Distinguishable and Indistinguishable Particles
Examples of distinguishable particles are immobilized particles and particles wherein a position vector,
[ \vec r ]
, does not change. Sum over all the states and sum over the indices independently. An expression of the partition function,
[ Q ]
, is below.
[ Q=q^N ]
There is an unresticted sum in the case of indistinguishable particles. There would be overcounting without a restricted sum. The number of states is much greater than the number of particles. One method of compensation is to divide by a factor of
[ N! ]
. Consider an example to show that the Boltzmann approximation makes sense.
Example
Consider an ideal gas in three dimensions. An expression of
[ \epsilon ]
is below, and the number of single-particle states with energy less than
[ \epsilon ]
is
[ \phi ( \epsilon) ]
. In this example, decouple each particle of the ideal gas and look at single-particle space.
<center>
[ \phi (\epsilon) = \frac
\left ( \frac
\right )^{\frac
{2}} ]
[ T = 300K ]
[ \epsilon \approx \frac
k_B T ]
[ m = 10^{-22} \mbox
]
[ a = 10 \mbox
]
[ \phi ( \epsilon ) = 10^
]
The Boltzmann approximation holds under normal conditions. Generally, the expression below of
[ Q ]
is valid at normal temperatures, low density, and large mass. The number of available states scales with
[ m^{\frac
{2}} ]
, and a lower density corresponds to more available states. These conditions are indications to use the Boltzman approximation. When there are lots of states to choose from, it is rare that two or more particles share the same state. This takes care of the Fermi violation rule.
Thermodynamics for a Monoatomic Ideal Gas
Decouple the particles and the Hamiltonian. There is a multiplication of single-particle partition function and a sum over individual interactions.
Gas atom
There different degrees of freedom of a gas atom. These degrees of freedom include translational or kinetic, electronic excitations, and nuclear. These are independent of each other. They can be decoupled and written as different types of Hamiltonians. Consider single particle energy and single partition function.
[ \hat H_{\mbox{single}} = \hat H_{\mbox{translational}} + \hat H_{\mbox{electronic}} + \hat H_{\mbox{nuclear}} ]
[ \epsilon_{\mbox{single}} = \epsilon_{\mbox{translational}} + \epsilon_{\mbox{electronic}} + \epsilon_{\mbox{nuclear}} ]
[ q_{\mbox{single}} = q_{\mbox{translational}} + q_{\mbox{electronic}} + q_{\mbox{nuclear}} ]
At normal temperatures, the nuclei are not excited. The number of states available is one, which means that the energy is constant. The value of
[ q_{\mbox{nuclear}} ]
is constant, so the derivative is zero. With regard to
[ q_{\mbox{electronic}} ]
, the first excited state could be included, and this term is considered in McQuarry. Consider the translational term here.
[ q_{\mbox{atom}} = q_{\mbox{translational}} ]
[ q_{\mbox{atom}} = sum_l e^{-\beta \epsilon_l} ]
[ \epsilon_l = \epsilon_
]
[ \epsilon_l = \frac
\left (n_x^2 + n_y^2 +n_z^2 \right ) ]
[ \epsilon_l = \sum_
^
\sum_
^
\sum_
^
e^{-\frac
\left (n_x^2 n_y^2 n_z^2 \right ) } ]
Assume that the states are dense, and the sums become integrals. Sum independently. Numbers related to the gamma function may be provided during the test.
[ \left ( \int_0^
e^{-\frac
, dn \right )^3 = \left ( \frac
\right )^{\frac
{2}} V ]
[ \Gamma (z) = z \int_0^
e^{-t^2} t^
dt ]
[ t^2=\frac{\beta h^2 n^2}
\frac
= \frac
\sqrt{\frac
{2 m}}\Gamma \left ( \frac
\right ) = \sqrt
]
{latex [ Q = \frac
]
[ Q = \frac
\left ( \frac
\right )^{\frac
{2}} V^N ]
Derive quantities from the partition function
The last relation is between the characteristic potential and the partition function. Quantities can be derived from this relation. Use Stirling's approximation.
F=-k_B T \ln Q
F = -k_B T \ln \left (\frac
\right ) - Nk_B T \ln q + k_B T \ln N!</math><br><math>F = N k_B T \ln q + N k_B T \ln N -N k_B T</math><br><math>F = -N k_B T \ln \left [\left ( \frac
\right )^{\frac
{2}} V \right] + N k_B T \ln N - Nk_B T
Equations of state can be derived. Use the differential.
dF = -Sdt - pdV + \mu dN
Look at the pressure in a monoatomic ideal gas.
p = - \left ( \frac
\right )_
p = \frac
p V = N k_B T
This is the ideal gas equation of state. Assume that only degrees of freedom are translational degrees of freedom. Consider the chemical potential. There is an explicit expression of\mu_0derived from first principles.
\mu = \left ( \frac
\right )_
\mu = - k_B T \ln \left [\left ( \frac
}
\right )^{\frac
V \right] + k_B T \ln N + \frac
- k_B T
\mu = - k_B T \ln \left [\left ( \frac
}
\right )-^
k_B T \right] + k_B T \ln \left( \frac
\right )
\mu = \mu_0 + k_B T \ln \left( \frac
\right )
Calculate the average kinetic energy of particles in a system.
