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I Fermi sphere and density of states

  1. Jun 6, 2016 #1
    Hello!
    When computing the density of states of electrons in a lattice, a material with dimensions [itex]L_x[/itex], [itex]L_y[/itex], [itex]L_z[/itex] can be considered. The allowed [itex]\mathbf{k}[/itex] vectors will have components

    [itex]k_x = \displaystyle \frac{\pi}{L_x}p[/itex]
    [itex]k_y = \displaystyle \frac{\pi}{L_y}q[/itex]
    [itex]k_z = \displaystyle \frac{\pi}{L_z}r[/itex]

    with [itex]p, q, r \in \mathbb{Z}[/itex].

    The only values of the wavevector [itex]\mathbf{k}[/itex] to be considered belong to just one out of the 8 octants of the Fermi sphere. This is because the values of [itex]\mathbf{k}[/itex] in the remaining 7 octants are equivalent to the ones in the chosen octant. Why?
    Thank you anyway,

    Emily
     
  2. jcsd
  3. Jun 6, 2016 #2

    marcusl

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    Was this statement made for a particular crystal lattice, or a general rule for all crystals?
     
  4. Jun 6, 2016 #3
    Hello Emily,

    I think is related to symmetry properties of the Fermi surface.
    Check out the Fermi surfaces of individual elements here: http://www.phys.ufl.edu/fermisurface/
     
  5. Jun 6, 2016 #4
    I think this confusion might arise from mixing two conventions for writing the wavefunction in a fourier series: When exponentials are used as a basis, the allowed wavevectors for a particle in a box are z*(2pi)/L, where z can be positive or negative. When using sines, which I assume is true in your case, the allowed values are only n*pi/L for positive n, as in https://en.wikipedia.org/wiki/Particle_in_a_box. This is because sin(x)=sin(-x).
     
  6. Jun 6, 2016 #5

    marcusl

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    You can see that many of these do not have 8-fold symmetry, hence my question.
     
  7. Jun 7, 2016 #6

    DrClaude

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    Additional remark: The first case corresponds to periodic boundary conditions, to second to particle in a box (infinite potential walls). In the end (V → ∞), both approaches give identical results.
     
  8. Jun 7, 2016 #7
    To marcusl: no, it is general, but referred to semiconductor materials.
    To navrit: I think it is related to the wavefunctions more than the symmetry of the Fermi sphere.
    To thephystudent: no, [itex]\sin(x) = - \sin(-x)[/itex] and so they seem to be not equivalent.

    I think the approach to be followed is the "particle in a box", a 3D box with [itex]L_x, L_y, L_z[/itex] dimensions.
    Here (slide 7) this approach is followed and the division by 8 is made as a "Correction factor for redundancy in counting identical states" [itex]\pm p, \pm q, \pm r[/itex] referring to the end of slide 5, where the wavefunction is defined. I can't see that redundancy.
    Again, here the division by 8 is made because "wavefunctions that differ only in sign are indistinguishable. Hence we should count only the positive" [itex]p, q, r[/itex] "states to avoid multiply counting the same quantum state". Why?
     
  9. Jun 7, 2016 #8
    In the Wikipedia page you linked, it is stated that negative values of [itex]p, q, r[/itex] are neglected because "they give wavefunctions identical to the positive" [itex]p, q, r[/itex] "solutions except for a physically unimportant sign change". It is what thephystudent mentioned: [itex]\sin(x) =? \sin(-x)[/itex]. So I would like to ask why is this sign physically unimportant.
     
  10. Jun 7, 2016 #9
    ok, I was a bit clumsy with minus signs. So sin(x)=-sin(-x) indeed, but that means sin(x) and sin(-x) are not linearly independent, as they are the same apart from a constant factor being -1 and thus only one of the two has to be considered a basis vector.
     
  11. Jun 7, 2016 #10
    Remember that we are counting states in the Fermi sphere in order to derive the state density. States are acceptable solutions of the Schrödinger equation for this problem. So, should they also be linearly independent?
     
  12. Jun 7, 2016 #11

    DrClaude

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    Yes, otherwise you are overcounting states. sin(x) and sin(-x) are the same physical state.
     
  13. Jun 7, 2016 #12

    marcusl

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    I get the symmetries of the box (the Schrodinger equation is separable in x,y,z), but why is the box a good representation of a lattice? Doesn't the Fermi surface incorporate symmetries of the crystal, whose lattice vectors may not line up with, nor are separable along, x,y,z?
     
    Last edited: Jun 7, 2016
  14. Jun 8, 2016 #13
    To DrClaude: ok, now it is more clear, thank you.
    To marcusl: I think the "particle in a box" is the simplest representation of an electron into a lattice; it is good as a first approximation. The effective mass includes the effects of lattice and so the electrons can be treated as free particles. We want they remain bound to the material, so we impose that they remain inside a box with the dimensions of the material.
    About the relations between the Fermi surface and the symmetries of crystal I can't answer, I'm sorry.
     
  15. Jun 8, 2016 #14
    To marcusl: maybe this can help you. Go to page 86 and to the beginning of paragraph 6.1.
     
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