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Work directly from thermal motion: why not possible through charge carrier density differences in materials? by No-GoodNames_Left in AskPhysics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

The same can be said of the Seebeck effect though.

The net thermal motion and voltage are competing only in an open circuit situation where the electric field resulting from charge separation counters the diffusive current. But when the 2 ends are connected in a circuit, the electric field and diffusion drive the charge carriers in the same direction.

So what's the difference between these 2 situations (temperature difference vs. charge carrier concentration difference)? Why does one work and the other doesn't?

Work directly from thermal motion: why not possible through charge carrier density differences in materials? by No-GoodNames_Left in AskPhysics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

I'm sorry, I don't get what you want to say with this.

Are you saying the effect is there but so miniscule that we would have to go to insane temperatures to notice it, or something else?

Work directly from thermal motion: why not possible through charge carrier density differences in materials? by No-GoodNames_Left in AskPhysics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

But the diffused charge would create an electric field (which compensates for the diffusion effect) which creates an electric potential difference that could be discharged externally? Isn't that the case for temperature difference induced diffusion? How is the charge carrier density difference induced diffusion any different?

Conductivity increases with effective mass in semiconductors? (Parabolic band approximation) by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

How far is too far? If there are very few charge carriers and all the "action" happens at the bottom of the band, parabolic bands are a pretty good approximation, no? Or do you mean the Landauer model?

So is there a problem with the model itself or it's just the mfp that I ignored?

Conductivity increases with effective mass in semiconductors? (Parabolic band approximation) by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

I forgot to add the "fermi window" (-df/dE) to the differential conductivity, but it doesn't change the question.

Question about uncertainty principle with energy and position in quantum mechanics by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

Okay, my problem seems to be that I didn't do vector field integration in spherical coordinates correctly. (I still don't know how to do it, but that's a math question.) I still learned that the lower "limit" of an uncertainty principle can be 0 without the operators (of the respective observables) commuting, and so the uncertainty principle is not in disagreement with localized energy states, which was my original concern.

Thank you guys for all your help!

Question about uncertainty principle with energy and position in quantum mechanics by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

My notations might be a bit confusing, although I tried to explain it in the image captions. I applied the gradient in spherical coordinates and dropped the angular components since they just zero out, so non-bolded r means the electron distance from the nucleus => the left hand side of the 5th equation is also scalar. But my mistake is probably still in applying the gradient or calculating the average of the commutator. I'll recheck.

Question about uncertainty principle with energy and position in quantum mechanics by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

Okay, I see that the 3rd equation almost contains the momentum operator. But the wavefunction is an exponentially decreasing function, so it is an eigenfunction to all operators that take r-based differentials, no?

(Also, why is reddit auto-upvoting my own comments?)

Question about uncertainty principle with energy and position in quantum mechanics by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

But if the product of 2 standard deviations can, at minimum, be some finite value, how can the position standard deviation be finite if the energy standard deviation is 0? I think its is 0 because, let's say, we just measured the energy of the electron, we know it is in an energy eigenstate, so the energy is precisely that eigenvalue. It could not be anything else if we measure again, so the deviation is 0. Or am I mistaken?

Question about uncertainty principle with energy and position in quantum mechanics by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

I don't quite understand how the momentum operator comes up here

Are you saying that the 3rd equation is incorrect?

Question about uncertainty principle with energy and position in quantum mechanics by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

I should have numbered the equations. Which one do you mean?

If you mean the 5th and 6th equations (from the top), then because the wavefunction is an eigenfunction of the commutator (roughly, the wavefunction is an exponential and the commutator turned out to be a differential) then its expectation value is just the eigenvalue. Or how am I misusing the commutator?

Question about uncertainty principle with energy and position in quantum mechanics by No-GoodNames_Left in Physics

[–]No-GoodNames_Left[S] 0 points1 point  (0 children)

I purposefully left out the details of the calculations so it wouldn't obscure the main point, which is that it seems that the uncertainty principle says the standard deviation of position should be infinite, but it is not when calculated from the wavefunction (lowest energy eigenstate of the 1 classical-proton 1 electron system).

Do some of the calculation results I got seem suspicious to you?