Posts Tagged ‘Grand Canonical Partition Function’

Physics Friday 91

October 2, 2009

When we modeled the ideal Fermi fluid (previous posts here, here, here, here, and here), we used the fact that the grand canonical partition function can be factored over the individual states:
,
where the individual state partition is
,
where the sum is over all possible values of n, the number of particles in the state. For fermions, the Pauli exclusion principle meant that n=0 and n=1 were the only allowed values, and the above series had only two terms. Bosons, however, can be placed in the same quantum state in any number, so for an ideal Bose gas, the above sum becomes an infinite (geometric) series:
.

Next, let us find the occupancy (mean occupation number) fk,ms:
.
Now, we note that for |x|<1,
.
This tells us that the sum in the numerator is thus , and so
.
Compare this to the occupancy for a fermion fluid ; the difference is a change of sign in the denominator. Note that unlike the fermion case, for a boson fluid, the occupancy can be greater than one.

For a boson fluid, we generally choose our energy scale so that our zero energy is the ground state. Now, note that the occupancy has a singularity at , where it goes to infinity. This means that for a Bose gas with a finite number of particles, the Gibbs potential μ must always be less that all our energy values; thus, with the above choice of energy scale, we see that μ must always be negative.

We can, as in the fermion case, use the quantum particle in a cubical box to develop the concept of the “density of states” D(ε):
,
which allows us to approximate the sum over all states as an integral over energy:

(where g0 is the number of different spin states: g0=2s+1, where s is the particle spin, here an integer). Using the same integration by parts as in the fermion case, we can change this to:
.
Similarly,
,
and
.
Note the difference in signs in the denominators in the integrals from the fermion case, and that still holds.
Using the math from this addendum, we can rewrite the above integrals in terms of polylogarithms:

and
.

We also note from our equation for Ñ, that if the number of particles is held constant, we see that the integral must be constant, and so as the temperature increases, the GIbbs potential μ must decrease (as it also does for a Fermi gas).

When examining the ideal Fermi fluid in the classical limit, we noted that when the fugacity is small (), we see that , and so the fermion occupancy can be approximated via
.
Now, note that means that for the boson occupancy,
,
the exact same approximation. Thus, the results from applying this approximation is the same (except for the different value of g0), and thus, the ideal Bose gas has the same condition for classical behavior, and, as expected, also approaches the classical ideal gas. Thus, the difference between Bose gases and Fermi gases is only significant in the quantum regime.

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Physics Friday 88

September 4, 2009

Previously, I used the grand canonical formalism to derive the fundamental thermodynamic relation for a system of spin-½ fermions in a quantum system with a small number of orbital states. Now, let us extend this analysis to the ideal Fermi fluid, the quantum analog to the classical ideal gas for fermions; a system of fermion particles with no (or negligible) interaction forces, such as a collection of neutrons in the interior of a neutron star, or a low-temperature gas of 3He atoms.
As with our previous work, we will use the grand canonical formalism; the fundamental relations will be independent of these particular boundary conditions. We consider spin-½ fermions, so that for each orbital state we have two states, given by ms=±½. The orbital states available can be specified by the wavenumber vector k of the wavefunction (see here). As before, our grand partition sum factors: . As we are considering fermions, each orbital state is either empty or singly occupied. The energy of an empty state is zero, and the energy of an occupied state (k,ms) is (independent of the spin direction). This means the partition sum for a state (k,ms) is

In turn, we can consider the product of the two states of orbital k and opposite spin directions as the “partition sum of mode k” , with terms for the empty mode, singly occupied mode (with two possible orientations), and 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 volume V=L3; I demonstrated the quantum particle in a cubical box here; we have the allowed orbital states given by , , where nx, ny, and nz are positive integers.
Thus, the number of states of energy less than or equal to E is the number of triplets of positive integers (nx, ny, nz) for which , where . Considering the three dimensional abstract space with coordinates (nx, ny, nz), we see our points are those contained in the eighth of the sphere of radius n in the first octant. As each point corresponds to a cube of unit volume in this space (visualize the lattice of integer points), for large n, we can approximate the number of orbital states of energy less than or equal to E by the volume of this eighth of the sphere:
.
Differentiating this with respect to energy, we can obtain a “density of (orbital) states”:
.
If we multiply this by two, since we have two spin orientations per each orbital state, we can approximate our sum in the grand potential with an integral:
.
There is no closed-form expression for the integral (though it can be expressed in terms of polylogarithms), so the physical properties derived from Ψ must also be expressed in terms of integrals (or polylogarithms), which can be computed numerically or via approximation methods.
We can, though, immediately determine the number of particles Ñ and internal energy U:
,
(which we could also have obtained via and our integral expression for Ψ),
and
.