F = k T^2 \left ( \frac
\right )_
F = k T^2 \frac
\left ( \frac
\ln T \right )</math><br><math>F = \frac
\bar \epsilon = \frac
</math></center>A way to approach problems is to make assumptions, make approximations, decouple interactions, use the equations of state, and derive single-particle relations.Boltzman approximationsConsider the Boltzman approximation. It is a good example to check. When considering distinguishable particles, the number of particles does not affect the average energy or pressure, but it does affec the chemical potential. Considder low density and low temperature.When does Boltzman break down?Let's consider <math>N</math> indistinguishable particles and the canonical ensemble. There are <math>n_k</math> particles in state <math>k</math> with energy, <math>\epsilon_k</math>.<center><br><math>E_j = \sum_k \epsilon_k n_k</math><br><math>N = \sum_k n_k</math><br><math>Q(N, V, T) = \sum_j e{ \beta E_j}</math><br><math>Q(N, V, T) = \sum_
^* e^
</math><br></center>The last sum above is restricted. The sum is restricted by the condition below<center><br><math>\sum_k n_k = N</math><br></center>Consider the system energy. The particles cannot be decoupled. The problem must be approached differently. Sum over all system states. The sum isn't hard to do. Use the grand-canonical ensemble. In this ensemble, the number of particles, <math>N</math> can fluctuate. Sum over all <math>N</math>.<center><br><math>\theta (T, V, \mu) = \sum_
^
\sum_j e^{-\beta (E_j - \mu N )
Sum over all the allowed states. The restriction is removed from the system.
E_j - \mu N = \sum_k n_k ( \epsilon_k - \mu )
E_j - \mu N = \sum_
^
\sum_
e^{-\beta \sum_k (\epsilon_k - \mu )
There is still a dependence. There is a restriction by the first sum. The sums are dependent. Write as an easier sum.
\sum_
^{infty}} \sum_
* \pi_k \left (e{- \beta ( \epsilon_k - \mu)} \right )^
Sum over the exponential. Since the number of particles goes to infinity, eachnkranges over all possible values. Write as independent. Below is a general formula of the grand-canonical ensemble. Assume particles in each state.
\sum_
^{n_{\mbox
}} \sum_
^{n_{\mbox
}} \sum_
^{n_{\mbox
}} \pi_k \left (e^{-\beta (\epsilon_k - \mu) \right )^
\Pi_k \sum_
^{n_{\mbox
}} \left ( e^{- \beta ( \epsilon_k - \mu ) \right )^
Particle specificsConsider fermions and bosons.FermionsThe maximum number of fermions that can be in the same state is one. Consider the grand-canonical partition function. The summation of fermions is below.
[ \sum_
\left ( e^{- \beta ( \epsilon_k - \mu ) } \right )^
]
[ 1 + e^{- \beta ( \epsilon_k - \mu )} ]
Bosons
An infinite number of bosons that can be in the same state. Consider the partition function. The sum can be found by using the equation relating to a geometric series.
[ Q = \sum_
^
\left ( e^{- \beta ( \epsilon_k - \mu ) \right )^
]
[ \mbox
]
[ x < 1 ]
[ \sum_
^{infty}} x^n = \frac
]
[ Q = \frac
{1 - e^{-\beta (\epsilon_k - \mu)}} ]
Restrictions - Fermi-Dirac and Bose-Einstein Statistics
Take into account how many particles can be in the same state. Below is the partition function considering Fermi-Dirac and Bose-Einstein statistics.
[ \mbox
]
[ \theta_
= \Pi_k \left ( 1 + \lambda e^
\right ) ]
[ \mbox
]
[ \theta_
= \Pi_k \left ( 1 - \lambda e^
\right )^{-1} ]
Consider the differential of
[ \phi ]
and derive quantities.
[ d \phi = -S dT - p dV - N d \mu ]
[ \bar N = - \left ( \frac
\right )_
]
[ \bar N = k_B T \frac
|_
]
[ \mbox
]
[ \bar N = \sum_k \frac{ \lambda e^{- \beta \epsilon_k}}{1 + \lambda e^{-\beta \epsilon_k}} ]
[ \mbox
]
[ \bar N = \sum_k \frac{ \lambda e^{- \beta \epsilon_k}}{1 - \lambda e^{-\beta \epsilon_k}} ]
The total number of particles is equal to the summation of the number of particles in different states over all states. The average number of particles in statekis
[ \bar n_k ]
. This term is called the average occupation number for state
[ k ]
.
[ N = \sum_k n_k ]
[ \bar N = \sum_k \bar n_k ]
[ \mbox
]
\bar n_k = \frac{ \lambda e^{- \beta \epsilon_k}}{1 + \lambda e^{-\beta \epsilon_k}}
\mbox
</math><br><math>\bar n_k = \frac{ \lambda e^{ \beta \epsilon_k}}{1 - \lambda e^{-\beta \epsilon_k}}</math><br></center>A relation is below considering Fermi-Dirac statistics and the chemical potential.<center><br><math>\bar n ( \epsilon ) = \frac
{1 + e^
\mu \approx \epsilon_F
\bar n ( \epsilon ) = \frac
Below is a plot of the occupation number versus energy. There are changes with energy and temperature.
Unable to render embedded object: File (Occupation_versus_energy.PNG) not found.
Consider the average number of particles with regard to Bose-Einstein statistics. Below is an equation and a graph
[ \bar n (\epsilon) = \frac
{e^
) } - 1} ]
Derive curves by considering the maximum number of particles in each state and the grand canonical ensemble. Theory and symmetry of the wavefunction were considered. There were no assumptions about interactions. Bose-Einstein condensation results from decreasing temperature. Push particles into ground state.
At high temperatures, the two functions approach the same curve. Both turn into Maxwell-Boltzman statistics.
[ \bar n(\epsilon) \right \alpha e^
]
[ P_
= \frac{ e^{- \beta \epsilon}}
]
At low temperature and high density, Boltzman statistics of a monoatomic gas turn into Fermi-Dirac statistics.
Diatomic Molecule
Use Boltzman statistics at room temperature. Decouple interactions. Vibrations are small with the approximation of a rigid rotor