# Are conduction electrons localized in space?

1. Apr 20, 2009

### JoAuSc

Let's say we use a very simple model consisting of non-interacting electrons in a 3D infinite square well, perhaps a cube of a single metal crystal. If an electron is in a particular energy state, then its wavefunction is spread across the entire crystal.

However, electrons are said to travel through conductors like particles, implying that their position is at least somewhat localized. This would imply that an electron with a somewhat well-defined position must be in a superposition of energy states whose wave functions cancel out everywhere but near the position of the electron.

So, which is it? I'm betting it's the latter one, and that the interactions between an electron and its environment act more as a measurement of its position than its energy, but I'm not sure and I'd like to hear from the experts.

2. May 22, 2009

### crazy_photon

I'm by no means an expert, but i'll put my 2 cents worth.

In my opinion its the former, i.e. electrons are de-localized in space (localized in momentum space). This is the only way that the notion of energy-band picture makes sense to me, along with Fermi-Dirac statistics. The electron 'moves' by changing its wavevector, i.e. in a case of infinite square well the excitation hops between the quantum number n describing the ladder of states.

Lol, I might get myself in trouble right now, but at least it hopefully will generate interesting discussion...

3. May 22, 2009

### ZapperZ

Staff Emeritus
This makes no sense!

The potential you're describing has nothing to do with the potential of a solid, much less, a metal. So why do you think you can use that to represent a metal?

And why are you inventing such a thing for a metal? What's wrong with the periodic potential that gives you the Bloch wavefunction?

Zz.

4. May 22, 2009

### crazy_photon

What Zz says is definitely true -- i.e. one has to take into account crystalline lattice. However, one can think of zeroth order model of the solid is that of a potential box, corresponding to the edges of the solid. I thought the question wasn't really about that, but whether the conduction electrons (say they are in this box if thats what we agree to use as a model) are localized in space or not.

I will have to go with 'no', because of the reasons provided in my previous post.

I'm new on this forum, but I think i can 'smell' a good discussion coming...

5. May 22, 2009

### ZapperZ

Staff Emeritus
I am not sure there was ever a infinite-potential well that has ever been used to "model" a metal. I can't recall seeing such a thing in any intro QM classes (or did I slept through it?).

Now, if one were to use the plane-wave "free" particle description to model the conduction electrons, THAT I can understand. But then, one then argues "But ZapperZ, what if we make the boundary of the infinite potential infinitely far away?"

Viola! Plane-wave free particle! So why go through the hassle of infinite potential set up?

Zz.

6. May 22, 2009

### jostpuur

The free electron model is quite common thing to see in introductions to solid state physics.

The Bloch waves are as delocalized as plane waves (or as eigenstates in big box potential), so replacing plane waves (or eigenstates in big box potential) with Bloch waves does not change the original question essentially. I'll ask the original question again, this time in a form demanded by ZapperZ:

"Do the electrons exist as localized wave packets, or as delocalized Bloch waves?"

7. May 22, 2009

### xepma

The Bloch waves just form a basis of the (relevant part of the) Hilbert space. If we had a single electron it would be in some superposition of these Bloch waves.

But it's not even correct to talk about a single electron inside a metal. There is one, multiparticle state/wavefunction describing all electrons. This wavefunction is antisymmetric with respect to all the electrons. And we use a basis of Bloch waves to describe the part of this wavefunction which is responsible for the conducting properties.

8. May 22, 2009

### ZapperZ

Staff Emeritus
Actually, we CAN. That's the whole point of Fermi Liquid Theory.

You change a single many-body problem and transpose it to many one-body problem. In doing that, you renormalize the charge carrier into a quasiparticle whereby the many-body interaction has been simplified via a mean-field potential.

Zz.

9. May 22, 2009

### xepma

Yes, ofcourse. But that doesn't get rid of the statistics. In the end you're still describing a completely antisymmetrized wavefunction. And not a wavefunction which is "just" a product of single-particle wavefunctions. Although you're right that this step is, in some sense, only taken at the end.

10. May 23, 2009

### hiyok

I'd like to add several remarks:
(1) It is surely not proper to model metals using infinite high well potential. Because, in this case, the electron can not be in Bloch states, which is based on translational symmetry. And thus, the electrons can not have any values of averaged velocity than zero. In other words, these electrons cannot move at all !

(2) In a real metal, I believe the conduction electrons are better visualized as wave packets. The reason is the interactons of the electron with its environment, and such interactions are random. Actually, this is exactly the picture adopted in the semi-classical theory that deals with the transportation properties of a metal (see e.g., D. Mermin)

hiyok

11. May 23, 2009

### sokrates

Many wrong things have been stated in this thread but this particular one from Zz is a jewel. Read it over and over again, and try to absorb it EXACTLY the way Zz put it.

Very good post.

12. May 23, 2009

### sokrates

With a very big, important, impossible to miss CAVEAT.

That the free electron model could be used ONLY IF the effective mass approximation is valid and that is true only for small energies around the chemical potential.

13. May 23, 2009

### sokrates

Wrong.

You don't definitely need Bloch waves to model conduction band electrons. Some kind of an effective mass approach is most frequently used in practical calculations.

And you can assume WHATEVER boundary condition you like (infinite well, PBC, Open, Dirichlet) AS LONG AS the boundaries are far, far away from the region you are interested in. This ensures that whatever you assume for the boundary condition DOES NOT affect the actual transport properties.

It turns out that the math is simplest when PBC (periodic bounary conditions are assumed) which gives $$k_x= \frac{2\pi}{L_x}$$ kind of periodicity ( allowing negative k-values). But you could just as well go about assuming an infinite well, and in this case
$$k_x = \frac{\pi}{L_x}$$ but this time negative k-values yield identical states and you end up with the SAME number of states you'd get if you had assumed PBC initially. This is re-assuring.

