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Fermions, density of states

  1. Jun 20, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider a gas of non interacting electrons in two dimensions with electronic density n by unit of area and mass m. The gas forms a square of sides L.
    1)Assume periodic boundary conditions, find the density of states by unit of area.
    2)Find the Fermi energy in function of m and n.
    3)Calculate the increase of energy that N particles produce (assuming [itex]k_BT << E_F[/itex] where [itex]E_F[/itex] is fermi's energy) and then calculate the specific heat for T~0K.



    2. Relevant equations
    Problably a lot, I have no reference textbook so all on the internet I guess.


    3. The attempt at a solution
    Totally stuck on part 1).
    Let me see if I understand the question correctly, they are asking me the "density of state" function divided by a unit of area? In other words, [itex]\frac{g(E)}{1au}[/itex]?
    I'm reading stuff at hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/eedens.html#c1), apparently [itex]n=\frac{g(E)}{e^{(E-E_F)}/(k_BT)+1}[/itex]. But I must determine E in function of L I guess.
    For a 2d quantum well, the allowed energies are [itex]E_m=\frac{m^2h^2}{8mL^2}[/itex] if I'm not mistaken. What bothers me here is the natural number m and the fact that I expressed a formula involving [itex]E_F[/itex], something I'm asked to find in the next question, so that I'm guessing I'm not going in the right way.
    I'd love some help, I'm really struggling with this course (I have no textbook).
     
  2. jcsd
  3. Jun 20, 2012 #2
    Have you calculated any density of states things before? Starting from the definition, you'd have
    [tex] \omega(E) = \sum_{\lbrace n \rbrace} \delta(E-E_n) [/tex]
    where the sum is over all microstates of the system. Now you're dealing with a continuous gas so you definitely want to switch from sum to an integral. Convince yourself that you get:
    [tex] \omega(E) \simeq \frac{g A}{(2 \pi \hbar)^2} \int d^2 p \delta(E - \frac{p^2}{2m}) [/tex]
    where g is the degeneracy factor of each momentum state (electrons have spin 1/2 so for them, g=2) and A is the area of the box. Then do the integral and don't get scared if the answer seems a bit trivial -- this is a special case.
     
  4. Jun 20, 2012 #3

    fluidistic

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    No I have not calculated any density of state before. What is exactly the delta function you are using? Also what is omega? We haven't taken a statistical mechanics course yet (scheduled for the next semester).
     
  5. Jun 21, 2012 #4
    So in the formulas, omega is the density of states and delta is Kronecker delta when summed and Dirac delta when integrated.

    The density of states counts the number of microstates corresponding to a given macrostate (with energy E). This is exactly the meaning of the first expression.

    As you have a gas of particles, the degrees of freedom of the microstates are simply the positions and momenta of individual particles. These form up the "phase space" of the system. The second integral is just an integral of the delta function over the entire phase space. You can calculate this integral directly:
    [tex]
    \omega(E) \simeq \frac{g A}{(2 \pi \hbar)^2} \int_0^{2\pi} d\theta \int_0^{\infty} dp p \delta(E - \frac{p^2}{2m}) = \frac{g A}{2 \pi \hbar^2} \int_0^\infty m dx \delta(E-x) = \frac{gmA}{2\pi \hbar^2}
    [/tex]
    so the density of states is does not depend on the energy! (this holds for two dimensions and nonrelativistic particles)

    Where did you get this problem from? Do you have any reference on how you're supposed to be able to do it? :-)
     
  6. Jun 22, 2012 #5

    fluidistic

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    Ok thanks a lot. It appeared in a test (that I failed). It's the only course which allows us to use any book when in a test.
    I went to a friend's house to study (he has all the books, I have none) and we found some help in Brehm's book. We also found out that omega doesn't depend on E, though I don't remember having the "g factor". I'll check this out and post here later, I'm totally burned out.
    I had a test today and I'm sure I failed it too, I thought "[itex]^{16}O^{12}C[/itex]" was a complex molecules with 16+12 atoms; I totally over looked it. Now it's obvious that it means the CO molecule, etc. This prevented me to calculate a data used for part 2), 3) and 4) of the first exercise of the test, which lasted more than 4 hours.
    The last exercise dealt with something I had never heard of before, even Brehm and Alonso-Finn's books didn't save me. It was about the magnetization of N carbon atoms with sping 1/2, the specific heat, etc.
    The semester ended today (at last!), I feel like I've learned absolutely nothing in this course. It was all about getting the books in time (1 copy for 2 main books in the university), search a formula between 10 thousands formulae here and there without understanding anything and know where and when to apply it. My 2 friends feel the same way though they didn't fail the tests.
    So I'll have to take the final exam unlike my friends (it's worth 100% of the grade for the course), in mid-July. I'll study hard and hope I'll understand something. So please give me 1 day and I'll come back to the problem of this thread :) (by the way we were stuck on part 3). Sorry for all this talk, I felt I had to talk about it a bit.
     
  7. Jun 23, 2012 #6

    ehild

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    The formula is not correct. It is the energy of a particle in a three dimensional box, in quantum state m=(n1,n2,n3) (Consider m as vector, it is not the mass m)
    Read the Hyperphysics page: The energy depends on three quantum numbers.
    You have to find the allowed energies for the particle in a two-dimensional box. What are the allowed energies then?


    ehild
     
  8. Jun 23, 2012 #7

    fluidistic

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    [itex]E=\frac{\pi^2 \hbar ^2 }{2mL^2} (n_1^2+n_2^2)=\frac{\pi^2 \hbar ^2 }{2mL^2} R^2[/itex].
    [itex]N(E)=\frac{\pi R^2}{4}[/itex], [itex]R^2= \frac{2mL^2E}{\pi^2 \hbar ^2} \Rightarrow R= \frac{L \sqrt {2mE}}{\pi \hbar}[/itex] thus [itex]N(E)=\frac{mEL^2}{2\pi \hbar ^2}[/itex].
    Now g(E) (same as omega (E)) [itex]=\frac{dN(E)}{dE}=\frac{mL^2}{2\pi \hbar}[/itex].
    To answer part 1) apparently one had to assume that [itex]L^2[/itex] was a "unit of area", sigh, and then [itex]g(E)=\frac{m}{2\pi \hbar ^2}[/itex]. I don't know if my friend made the exact the same steps but he got full credit for this answer (that's what he told me by looking at this result).
    2)[itex]N=2\int _0 ^{E_F} g(E)dE = \frac{2mE_F L^2}{2 \pi \hbar ^2} \Rightarrow E_F= \frac{N\pi \hbar ^2}{L^2m}[/itex] where N is the number of electrons with energies between 0 and [itex]E_F[/itex]. So [itex]N=nL^2[/itex] and thus [itex]E_F= \frac{n \pi \hbar ^2}{m}[/itex]. I don't know if the result is right here. I stolen the "idea" in Brehm's book.
    3)[itex]c=\frac{\partial E }{\partial T }[/itex], no idea how to do this part. Well obviously I must find E in function of T...
     
  9. Jun 24, 2012 #8

    ehild

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    You need to find the total energy of the gas, and its change with temperature. The total energy is the integral of the number of electrons with energy E multiplied by E. use the formula for n(E) that you wrote in the original post.
    You certainly find derivation for Cv of the electron gas in three dimension. For example:http://www.cmmp.ucl.ac.uk/~ikr/3225/Section 6.pdf
    Do the same with this two-dimensional problem.

    ehild
     
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