Does speed=frequency*wavelength for electrons?

[AdrianMay]

Hi folks,

Does the above common-sense equation still hold for electrons? If so, it looks problematic to me.

Consider an electron heading up a potential gradient. It's going to slow down, so either the wavelength or the frequency or both must decrease. But momentum = h/wavelength, and the momentum is decreasing, so the wavelength must increase and the frequency must decrease even faster to make up for it.

When I read the derivation of the Schroedinger equation:

i(dY/dt) = -(1/2m).(d^2Y/dx^2) + VY (h=1 by units)

the E in HY=EY seems to be total energy, i.e. kinetic + potential, and that's the E that's proportional to frequency, but total energy must be constant in this situation because it's just swapping kinetic for potential as it climbs the hill. That already seems silly because potential energy is a relative quantity but the SE seems to be feeding it into an absolute quantity like frequency. If I redefine my null point of potential energy then nothing should change, but the SE seems to say something else.

So what on Earth is going on?

Adrian.

 

[jtbell]

v = f \lambda works for electrons, but v isn't the particle velocity. It's the phase velocity of a wave that is associated with the electron. The electron is actually a sum of waves with slightly different momenta (and wavelengths), which forms a wave packet. The particle velocity turns out to be the group velocity of the wave packet.

 

[AdrianMay]

Thanks, but what's with the null point of V? I mean, if I charge up the whole room to 10kV and repeat the experiment, nothing should change, right? But the SE would seem to imply that the frequency would increase.

Adrian.

 

[alyflex]

Thanks, but what's with the null point of V? I mean, if I charge up the whole room to 10kV and repeat the experiment, nothing should change, right? But the SE would seem to imply that the frequency would increase.

Adrian.

It would seem that way, but it can be shown that any constant potential can be absorbed by the phase if I remember correctly.

Just try to redefine the energy yourself and see if anything changes.

 

[AdrianMay]

It would seem that way, but it can be shown that any constant potential can be absorbed by the phase if I remember correctly.

Just try to redefine the energy yourself and see if anything changes.

I think it does. The SE was derived from E = T + V, so it's evidently trying to bring V into E, but E = hf, so it actually wants f to depend on V.

 

[AdrianMay]

I still don't get this. I just tried the infinitely deep square well thing thing with a constant finite potential at the bottom of the well and it seems to come out different, although I'm not sure what the result means. The usual thing is to solve:

ih.d/dt Y = -h^2/2m . d^2/dx^2 Y = VY

with V=0 using:

Y = Aexp[ i/k(px-Et) ]

which gives h=k. I suppose that should be hn=k or h=nk. Anyway, if V is non-zero I end up with:

(1-h/k).Eh/k = V

With V=0 that reduces to h=k as above, but with different Vs I get different Es which means different frequencies. So, my initial concern that shifting the zero point of V results in the silly effect of changing the frequency seems confirmed. I think that was already obvious from E=K+V and E=hw but alyflex asked me to try it.

Can anybody explain what's going on?

Adrian.

 

[AdrianMay]

Or, if I insert

Y = Aexp[in/h . (px-Et)]

I get

nE = (np)^2/2m + V

Same problem.

Adrian.

 

[PhilDSP]

You might want to read de Broglie's material where he derives his equations and theory. One textbook is called "Introduction to the Principles of Quantum Mechanics" if I remember correctly. His approach is based on the idea that differences in energy potential seek equilibrium and therefore travel through space as waves where phase is always conserved.

So I'd guess that the potential you describe should be interpreted as the energy potential between the object (electron) and the space it is traveling through. With V = 0 there may be no potential difference between the two but the electron still has an internal kinetic energy where presumably something within the electron is modulating and still produces a wave.

 

[AdrianMay]

So I'd guess that the potential you describe should be interpreted as the energy potential between the object (electron) and the space it is traveling through.

Not sure what you mean by that. With or without interpretations, the maths still has to add up. The maths is saying that if you make an infinitely deep square well with a potential of 1 in the well, the frequency comes out different from if it's zero. It might seem obvious to set the bottom of the well to zero, but there's no such obvious choice in the case of a harmonic oscillator. There's other maths to say how the potential is given by the charges around the neighbourhood.

In the mean time, I think I decided to just believe it. It would mean that if your square well experiment was inside the ball of a van der Graff generator, and you turned it on, the frequency would gradually increase. Is that true?

Adrian.

 

[The_Duck]

Note that the phase of the wave function is not observable! So the frequency of an eigenstate is not something you can measure. Here are a couple observables for the infinite square well:

-the difference in energy between the nth energy level and the mth
-the probability density of the nth state

I claim that neither of these change if you shift the floor of the potential by a constant.

Here is a frequency that is measurable. Consider the infinite square well with a floor at V=0. The energies are

E_n = \frac{\hbar^2 \pi^2}{2 m L^2}n^2

which are all multiples of the energy of the ground state

E_1 = \frac{\hbar^2 \pi^2}{2 m L^2}.

So the frequency of any eigenstate is a multiple of the ground state frequency. So if you let any eigenstate evolve for a time T = \hbar / E_1, it will return to its original phase. Since any wave function can be expressed as a sum of eigenstates, if you let /any/ wave function evolve for a time T, it will return to its original state, and thus its original probability density.

This is a frequency that is measurable, because if I make a state that is the sum of a few different eigenstates, then the probability density is not constant in time, but sloshes around the well with period T. I claim that the motions of the probability density have the same period T even if you shift the floor of the well. Nothing interesting changes if you shift the potential by an overall constant.

This also happens in the harmonic oscillator. There the period T is the natural period of the harmonic oscillator. You certainly don't want this to depend on the zero of potential, and if you work it out you find that it doesn't.

Edit: http://www.falstad.com/qm1d/ has an applet where you can see this "revival" phenomenon that happens in the infinite square well. It's pretty neat: you set up a wave function and it spreads out and bounces around for a while, until at a later time it magically coalesces into a perfect copy of the original state.

 

[AdrianMay]

I think that clears it up then. Thanks!

 

[DrDu]

There are several problems here: In the case of photons the frequency that is observable is the change e.g. of electric field with time. This does however not refer to a single photon but to a coherent superposition of photon number states. Hence what determines the frequency is not some absolute energy (which does not exist) but the energy difference between states with n and n+1 photons. The same holds true for the momentum.
For the electron the situation is different in two respects: 1. An electron is a fermion and states with different particle number cannot be superposed to form a coherent state which would give rise to an observable field. Hence the frequency cannot be observed directly.
2. Electrons have a rest mass, hence the dependence of E (or f) on p (or lambda) is nonlinear. Hence one has to be more precise about how to define velocity. The usual velocity of a particle is the group velocity which is dE/dp. The velocity E/p is the phase velocity and can be seen to diverge for p->0 as E=mc^2 in that limit.

 

[AdrianMay]

I just got to the relevant bit of Sakurai. I think the attachment is what DrDu is referring to. Right?

Adrian.

 

[DrDu]

Yes, that is exactly what I am referring to.