Partition function of 2 bosons in two energy level 0 and E

[Apashanka]

Homework Statement

Partition function of 2 bosons in two energy level 0 and E

Relevant Equations

Partition function of 2 bosons in two energy level 0 and E

For 2 bosons each of which can occupy any of the energy levels 0 and E the microstates will be 3
0 E
a a
aa -
- aa
the partition function is therefore $$z=1+e^{-\beta E}+e^{-2\beta E}...(1)$$
Another approach to do..
The single particle partition function is
$$z=1+e^{-\beta E} $$ and therefore for 2 identical particle the partition function becomes $$z^2/2!..…(2)$$ and as such (1) and (2) doesn't match……
Am I wrong somewhere??
Thanks,
Apashanka

 

[mfb]

The second approach only works if you can neglect the cases where more than one particle is in the same state.

 

[Apashanka]

The second approach only works if you can neglect the cases where more than one particle is in the same state.

Therefore here the first one is the correct isn't it??

 

[Apashanka]

@mfb But for distinguishable particle the second approach works ...

 

[mfb]

Yes, because they can never occupy the exact same state together. Just states at the same energy.
If the problem statement says "2 bosons" you can assume that they are indistinguishable, otherwise it would be pointless to say that they are bosons.

 

[Apashanka]

Yes, because they can never occupy the exact same state together. Just states at the same energy.
If the problem statement says "2 bosons" you can assume that they are indistinguishable, otherwise it would be pointless to say that they are bosons.

What I mean to say is if a and b be two distinguishable particle
Then the microstates will be
0 E
a b
ab -
- ab
b a
In this $$z=1+2e^{-\beta E}+e^{-2\beta E}$$
And the single particle partition function is $$z_{sing}=1+e^{-\beta E}$$
Therefore $$z=z^2_{single}$$ matches beside two particle occupies the same energy state...

 

[mfb]

Yes, that scenario is different from your original problem.

 

[Apashanka]

Actually I got stuck doing this problem

What I am here getting is the if considering single particle partition function z and then for 2 identical particles the partition function becomes $z^2/2! =(1+e^{-\beta E})^2/2!$ and then taking the mean energy as $-\frac{\partial}{\partial \beta}ln z$ then none of the answer matches
Otherwise if finding the microstates and then the partition function of these 2 bosons $z=1+e^{-\beta E}+e^{-2\beta E}$ then also none of the answer matches...
Can anyone suggest where is the mistake??

 

[mfb]

Well, the approach with $z^2/2$ can't work, it needs distinguishable particles and/or temperatures so high that the difference between bosons and fermions doesn't matter because states are unlikely to have more than one particle.

None of the answers make sense. (d) is the only one where the numerator is the derivative of the denominator, but the 2 in the denominator makes no sense. I guess either (a) or (d) were supposed to be the answer and the 2 shouldn't be there in (a) or the 2 should be a 1 in (d).

 

[Apashanka]

Well, the approach with $z^2/2$ can't work, it needs distinguishable particles and/or temperatures so high that the difference between bosons and fermions doesn't matter because states are unlikely to have more than one particle.

None of the answers make sense. (d) is the only one where the numerator is the derivative of the denominator, but the 2 in the denominator makes no sense. I guess either (a) or (d) were supposed to be the answer and the 2 shouldn't be there in (a) or the 2 should be a 1 in (d).

Yes sir thanks a lot