H problem in statistical mechanics

[evilcman]

Is there some kind of resolution to the Hydrogen atom problem in statistical physics, that is the fact that canonical partition function diverges for $$E_n = - E_0/n^2$$ with degeneracy $$n^2$$ since $$Z = \sum n^2 exp(-\beta E_0/n^2) > \sum n^2 exp(-\beta E_0)$$ , which makes the H atom problem seem ill-posed in statistical physics. I am sure this problem is treated in detail somewhere, can someone point me to a reference?

 

[DrDu]

If you calculate all energies with respect to the ground state, the problem disappears.

 

[evilcman]

Calculating energies with respect to the ground state just means adding $$E_0$$ to all energies, that is multiplying the partition function with $$exp(\beta E_0)$$ and the sum will still diverge.

 

[DrDu]

Ok, I see the problem now.

 

[evilcman]

Simple regularization doesn't help either, if we sum up to state N only, then we get:
$$-<E> = \frac{\sum E_0 exp(-\beta E_0 / n^2 )}{\sum n^2 exp(-\beta E_0 / n^2 )} < \frac{N E_0 exp(-\beta E_0)}{N^3} \to 0$$
which would mean that at ANY temperature, all hydrogen atoms are at a highly excited state($$n \to \infty$$)

 

[DrDu]

I suppose the problem to be due to the long range nature of the Coulomb potential.
Landau Lifshetz, Quantum mechanics discuss the Coulomb problem quite at length. So I would give it a try.
In practice, the Hydrogen atom only has a finite number of bound states the highly excited ones being destroyed by interaction with other atoms.

 

[DrDu]

which would mean that at ANY temperature, all hydrogen atoms are at a highly excited state($$n \to \infty$$)

Btw, this is in deed a general feature. Isolated bound states are never stable thermodynamically as there are an infinite number of competing non-bound states. Remember that also for the H atom you also have the continuum of states with E>0 which also contribute to Z.

 

[kanato]

Wait, when you start you write down $$E_n = -E_0/n^2$$ but your partition function should be $$Z = \sum_n g_n \exp(-\beta E_n)$$ (where g_n is the degeneracy) so you should have $$Z = \sum_n n^2 \exp(\beta E_0/n^2)$$ (no negative in the exponential.) This should change the sign of your inequalities. Just a little testing in excel shows me that this is the source of the divergence in your partition function. If I correct the sign, at low temperature the partition function converges rapidly.

Btw, this is in deed a general feature. Isolated bound states are never stable thermodynamically as there are an infinite number of competing non-bound states. Remember that also for the H atom you also have the continuum of states with E>0 which also contribute to Z.

I've never heard this. Do you have a reference which explains this in detail? An infinite number of non-bound states should only be a problem if the number of them grows exponentially with energy, so that the degeneracy factor in the partition function beats out the rapidly decaying exponential. Even then, one can always make the decaying exponential decay faster by going to lower temperature, but that should have no effect on the number of unbound states.

In any case, this doesn't explain why his model is unstable, since the non-bound states are neglected.

 

[kof9595995]

Wait, when you start you write down $$E_n = -E_0/n^2$$ but your partition function should be $$Z = \sum_n g_n \exp(-\beta E_n)$$ (where g_n is the degeneracy) so you should have $$Z = \sum_n n^2 \exp(\beta E_0/n^2)$$ (no negative in the exponential.) This should change the sign of your inequalities. Just a little testing in excel shows me that this is the source of the divergence in your partition function. If I correct the sign, at low temperature the partition function converges rapidly.

I doubt it, since clearly
$$\mathop {\lim }\limits_{n \to \infty } {n^2}\exp (\frac{{\beta {E_0}}}{{{n^2}}}) = \infty$$

 

[kanato]

hmm you're right, it only seems to converge initially. In any case, the sign error still significantly impacts things, especially at low temperature, and especially for the finite sums mentioned..

 

[DrDu]

I've never heard this. Do you have a reference which explains this in detail? An infinite number of non-bound states should only be a problem if the number of them grows exponentially with energy, so that the degeneracy factor in the partition function beats out the rapidly decaying exponential. Even then, one can always make the decaying exponential decay faster by going to lower temperature, but that should have no effect on the number of unbound states.

In any case, this doesn't explain why his model is unstable, since the non-bound states are neglected.

