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Quantum statistics: density of states problem

  1. Jun 18, 2012 #1
    When you consider a electron L×L×L box, I think I understand how to derive the DOS-spectrum.
    Unfortunately, when a small change is made to the problem, I really don't understand what to do, so I probably don't understand the theory at all..

    This is the question:

    1. The problem statement, all variables and given/known data
    Consider a particle with mass M in three dimensions, confined to an L × L square in the x,y-direction, and held in a parabolic potential (1/2)kz^2 in the z-direction.

    Q1: Determine the one-particle spectrum for the system
    Q2: Calculate the density of one-particle states

    2. Relevant equations

    D_n(E) = (dΩ_n(E))/d(E) (1)
    U = 2 ƩnxƩnyƩnzε(n) (2)
    λn = 2L/n (3)
    pn = h/λ (4)

    3. The attempt at a solution
    I take formula (2) and change it to an integral (I don't know why I can do this, but I think it is for large n, but then it's also strange because in Q1 I have to determine the one-particle spectrum! (or isn't that the spectrum for one-particle, and does it mean something else?)

    then for nx and ny I can use (3) and (4) to show that te allowed energies are something like

    ε = ((h^2)/(8mL^2))*((nx)^2 + (ny)^2)
    then how do I plug in the parabolic potential? I have absolutely no clue at all. Catastrophe in my brain!
    I would be very grateful if you would help me with this problem!

    Suske
     
  2. jcsd
  3. Jun 18, 2012 #2

    TSny

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    The Schrodinger equation for this system is separable in the coordinates x, y, z. So, the energy of a single particle state will be a sum of an energy for each variable: E = Ex + Ey + Ez. You've already got the first two terms taken care of. Considering that the motion in the z direction occurs under the potential (1/2)kz2, what would be the appropriate expression for Ez?
     
  4. Jun 19, 2012 #3
    Thank you very much!

    So then I put V(z,t) = is (1/2)kz^2,

    Which gives me Ezn = (h/2pi)&sqrt(k/m)*(n + 0.5)?
    And can I then transform that into (h^2)(k^2)*(n+0.5)/(2m)? <-- this is probably wrong?

    Usually with just a particle in a LxLxL box I can 'count' the number of states with a specific energy with a sphere, but how does this apply here?

    Thanks!

    Suske
     
  5. Jun 19, 2012 #4

    TSny

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    Good. Ez = [itex]\frac{h}{2π}[/itex]ω(nz + ½) where ω = [itex]\sqrt{k/m}[/itex]. (What are the allowed values of nz?)

    Write out the total energy E = Ex + Ey + Ez in terms of the quantum numbers nx, ny, and nz. That gives the energy spectrum of the one-particle states. You can express it as E = a(nx2 + ny2) + b(nz + ½) where a and b are just constants.

    When finding the density of one-particle states, I believe you treat the quantum numbers as large compared to 1. So, I think it’s safe to drop the ½ compared to nz. Then, E = a(nx 2+ ny2) + bnz

    As you noted, if you were dealing with a free particle in a cube, you can count the number of states by considering a sphere. This is because in that case E = a(nx 2+ ny2 + nz2). So, if you construct a 3D Cartesian coordinate system with axes denoting nx, ny, nz then a specific value of E corresponds to the surface of a sphere in this coordinate system.

    But now you have E = a(nx 2+ ny2) + bnz . For a certain value of E, you’ll need to think about the shape of the surface defined by this relation. Also, the allowed values of nx, ny, and nz will restrict you to only a portion of that surface. Then you need to think about how to use the surface to find the total number of states with energy less than or equal to E.
     
  6. Jun 19, 2012 #5
    Thanks again!

    Oke. Only positive integers are allowed for nx, ny, nz
    Therefor, the surface of the paraboloid (right?) is four times 'too much'

    so for a certain E, the amount of states (nx, ny, nz)

    equals 1/4 [itex]\int\int[/itex] sqrt (1+(4(a/b)^2)*r^2) r d \Theta dr

    I put it in polar coordinates.. but now I don't know what the limits of the integral are? And whether this is correct at all..

    I used [itex]\int\int[/itex] sqrt{1 + (derivative to x)^2 + (derivative to y)^2} dA = surface;

    Could you please help me find the integral-limits? and when you integrate this, i'm afraid it won't leave anything like (nx, ny, nz), so you can't just convert this back to a 'normal' n..!

    After that, you just rewrite your equation so you have n in terms of e, and then you differentiate to e to have your DOS(e)?

    Almost there?

    Thanks again!
     
    Last edited: Jun 19, 2012
  7. Jun 19, 2012 #6

    TSny

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    Right, a paraboloid that may be expressed as E = a(x2 + y2) + bz where, for convenience, we let x, y, z denote nx, ny, nz.

    To find the volume enclosed by the paraboloid, slice it in circular disks perpendicularly to the z axis. For a certain value of z, show that the radius of the disk is r = ([itex]\frac{E-bz}{a}[/itex])(1/2).

    Integrate to sum up the volumes of all of the disks from z = 0 to z = E/b (right?).

    The resultant volume is the total number of quantum states N with energies less than or equal to E except, as you noted, you need to take 1/4 of this volume because we only have quantum states in the region where x, y, z are positive. Also, for electrons we have to allow for two states of spin. So, this introduces a factor of 2.

    I got (might be wrong!) N = [itex]\frac{πE^{2}}{4ab}[/itex].

    From this you can get the density of states.
     
  8. Jun 19, 2012 #7

    TSny

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    I see that the problem just says you have a particle of mass M. So, we don't know if it has spin. If we ignore spin, then of course N will be half of what I stated.
     
  9. Jun 22, 2012 #8
    Thank you very much, I got the answer completely right! this is great
     
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