Calculating the momentum of a debroglie wave from its wave function

[mitch_1211]

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

 

[Matterwave]

Do you know what the wavelength of a sin function is?

 

[mitch_1211]

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

 

[Demystifier]

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.

 

[mitch_1211]

@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...

 

[Demystifier]

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.

 

[mitch_1211]

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...

 

[mitch_1211]

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]

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=\frac{h}{\frac{2\pi}{k}}

but this will give the SI units of momentum, not keV/c..

 

[mitch_1211]

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...

 

[mitch_1211]

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?

 

[mitch_1211]

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?

 

[mitch_1211]

i also should've specified that that wavefunction was for an electron...

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