Physics Friday 87

Continuing from last week with our simple fermion system, we now consider the system enclosed in walls which are impermiable to our fermions, so that the particle number Ñ is now constant. While this is no longer the grand canonical formalism, the fundamental equation is still valid, as it is an attribute of the system independent of its boundary conditions. However, the chemical potential μ, usually called the Fermi level when speaking of a system of indistinguishable fermions, will not be a constant, but will instead vary with temperature so as to keep Ñ constant. Using our work from last time, we have particle number . While this gives Ñ as a one-to-one function of μ (as Ñ(μ) is an increasing function), it is not one that has an analytic inverse, and solving for μ given Ñ and T must generally be done numerically.

We can also consider the low temperature limit kT→0 (β increasing without bound). Recall that the probability of occupation for a given state in our system is , and since we are dealing with fermions, this gives the occupancy (expectation value for the number of particles in the state): . We see that for μ=εn, the occupancy f(μ,T) of our state is exactly 1/2, independent of the temperature. For εn>μ, increases without bound as β does, so that the probability of occupation goes to zero; for εn<μ, as β increases, so that the probability of occupation goes to one. Thus, as the temperature approaches absolute zero, we find the states with energy below the Fermi level fully occupied and those with energy above the Fermi level empty.
This is what we should expect from a system of fermions. As the temperature approaches absolute zero, the particles should go to the lowest energy state allowed; however, being fermions, the Pauli exclusion principle applies, and thus the ground state can hold only a limited number of particles; when it fills, then the particles begin filling the next level. Then it fills, and so on, until we reach the total number of particles; the threshold between filled and empty states is thus our Fermi level. The Fermi level at a temperature of absolute zero is known as the Fermi energy: .
We also see that as the temperature begins increasing away from absolute zero, the transition at the Fermi level becomes less sharp; the occupancy for states of energy just below the Fermi level decreases and that for states just above the Fermi level increases.
FermiFig1
Figure 1: Occupancy of a state as a function of state energy at various low temperatures.

In particular, the energy range over which this transfer occurs is on the order of 4kT: expanding in a Taylor series about the point εn=μ, we get .
We should also note that the equation for occupancy as a function of energy possesses inversion symmetry about the point εn=μ, pn,m=1/2.

Now, suppose we have the special case ε1<ε2=ε3, so that ε1 is the ground state, and the excited state is degenerate. Let Ñ=2, so that at absolute zero, the ground state is filled, the excited states are empty, and the Fermi level lies somewhere between
ε1 and ε2. Now, for very low T, we use the approximation for occupancy:
.
In this case, our equation for Ñ becomes:
. Plugging in Ñ=2, we can solve the above for μ, to get

(the ellipsis indicates terms of higher order in kT, which are neglected in our approximation). This illustrates an important general principle: as the temperature increases, the Fermi level is ‘repelled’ by energy regions with higher densities of states.

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2 Responses to “Physics Friday 87”

  1. Physics Friday 90 « Twisted One 151’s Weblog Says:

    […] and zero for . Using our integral formula for the number of particles in an ideal Fermi fluid (see here), we see: , and thus: For our metal, this is The temperature TF for which μ0=kTF (the Fermi […]

  2. Physics Friday 88 « Twisted One 151’s Weblog Says:

    […] doubly occupied mode. Each state is independent, so we have mean occupation number (in analogy to here). Now, we can find the grand potential: . Now, consider our particle confined to a cubical box of […]

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