Well, in principle this was already known to our forefathers under the name of Ostwalds dilution law:
http://en.wikipedia.org/wiki/Law_of_dilution

 

[kof9595995]

Btw, this is in deed a general feature. Isolated bound states are never stable thermodynamically as there are an infinite number of competing non-bound states. Remember that also for the H atom you also have the continuum of states with E>0 which also contribute to Z.

So for example, even for H_2 gas at room temperature, there should be ionizations, though very weak. Am I interpreting you correctly?

 

[DrDu]

Yes. You may compare it to the vapor pressure of water. Although it is small for ice, it is non-zero and a piece of ice may sublime completely.

 

[genneth]

I doubt it, since clearly
$$\mathop {\lim }\limits_{n \to \infty } {n^2}\exp (\frac{{\beta {E_0}}}{{{n^2}}}) = \infty$$

The partition function itself is often non-convergent. The question is whether the various derivatives of the log of Z is. I haven't been convinced so far that there are any problems...

 

[kof9595995]

The partition function itself is often non-convergent. The question is whether the various derivatives of the log of Z is. I haven't been convinced so far that there are any problems...

Could you give an example?

 

[genneth]

Could you give an example?

This tends to happen generically in systems with large number of states. Even something like the Ising model gives an infinite answer if you don't have infrared cutoffs.

 

[kof9595995]

But if partition function is divergent, that means entropy S=klnZ+...will be infinite. Is an infinite entropy acceptable?

 

[DrDu]

To calculate the entropy you have to assume a fixed volume. Once you do this the continuum becomes a set of very narrowly spaced discrete levels and also there won't be any longer an infinite number of bound states. below E=0. However, increasing the volume you will finally find that the equilibrium solution for any temperature >0 is an ionized plasma when you consider only one atom (or when the concentration goes to zero in the limit V-> infinity).

 

[kof9595995]

It makes sense. So when considering reality, we must get rid of infinity in one way or another, right? So I still disagree with genneth about "The partition function itself is often non-convergent. The question is whether the various derivatives of the log of Z is. I haven't been convinced so far that there are any problems... " in post 14.

 

[DrDu]

You may try to calculate e.g. the mean energy, which is basically the derivative of log Z with respect to beta. Even when you only sum over the bound states, its value is 0 for all finite T as $$Z=\sum n^2 \exp(-\beta E)$$ in the denominator diverges more rapidly than $$d/d\beta Z=E_0 \sum \exp(-\beta E)$$. So the hydrogen atom really lives in the states with n=infinity.

 

[kof9595995]

Yes, but with n=infinity, entropy is infinite, so despite that the energy is finite, n=infinity is still problematic, isn't it?

 

[genneth]

But if partition function is divergent, that means entropy S=klnZ+...will be infinite. Is an infinite entropy acceptable?

The measurable entropy (in the thermodynamic sense) is S = -dF/dT, where F = -ln Z. You're confusing it with S=k ln W, where W is the number of microscopic states corresponding to the macroscopic state.

 

[kof9595995]

No I wasn't, for Boltzmann distribution you can show S=NklnZ+<E>/T, I just didn't write down the 2nd term.

 

[genneth]

What makes you think <E> isn't (minus) infinite also?

Remember that your statement is actually just the simple thermodynamic (! --- no statistical mechanics at all) statement that U - TS = F. Now, since U and F are both energies, their absolute values make no sense at all; their difference, as you've noticed, does make physical sense, and is in fact well defined by the statistical mechanics procedure.

However, I think you're confused about a fundamental statistical mechanics point. In statistical mechanics, the thermodynamic limit (infinite number of states) is the *last* step to be taken in any calculation. All thermodynamic quantities are the limits of finite, well-defined averages. Failure to do this will quite often yield complete nonsense. To be honest, this is a very subtle issue which tends to be overlooked in textbook treatments, because the actual calculation, with full rigour, tends to be essentially impossible. For a historical account, see the original papers by Yang and Lee on the lattice gas model, where they overturned a (then) some time held conviction that in fact statistical mechanics could not describe liquid phases.

 

[kof9595995]

What makes you think <E> isn't (minus) infinite also?

As OP showed in post 5

However, I think you're confused about a fundamental statistical mechanics point. In statistical mechanics, the thermodynamic limit (infinite number of states) is the *last* step to be taken in any calculation. All thermodynamic quantities are the limits of finite, well-defined averages. Failure to do this will quite often yield complete nonsense. To be honest, this is a very subtle issue which tends to be overlooked in textbook treatments, because the actual calculation, with full rigour, tends to be essentially impossible. For a historical account, see the original papers by Yang and Lee on the lattice gas model, where they overturned a (then) some time held conviction that in fact statistical mechanics could not describe liquid phases.

