Calculating the momentum of a debroglie wave from its wave function

In summary, the conversation discusses the calculation of momentum for a wave function given by psi = Asin(8.92e10 x), where x is in metres. The conversation covers concepts such as planks constant, wavelength, operators, eigenstates, and eigenvalues in relation to finding the momentum. It also explains how energy units can be converted to momentum units, such as MeV/c, and how this relates to the dimensions of mass, length, and time. The conversation also touches on using the kinetic energy equation to find momentum and whether it is applicable to quantum relations. Overall, the conversation highlights the complexity and confusion surrounding the concept of momentum in quantum mechanics.
  • #1
mitch_1211
99
1
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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  • #2
Do you know what the wavelength of a sin function is?
 
  • #3
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
 
  • #4
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.
 
  • #5
@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...
 
  • #6
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.
 
  • #7
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...
 
  • #8
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!
 
  • #9
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?
 

1. What is the Debroglie wave function?

The Debroglie wave function is a mathematical representation of the wave-like behavior of a particle, proposed by physicist Louis Debroglie in the 1920s. It describes the probability amplitude of a particle based on its momentum and wavelength.

2. How do you calculate the momentum of a Debroglie wave from its wave function?

The momentum of a Debroglie wave can be calculated by taking the derivative of its wave function with respect to position. This is represented by the formula p = h/λ, where p is momentum, h is Planck's constant, and λ is the wavelength of the wave.

3. What is the relationship between the Debroglie wavelength and the momentum of a particle?

According to Debroglie's hypothesis, the wavelength of a particle is inversely proportional to its momentum. This means that as the momentum of a particle increases, its wavelength decreases and vice versa.

4. Can the Debroglie wave function be used to describe all particles?

In quantum mechanics, the Debroglie wave function can be applied to all particles, including electrons, protons, and even larger particles like atoms and molecules. However, it is most commonly used to describe the behavior of subatomic particles.

5. How is the concept of Debroglie waves related to the wave-particle duality?

The wave-particle duality is the idea that particles can exhibit both wave-like and particle-like properties. The Debroglie wave function is one way of mathematically representing the wave-like behavior of particles, providing evidence for the wave-particle duality concept.

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