[These integrals may also be expressed in terms of polylogarithms:
,
and
. See here for more of the math involved.]

Now, if we perform integration by parts on the integral for Ψ, with and , we obtain
.
We also recall from the end of this post, that for simple systems, the equation of state is . Thus, we see that
. Note that this is exactly the result for a classical ideal gas whose molecules have only translational degrees of freedom available (see here and here).

Next week, I intend to show that this reduces to the classical ideal gas for high temperature, and extract the criterion separating the quantum and classical regimes.

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Physics Friday 86

August 21, 2009

Let us consider a model quantum system with three permitted spacial ‘orbital’ states; a particle in state n=1,2, or 3 has energies ε1, ε2, and ε3, respectively. Now suppose are particles are spin-½ fermions. Then the spin projection has two possible states, ms=½ and ms=-½; these are designated “up” and “down”. Thus, there are six possible states (n,ms), with n=1,2,3, and ms=±½. Since the particles are fermions, no more than one particle may be found in each of these states at a given time.
Now, let us consider this system in contact with a thermal reservoir with a fixed temperature T, and a reservoir of our fermions with Gibbs potential μ; this is to say, in the grand canonical ensemble. Compute the grand canonical partition sum. To compute it, we note that the grand canonical partition sum factors across non-interacting states, like the canonical partition sum does. Thus,
\mathcal{Z}=z_{1,-\frac{1}{2}}z_{1,\frac{1}{2}}z_{2,-\frac{1}{2}}z_{2,\frac{1}{2}}z_{3,-\frac{1}{2}}z_{3,\frac{1}{2}}. Now, each orbital state partition sum has only two terms: one for the empty state, with E=0 and N=0; and one for the occupied state, with E=εn and N=1. Thus, (assuming no magnetic field, so energy is independent of the spin direction):
z_{n,m_s}=e^{-\beta(0-\mu\cdot0)}+e^{-\beta(\epsilon_n-\mu\cdot1)}=1+e^{-\beta(\epsilon_n-\mu)}.
Now, we could also pair the states with the same spatial orbit n (but differing spin alignments):
z_{n}=z_{n,-\frac{1}{2}}z_{n,\frac{1}{2}}=\left(1+e^{-\beta(\epsilon_n-\mu)}\right)^2=1+2e^{-\beta(\epsilon_n-\mu)}+e^{-2\beta(\epsilon_n-\mu)}.
We can interpret this expanded product in terms of the four possible states of n: the empty state, two singly occupied states (of opposite spin), and one doubly-occupied state.
The probability that the state (n,ms) is empty is thus \frac{1}{z_{n,m_s}}, and the probability that it is occupied is p_{n,m}=\frac{e^{-\beta(\epsilon_n-\mu)}}{z_{n,m_s}}=\frac{1}{e^{\beta(\epsilon_n-\mu)}+1}.

