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Polarization of a Ideal Fermi Gas

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    At T=0, what is the largest density that a gas can be completly spin polarized by a magnetic induction of 10 telsas

    2. Relevant equations
    μn= 10^-26 J/T
    mass= 5*10^-27 kg
    spin= 1/2

    3. The attempt at a solution
    I am really not sure where to begin. The spin polarization is the degree which the spin is aligned with a given direction and at T=0, all of the particles fall into the lowest energy level.
    With this, in the ground state
    f(ε,τ)= 1/ e^(ε-μ/τ)-1

    Any help?!
  2. jcsd
  3. Feb 15, 2012 #2


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    I've seen similar questions with distinguishable particles (i.e. atoms), where they all go into the lower-energy polarisation state when T goes to zero. But, your problem is about a Fermi gas (which are indistinguishable particles). So I've not come across this type of question before.

    But I think I can guess what kind of answer they are looking for. To start with, Its a Fermi gas, right, so what would happen if the magnetic field wasn't there? You know about the Fermi energy and how it depends on the number density?
  4. Feb 15, 2012 #3
    The number of fermions in the system must be large enough so that adding one more fermion to the system has negligible effect on μ. Since the F–D distribution was derived using the Pauli exclusion principle, which allows at most one electron to occupy each possible state, a result is that 0<ni<1
    and that the number of fermions= 1/(e^(ε-μ/kT))+1
  5. Feb 15, 2012 #4


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    Did you just copy and paste that from wikipedia? You're going to need to do better than that! More useful would be this page: http://en.wikipedia.org/wiki/Fermi_energy because when the question mentioned density, I think it is talking about number density of the particles in physical space. (not the probability of a certain state being occupied). So reading through the webpage, do you see how the Fermi energy is related to the density of particles in physical space?
  6. Feb 15, 2012 #5
    Yes, I understand it now. The more particles that are present in a given system, the higher the Fermi energy will be and vice versa. Now that we have the number density equation, how does the chemical potential and temperature come into play?
  7. Feb 15, 2012 #6
    Does ε=uB?
    Becuase if that is so, I can set uB=( hbar ∏^2 / (2mL^2)) (N/2)^2 then solve for N and that would give me the number density I am looking for?
  8. Feb 15, 2012 #7


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    We're starting to think along the same lines now. But you've got the equation for the Fermi energy in 1d, I'd guess the question would want the 3d case. Also, why solve for N? Surely you should solve for N/L (or in 3d, N/V).

    And about the problem more generally: I'm not sure if this is the right method (as I said to begin with, I haven't done a problem like this). But it is my first guess. It's basically like saying that once the energy splitting due to magnetic field is greater than the Fermi energy, then the energy due to magnetic field becomes greater than the kinetic energy, so to minimise energy, all the particles go in the lowest spin state.

    That's the reasoning, so its not really a rigorous method. We could at least say that when the energy splitting due to magnetic field is much greater than Fermi energy, then all the particles will be in the lowest energy spin state.

    EDIT: Anyway, what do you think? The equation for the Fermi energy in terms of the number density uses the fact that for every state, there is a further degeneracy, because there is a spin up and spin down state, which are both occupied. And we now say that there is a magnetic field which is strong enough that for all the states, only one of the spin states is occupied. So what does this tell us about the new effective Fermi energy? And we're assuming that the total energy is minimised, so at what magnetic field strength will all the particles be in the same spin state, to minimise the energy?
    Last edited: Feb 15, 2012
  9. Feb 15, 2012 #8
    Does this take into account that T=0?

    Sorry... I'm having a little trouble wrapping my head around this
    Last edited: Feb 15, 2012
  10. Feb 15, 2012 #9


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    yeah, normally at T=0, the particles fill up all the states up to the Fermi energy. (In other words, there is low temperature, so they are very boring and are not excited and take the lowest possible energy levels).

    Now we say there is a magnetic field, causing splitting of the energy of spin states. So we might be able to minimise the energy even more by moving a particle to a state with higher 'kinetic energy', but with a lower 'magnetic potential energy'.

    Essentially, we know the system takes the lowest total energy because T=0. And if T was not zero, the situation would be far more complicated.
  11. Feb 15, 2012 #10
    Last edited: Feb 16, 2012
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