How Does the Grand Partition Function Compare in Fermi and Bose Systems?

[poiuy87]

Homework Statement

a) Suppose particles can be absorbed onto a surface such that each absorption site can be occupied by up to 6 atoms each in single-particle quantum state $$\psi_{\it i}$$ with an absorption energy $$\varepsilon_{i}$$. Write down the grand partition function for one site.

b) If $$\left( \varepsilon_{i} - \mu \right) / k_{B} T = 0.7$$

Show that the expression for the grand partition function is very close to that of a Bose system where any number of particles may occupy the site.

c) Find the probability that there are six particles on the site.

Homework Equations

None given

The Attempt at a Solution

For a fermi system only one particle can occupy each state so the following are possible:

Quantum state--------Number of configurations--------Energy--------Number of particles

|0,0,0,0,0,0>--------------------1----------------------0--------------------0
|1,0,0,0,0,0>--------------------6----------------------1--------------------1
|1,1,0,0,0,0>--------------------15---------------------2--------------------2
|1,1,1,0,0,0>--------------------20---------------------3--------------------3
|1,1,1,1,0,0>--------------------15---------------------4--------------------4
|1,1,1,1,1,0>--------------------6----------------------5--------------------5
|1,1,1,1,1,1>--------------------1----------------------6--------------------6

Grand partition function $$\Xi = 1 + 6e^ { - \left( \varepsilon - \mu \right) / k_{B} T} + 15e^ { -2 \left( \varepsilon - \mu \right) / k_{B} T} + 20e^ { -3 \left( \varepsilon - \mu \right) / k_{B} T} + 15e^ { -4 \left( \varepsilon - \mu \right) / k_{B} T} + 6e^ { -5 \left( \varepsilon - \mu \right) / k_{B} T} + e^ { -6 \left( \varepsilon - \mu \right) / k_{B} T}$$

--> $$\Xi = \left( 1 + {e^ { - \left( \varepsilon - \mu \right) / k_{B} T}} \right) ^ 6$$


Up to here I am fine, I know the question just says to write down the grand partition function but I thought I would show how I got to it. Obviously in the quantum states the 1 can take any position hence the number of configurations.

I have no idea how to do this for Bose particles, there are hundreds of different combinations of quantum states as each position can hold 0-6 atoms.

Please can somebody tell me how to find the GPF for bose particles, it must be fairly straight forward as the question is only worth 1 mark on the exam.

Thanks.

 

[Jhero]

Dear student,

Thank you for your detailed explanation of your thought process. Your approach for the fermi system is correct and you have reached the correct grand partition function. For a Bose system, the approach is slightly different.

In a Bose system, any number of particles can occupy a single quantum state. This means that for a single site, there can be 0, 1, 2, 3, 4, 5, or 6 particles. Therefore, the number of configurations for a single site is given by the sum of these possibilities:

Number of configurations = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7

This is because there are 7 possible ways to choose the number of particles on a single site. Now, using the same logic as for the fermi system, the grand partition function for a Bose system would be:

\Xi = 1 + e^{-\left(\varepsilon - \mu\right)/k_BT} + e^{-2\left(\varepsilon - \mu\right)/k_BT} + e^{-3\left(\varepsilon - \mu\right)/k_BT} + e^{-4\left(\varepsilon - \mu\right)/k_BT} + e^{-5\left(\varepsilon - \mu\right)/k_BT} + e^{-6\left(\varepsilon - \mu\right)/k_BT}

--> \Xi = \sum_{n=0}^6 e^{-n\left(\varepsilon - \mu\right)/k_BT}

This can be simplified to:

\Xi = \frac{1-e^{-7\left(\varepsilon - \mu\right)/k_BT}}{1-e^{-\left(\varepsilon - \mu\right)/k_BT}}

If we substitute the given value of \left(\varepsilon_{i} - \mu \right) / k_{B} T = 0.7, we get:

\Xi = \frac{1-e^{-4.9}}{1-e^{-0.7}} \approx \frac{1}{1-e^{-0.7}} \approx 1.43

As you can see, this is very close to the fermi system grand partition function that you had calculated earlier. This is because at high temperatures, the difference between fermi and