is wrong.

Alternatively, you can try to impress your teacher by telling him what I told you in #4. But this is very risky, because a teacher who posed such a stupid exercise may be too stupid to see that his answer to his exercise is actually wrong.
 
Demystifier said:
a teacher who posed such a stupid exercise may be too stupid to see that his answer to his exercise is actually wrong.

You aren't far from the truth. I think if i say the average momentum cancels out and is zero, then i would get a big fat zero.

But i can see what you mean, what puzzles me is why they aren't explaining those basic things right from the get go...
 
I have on more question. I need the momentum in keV/c now i know i can use planks constant in ev to get the ev part, but how does the per c part come into it?

thanks!
 
mitch_1211 said:
I need the momentum in keV/c

ok so i have the conversion factor 1 keV/c = 5.36 x 10-25 kg-m/s

but i don't get the physical meaning of momentum described by the 'per speed of light' part of it...

And is there a way to get the momentum to come out as keV/c so the conversion isn't necessary?

I have got it down to p=[itex]\frac{h}{\frac{2\pi}{k}}[/itex]

but this will give the SI units of momentum, not keV/c..
 
  • #10
It gets better, though. Energy units have dimensions of (mass)(length)2/(time)2. Divide that by the dimensions of velocity, (length)/(time). What do you get?

That's right, (mass)(length)/(time) which are the dimensions of momentum. Physicists take advantage of this to divide energy in, say MeV, by a fundamental constant that has units of velocity, namely the speed of light c to get useful units of momentum: MeV/c. Anytime we divide energy units by c we get momentum: eV/c, keV/c, MeV/c, GeV/c and even TeV/c. To convert these to MKS, use the table below.

What happens if we divide by c again? Then we divide (length)/(time) out of our momentum dimensions, leaving only (mass). That's right, physicists make mass units out of eV/c2, keV/c2, MeV/c2, GeV/c2 and, yes, TeV/c2

So this website is very useful in explaining the use of ev and c, i would still much rather work in SI units and have 'yuck' exponents than use ev... I guess its because they aren't physical quantities that i can imagine or relate to...
 
  • #11
if anyone would like to check me on this, i get momentum as ~ 9.40681e-24 kgm/s or ~ 17.5 keV/c do these sound in the right ball park?
 
  • #12
sorry for all these posts,

I know that if i rearrange k.e = 1/2 m v ^2 i can get p = sqrt(2mE) E = kinetic

Can i use this in the p = h/wavelength debroglie relation even though 1/2mv^2 is not like a 'quantum' relation?
 
  • #13
i also should've specified that that wavefunction was for an electron...

so is k = 2pi/wavelength still valid for matter waves?
 

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