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E versus K diagrams

  1. Jul 15, 2010 #1
    *Hope this is in the right forum*

    This upcoming semester I'll be taking my first course in Semiconductor physics and I'm studying out of the textbook independently this summer to prepare myself.

    I've been learning about energy bands and I wanted to clarify a couple of concepts that my text is rather vague on:

    1. What exactly are k and crystal momentum? I know ħ*k is the crystal momentum, with units kg*m/s, and k has units kg*m/(J*s^2), but I'm not sure if I'm interpreting them correctly. I've been more or less treating k as just the momentum of the electron when I consider E versus K diagrams but is that, for want of a better word, correct?

    2. How does (1/ħ)dE/dk give rise to the velocity of the particle? I've been staring at the derivation for days, and I see exactly where it comes from, but intuitively it doesn't make sense. I may be guilty of taking basic concepts of momentum too far into quantum mechanics, but isn't the velocity of the particle a component of momentum? Is the velocity given different then the velocity of the particle in k? Or do I need to readjust my concept of momentum?

    3. I can see that (1/ħ)^2*d^2E/dk^2 = 1/m. If we multiply that by the electric force of the electron (-e*E, where -e is the electron charge, and E is the applied field), we have acceleration (F/m). Is it reasonable to consider the second derivative of E with respect to k proportional to the acceleration of the particle?

    Any good introductory sources online I can check out to better familiarize myself with these concepts? Cheers!
  2. jcsd
  3. Jul 15, 2010 #2
    I find easier to think about k as the wave vector, the spatial variation of the electron wave function.
    The velocity then is just the group velocity. It would ve the velocity of a wave packet with an small bandwidth around k.
  4. Jul 17, 2010 #3
    Unfortunately my physics upbringing has been really shoddy in terms of waves. I've taken to going back to my old physics text to read chapters on wave mechanics we never covered in class. So I don't have a truly intuitive understanding of wave vectors and that doesn't quite help.
  5. Jul 18, 2010 #4
    To your first question, think about the position vectors that constitute the Bravais lattice. Simply repeating those vectors yields the full lattice structure. Well, the wave vectors k are just another way of reproducing the Bravais lattice, only this time in 'momentum space.' The E versus k diagrams are responsible for showing us electronic band structure, but the wave vectors k by themselves do not represent the momentum of the electrons.

    As for your second question, I'm a little a bit confused about what you mean but I'll give it a go anyway. You mentioned intuition so let's think intuitively. The equation for the energy levels of a free particle is given by:

    [tex]E = \frac{\hbar^{2}\kappa^{2}}{2m}[/tex]

    Take the derivative of that with respect to k and you're left with momentum divided by mass, which is just the velocity of the particle. I'm not quite sure what you mean by the "velocity of the particle in k." You are calculating the particle's velocity as a function of k.

    And I would say yes to 3.
  6. Jul 19, 2010 #5
    I think my first question is just going to require me to think on it a bit.

    As for the second question, see, I can perform that exact derivation, no problem at all, but then I recall this:

    k = p/ħ = mv/ħ

    Is this only true for the free electron? My text is rather vague on the matter. But assuming it isn't, then k itself is proportional to the velocity. Maybe I'm just letting myself get caught up in this definition, but it doesn't seem sensible to think that the rate of change of energy with respect to velocity should give velocity back.

    Of course it hits me now that my text is principally talking about 1D crystals here, and I know in the 3D case there are 3 k vectors. Are the individual velocities related to the various k vectors just different from the velocity given back by the first derivative with respect to k?
  7. Jul 19, 2010 #6
    I think your problem may be because you need to use effective mass if you're going to say p=mv, which of course depends on k. Near the bottom of the conduction band, where the band is nearly parabolic, the electron is nearly free, and then you end up with k proportional to v as you say.

    This whole semiclassical thing can get problematic because people talk about k being the momentum of the electron, which is related to the velocity, yet the electron is basically a standing wave whose wave function spans the whole crystal, so how can it move at all? I think the real picture is that you need a wave packet in order to describe charge actually moving in the crystal, but more often with semiconductor courses you just learn that you can derive these quantities from the dispersion relation, and then you move on to fermi distributions, band bending and so on without really resolving the quantum/classical disconnect.

    Just a disclaimer, some of the things I said may be wrong.
    Last edited: Jul 19, 2010
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