So don't get things all mixed up. Boundary conditions are completely independent of how you model the solid inside.

Inside the solid, you could choose a free-electron model (provided that you have a valid effective mass description) or you could do it exactly using the Bloch waves, but these have nothing to do with the kind of boundary condition you assume.

I know what bothers people when I say you can have a particle-in-a-box and current flow in that box at the same time, because they are thinking in terms of resonant energies ,and they assume discrete energy levels. But this boils down to one of the most difficult problems in physics:

You have the simplest conductor with one single energy (ignoring spin) and you make two contacts to it. How do you model the current flow through this thing?

The answer only became clear in the last 20 years with the rise of mesoscopic physics and some theoretical acknowledgement of the importance of contacts. And the current viewpoint in the community is that when you make contacts to a BOX the levels are not really resonant energies anymore - but they broaden out in infinite range.

And energy becomes and independent variable of the electrons because you can EXCITE the box from the contacts AT ANY ENERGY.

Quantum Transport
nanoHUB.org
Meaning of Resistance at the Nanoscale,

etc...

Last edited: May 23, 2009
14. May 24, 2009

### hiyok

Hi, sokrates,

I understand what you mean. But, actually I was not talking about the boundary conditions. I was either not talking about Bloch waves. What I really intended is that, maybe you can have many options of boundary conditions, but it is inevitable that these electrons are in delocalized states, otherwise there won't be any current and there would not be any disputes over the original post any more.

hiyok

15. May 24, 2009

### Manchot

This is a subject that I've often found confusing, as I've never been able to reconcile the semiclassical approximation that people use (e.g., the Boltzmann transport equation) with the quantum mechanical structure of a solid. That is, Bloch waves are obviously extremely delocalized, while the BTE requires that particles be localized in position and momentum space. Is it just a matter of constructing the "right" wavepacket? Is it mostly due to the decoherence that interaction with the environment provides? Or is it some combination of the two?

16. May 25, 2009

### hiyok

I believe the decoherence should be the sole cause. If one cools a metal to much low temperatures, the semiclassical picture should eventually break down.

17. May 26, 2009

### JoAuSc

Thank you, everyone. I didn't expect to go to physicsforums and find a month-old thread of mine with 15 replies.

I don't doubt that an infinite square well is a bad approximation for a solid, missing behaviors such as band gaps and such, but it seems like a good way to discuss whether electrons in solids are localized and thus in a superposition of basis states, or spread out across the solid. I'm using this model since it's the first one mentioned in my solid state physics textbook (Chapter 6 in Introduction to Solid State Physics by Kittel, described as a "free electron gas".) I'm not using the Bloch waves since the sine waves of an infinite square well are much simpler.

That's a good point. I can't imagine that the choice of boundary conditions would make that much difference, though, and it seems that a metal cube of a finite size would be more accurately described by something with boundary conditions that aren't infinitely far away.

That would only be true for eigenstates. For a superposition of, for example, the first and second energy states of the 1D infinite square well, <x> shifts back and forth.

That brings something to mind. I didn't think about how localized electrons would collide with, say, a phonon, where you start out with two incident wavevectors and end up with two different ones. A localized electron would necessarily have many wavevector components, so in a sense if it was involved in a collision, then part of its wavefunction would be reflected and the remaining part wouldn't be affected, which seems strange. Also, wavepackets spread out over time, so if an electron was indeed located in a certain region, then it'd either need to have its position measured on a regular basis, or it'd spread out.

Yeah, exactly my problem. There are large gaps in my knowledge of this sort of thing.

18. May 26, 2009

### crazy_photon

Thanks for reading my reply. Again, I'm not an expert, but I don't think many people on here are (not intended as an insult). So, i feel justified sharing 'my view' on your question, without fear of being wrong - afterall we're just having a scientific discussion.

The problem of electron-phonon scatter:

Back to the 'constraints' of the question, i.e. your particle(s)-in-a-box model. First off, as many have pointed out, that is not the correct way of describing the problem, but again, in the 'zeroth' order I don't see why not. Again, for discussion sake...

In such a model you get stationary solutions (i.e. standing waves). I'm talking for low enough temperatures that the modes are indeed orthogonal (If I say zero temperature, we don't have phonons). If a phonon 'comes along' (there's no lattice per se, but nonetheless), it will interact with the electron iff both momentum and energy are conserved (of course). So, if there's overlap in energy between the two (say you're interacting with acoustic phonon), the rest is conservation of momentum. Since standing wave has zero momentum I see that two phonons have to be emitted in the opposite directions (of half energy each). In that process electron would 'hop' up or down the ladder (emission versus absorption of phonon) corresponding to the gained energy.

19. May 27, 2009

### ZapperZ

Staff Emeritus
Er... Alpha Centauri can be considered to be infinitely far away when you consider gravitational forces acting on the earth. But yet, to our galaxy, it isn't. If you are the size of Angstroms, the edge of a metal cube that is centimeters long is infinitely far away such that it doesn't matter anymore. That is why in many description of the properties of solids, the boundary condition of the surface or the edge of the material is typically insignificant. You don't see the BCS ground state accounting for the size of the superconductor, do you?

Zz.

20. May 27, 2009

### JoAuSc

You're arguing that for bulk solids, it physically doesn't matter whether you choose the infinite square well's sine waves or the free particle models's complex exponentials, but that using the latter is much simpler to deal with. I'll concede that, though only the ISW has a non-zero fermi energy. The Kittel text I referred to above, strangely enough, mentions the infinite square well as the model used but uses the complex exponentials in calculations.

I'm gonna have to take your word for it.