- #1
Beer-monster
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Homework Statement
Optical tweezers have been used to control and manipulate atoms. For
simplicity, we model a very small quantum tweezer as a structure having quan-
tum levels with energies E n = n, where n = 0, 1, 2...N, and N 1.
A) Assume that the atoms are distinguishable and that there are ten atoms in the
tweezer. Calculate the partition function for this quantum tweezer as a function
of temperature. At temperature T , what is the average occupation number of
atoms in the ground state (i.e. n = 0 level) following this assumption?
B) Fermionic atoms are indistinguishable and obey the Pauli exclusion principle.
(For Parts B-D of this problem you should assume that the atoms are all spin
polarized—in other words, assume the fermions are all identical spin states.)
Find out in which limit the results in part A are approximately correct for
fermionic atoms.
C) Use the Pauli exclusion principle and derive the exact partition function for
a quantum tweezer weakly connected to an atom reservoir (so that atoms can
be exchanged) with the chemical potential set to µ.
D) By varying the chemical potential, one can control the number of atoms in
the tweezer. For what value of µ can you be certain to load exactly 3 atoms
into the tweezer in the low T limit?
Homework Equations
Boltzmann distribution [itex] \bar{n} = \frac{e^{\frac{\epsilon_{n}}{kT}}}{\sum^N_{n=0} e^{\frac{\epsilon_{n}}{kT}}} [/itex]
Fermi-Dirac distribution [itex] \frac{1}{e^{\frac{\epsilon_{n}-\mu}{kT}}+1} [/itex]
The Attempt at a Solution
To be honest I'm somewhat confused about what this question is actually asking for in parts A and B. It seems to be that's its almost asking for a derivation if quantum statistics. First its asks for the partition function and average occupation number (Boltzmann distribution).
Then it asks how would this change if we took into account the Pauli exclusion principle and indistiguishability which seems like it would mean a derivation of the Fermi-Dirac distribution.
Does this seem like I'm reading this right?