Now, our grand canonical partition sum is
\mathcal{Z}=\left[1+e^{-\beta(\epsilon_1-\mu)}\right]^2\left[1+e^{-\beta(\epsilon_2-\mu)}\right]^2\left[1+e^{-\beta(\epsilon_3-\mu)}\right]^2. Thus, the grand potential is
\Psi=-\frac{\ln\mathcal{Z}}{\beta}=-\frac{2}{\beta}\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right].
Now, the mean number of particles ⟨N⟩ can be found from Ψ:
\begin{array}{rcl}\langle{N}\rangle&=&-\frac{\partial\Psi}{\partial\mu}\\&=&\frac{2}{\beta}\frac{\partial}{\partial\mu}\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]\\&=&\frac{2}{\beta}\left[\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_1-\mu)}\right)+\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_2-\mu)}\right)+\frac{\partial}{\partial\mu}\ln\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]\\&=&\frac{2}{\beta}\left[\frac{\beta{e^{-\beta(\epsilon_1-\mu)}}}{1+e^{-\beta(\epsilon_1-\mu)}}+\frac{\beta{e^{-\beta(\epsilon_2-\mu)}}}{1+e^{-\beta(\epsilon_2-\mu)}}+\frac{\beta{e^{-\beta(\epsilon_3-\mu)}}}{1+e^{-\beta(\epsilon_3-\mu)}}\right]\\&=&\frac{2}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{2}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{2}{e^{\beta(\epsilon_3-\mu)}+1}\end{array}
Note that this is the same result we get were we to sum the probability of occupation over all six states, as we should expect.
Similarly, we could find the energy via U=-\frac{\partial}{\partial\beta}(\ln\mathcal{Z})+\mu\langle{N}\rangle, or we can use the occupation probabilities to get:
U=\sum_{n,m}E_{n,m}p_{n,m}=\frac{2\epsilon_1}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{2\epsilon_2}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{2\epsilon_3}{e^{\beta(\epsilon_3-\mu)}+1}.
We can also find entropy via S=k\left(\ln\mathcal{Z}+\beta{U}-\beta\mu\langle{N}\rangle\right) (see here), getting:
S=2k\left[\ln\left[\left(1+e^{-\beta(\epsilon_1-\mu)}\right)\left(1+e^{-\beta(\epsilon_2-\mu)}\right)\left(1+e^{-\beta(\epsilon_3-\mu)}\right)\right]+\frac{\beta(\epsilon_1-\mu)}{e^{\beta(\epsilon_1-\mu)}+1}+\frac{\beta(\epsilon_2-\mu)}{e^{\beta(\epsilon_2-\mu)}+1}+\frac{\beta(\epsilon_3-\mu)}{e^{\beta(\epsilon_3-\mu)}+1}\right].

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Physics Friday 84

August 7, 2009

The canonical ensemble is a powerful tool for statistical mechanics. Let us now illustrate another powerful formalism, an extension known as the “grand canonical” formalism.

In the canonical formalism, our system is in contact with a thermal reservoir with which it may exchange energy; the reservir is large enough that it’s temperature be treated as a constant (thus giving it an entropy which varies linearly with energy). In the grand canonical formalism, we add the ability to exchange particles, as well as energy, with the particle reservoir large enough that the chemical potential is treated as a constant.

If we treat the combination of our system and the reservoir as a closed system with every microstate equally probable, we have for a state where our system has energy Ei and particle number Ni, the fractional occupation is the number of microstates of the reservoir for which its energy is Et–Ei and it’s particle number is Nt–Ni divided by the total number of microstates for the system and reservoir combination with total energy Et and a total of Nt particles:

However, using the entropy definition , we can rewrite this in terms of the reservoir and total entropies (as functions of energy and particle number):
.
As with our canonical formalism, we can expand the entropy as a Taylor series in energy about Et–U and in particle number about the average particle number N
, (using ).
And via additivity of the entropy, .
Thus
,
where is the thermodynamic beta and is the “grand potential”. As with the canonical ensemble, we normalize this: , where
is the “grand canonical partition function”, and the sum is over all states, with state s having energy Es and Ns particles.

We see that just as , we can solve for the grand potential:
.
Let us define . Now, the expectation value for the number of particles is
,
or, using the chain rule,
.
Similarly, one can show that
,
the variance in the particle number.

Now, consider the logarithmic partial derivative of Ƶ with respect to β:
,
and we have our internal energy. We can also show by a second differentiation that
.

From our equation for the grand potential, we see
.

Next, consider the fundamental thermodynamic relation . Soving for the mechanical work term,
. Now, if we continue to hold β (and thus T) and μ constant, the right hand side of the previous relation is just –dΨ:
.
From this, one can obtain the equation of state
.

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