Yes you're right, and I've also learned these before, but I don't think this can help us with this problem.

 

[genneth]

As OP showed in post 5



Yes you're right, and I've also learned these before, but I don't think this can help us with this problem.

I mean that the entropy is actually finite and well-defined. You should try and show this explicitly if you have not so far --- regularise the partition function and calculate S for that finite system, then take the limit and observe that it remains finite.

 

[kof9595995]

regularise the partition function and calculate S for that finite system, then take the limit and observe that it remains finite.

If we do a cut off to the energy level to regularise the partition function, there's surely no problem anymore, as discussed in previous posts;but whether taking the thermodynamic limit is irrelevant to the convergence problem.

 

[genneth]

Perhaps this will clear things up: http://deepblue.lib.umich.edu/bitstream/2027.42/69464/2/JMAPAQ-36-3-1208-1.pdf

It points out that fundamentally the problem with a single hydrogen atom is actually that it has infinite extent --- a single atom is in fact in the thermodynamic limit. A physical cutoff should be to put things inside a conducting box, which would regularise both the bound and unbound states, but tends to make explicit calculations much harder.

 

[kof9595995]

Thanks for the resource, so by "a single atom is in fact in the thermodynamic limit" you mean V=infinity for a single atom? Emm...perhaps it's just semantics but I would not call it "thermodynamic limit", because I think it's more proper to define it by letting particle number N=infinity with specific volume fixed, and V=infinity is a consequence of this definition.

 

[genneth]

Thanks for the resource, so by "a single atom is in fact in the thermodynamic limit" you mean V=infinity for a single atom? Emm...perhaps it's just semantics but I would not call it "thermodynamic limit", because I think it's more proper to define it by letting particle number N=infinity with specific volume fixed, and V=infinity is a consequence of this definition.

I agree, because intuitively it seems like a thermodynamic limit should be a many-body limit. However, from a purely mathematical (and therefore precise) point of view, neither is entirely correct. If we are dealing with a canonical ensemble, the extensive variables are S and V; the thermodynamic limit corresponds to holding the conjugate variables T and p constant and letting S and V tend to infinity. The particle number N, is held fixed by definition. If we use a grand canonical ensemble, N is also extensive, and the thing which is held constant is the chemical potential \mu. Your statement about holding the specific volume (i.e. density) constant is equivalent to holding \mu constant in the non-interacting limit, and approximate the same in the weakly interacting limit.

In our current problem, N=1, and we are discussing a canonical ensemble --- that was what was used for the definition of the partition function. If we wanted, we could have thrown in the many-body version with \mu and N, but without interactions it would have factored into N copies of the canonical one.

 

[kof9595995]

If we are dealing with a canonical ensemble, the extensive variables are S and V; the thermodynamic limit corresponds to holding the conjugate variables T and p constant and letting S and V tend to infinity. The particle number N, is held fixed by definition. If we use a grand canonical ensemble, N is also extensive, and the thing which is held constant is the chemical potential \mu. Your statement about holding the specific volume (i.e. density) constant is equivalent to holding \mu constant in the non-interacting limit, and approximate the same in the weakly interacting limit.

I'm not really familiar with these, because all discussions about thermodynamic limit I've read are based on grand canonical ensemble, do you have any online resource about it so that I could read in more details?

 

[genneth]

I'm not really familiar with these, because all discussions about thermodynamic limit I've read are based on grand canonical ensemble, do you have any online resource about it so that I could read in more details?

I'm afraid nothing really comes to mind at the moment. This is the sort of detail which tends to develop in people independently with time (relative to when they first learn statistical mechanics), because the usual textbook path through the topic will avoid it. A suggestion, which I learned from our local Soviet expert, is Landau and Lifschitz. The Soviet era books tended to have been written completely independently and are therefore most likely to provide orthogonal information to the usual textbooks. Didn't someone in this thread already mention that this pathology of hydrogen atom statistical mechanics is discussed in somewhere Russian?

 

[DrDu]

Yes, I recommended Landau Lifshetz too. I think the divergence of S and Z is unavoidable for infinite volume limit.
The thermodynamic limit is taken as V to infinity at fixed particle density.
Total energy and entropy diverge in that limit but energy density and entropy density approach a constant value.
However, the case of a single atom in an infinite volume is still special as it corresponds to vanishing energy and particle density.

 

[kof9595995]

Ok, thank you guys, I'll have a look at it.