Number of particles in canonical ensemble

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In summary, to include the constraint (1) in the calculation of the quantum partition function, one can use the Lagrange multipliers.
  • #1
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TL;DR Summary
Is there an efficient way to count the number of particles in a quantum canonical ensemble?
Consider ##N## identical atoms, with negligible interaction, in a thermal equilibrium. How to encode the number ##N## in the partition function and other related statistical quantities?

I know how to do it in the grand-canonical ensemble, either quantum or classical, with the aid of chemical potential ##\mu##. But in the grand-canonical ensemble the number of particles fluctuates, while I want to describe a system (say a gas in small ampule, surrounded by a thermal bath) in which the number ##N## of atoms is fixed. One practical method is to compute everything in the grand-canonical ensemble anyway, because, due to the law of large numbers, the fluctuations around the average value are negligible, so for practical purposes one can treat the average value as the actual fixed value. But suppose that, as a matter of principle, I do not want to do that. I want to use the canonical ensemble, and not the grand-canonical one. What can I do?

In general, the partition function in canonical ensemble is
$$Z=\sum_s e^{-\beta E_s}$$
where ##s## labels different possible states. Classical and quantum physics differ in how we define "different possible states". In classical physics
$$\sum_s=c\int d^3x_1\cdots\int d^3x_N\int d^3p_1\cdots\int d^3p_N$$
where ##c## is an arbitrary normalization constant, while in semi-classical physics the ##c## is fixed as
$$\sum_s=\frac{1}{N!}\int d^3x_1\cdots\int d^3x_N\int \frac{d^3p_1}{(2\pi)^3}\cdots\int\frac{d^3p_N}{(2\pi)^3}$$
in units ##\hbar=1##. In quantum physics, however, the states are counted in the Hilbert space, rather than in the phase space. Since the interactions are neglected, the most convenient basis of the 1-particle Hilbert space is the momentum space, so
$$\sum_s=\prod_{{\bf p}}\sum_{n_{\bf p}}$$
where ##{\bf p}## is a 1-particle momentum and ##n_{\bf p}## is the number of particles having the same momentum ##{\bf p}##. For fermions ##n_{\bf p}=0,1##, while for bosons ##n_{\bf p}=0,1,2,\ldots,\infty##.
However, since the total number ##N## of particles (atoms) is finite, there is an additional constraint
$$\sum_{{\bf p}} n_{\bf p}=N \;\;\;\;\;\;\;\;\;\;\; (1)$$

We see that quantum ##\sum_s## is very different from the classical and semi-classical one. In particular, the dependence on ##N## appears mathematically in very different ways. This has consequences for the computation of the partition function. In the classical and semi-classical case we write
$$E_s=\epsilon({\bf p}_1)+\ldots +\epsilon({\bf p}_N)$$
where ##\epsilon({\bf p})= {\bf p}^2/2m## is the 1-particle energy. The semi-classical partition function is hence
$$Z=\frac{1}{N!}Z_1^N$$
where
$$Z_1=V\int\frac{d^3p}{(2\pi)^3}e^{-\beta\epsilon({\bf p})}$$
is the 1-particle partition function, and similarly for the classical partition function. The dependence on ##N## in the classical and semi-classical partition function is clear.

The quantum case, however, is problematic. The energy is simple
$$E_s={n_{\bf p}}\epsilon({\bf p})$$
which leads to
$$Z=\prod_{{\bf p}} h(\beta \epsilon({\bf p}))$$
where
$$h(\beta \epsilon({\bf p})) \equiv \sum_{n_{\bf p}} e^{-\beta
n_{{\bf p}} \epsilon({\bf p})}$$
The problem is that I don't know a simple way to explicitly include the constraint (1) in the computation. But if I try to simply ignore the constraint (1), I get the familiar expression
$$h(\beta \epsilon({\bf p}))=1+e^{-\beta\epsilon({\bf p})}$$
for fermions and
$$h(\beta \epsilon({\bf p}))=\frac{1}{1-e^{-\beta\epsilon({\bf p})}}$$
for bosons, which are problematic because they do not depend on ##N##.

To transform the product over momenta into a sum, one can write
$$Z=e^{{\rm ln} Z}=e^{\sum_{{\bf p}} {\rm ln} h(\beta \epsilon({\bf p})) }$$
so using the approximate quantum formula
$$\sum_{{\bf p}}=V\int\frac{d^3p}{(2\pi)^3}$$
one writes the quantum partition function as
$$Z={\rm exp}\left( V\int\frac{d^3p}{(2\pi)^3} {\rm ln} h(\beta \epsilon({\bf p})) \right) $$
which can be compared with the classical and semi-classical partition function. However, since the above expressions for ##h## do not depend on ##N##, we see that the quantum ##Z## does not depend on ##N##, while the classical one does.

So to resolve the problem, one should somehow include the constraint (1) in the calculation of ##h##. Does anybody know how to do it explicitly?
 
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  • #2
The constraint (1) can be included in such considerations by using the Lagrange multipliers - with the appropriate multiplier being the chemical potential ##\mu##. If you consider, e.g., the variational principle of the form $$\langle \hat{H} \rangle = \text{min}$$ for the Hamiltonian operator, with the additional constraint $$\langle \hat{N} \rangle = \text{const}$$ for the particle-number operator, you can transform between the "variables" ##\mu## and ##N## using $$\langle (\hat{H}-\mu\hat{N}) \rangle = \text{min}$$ where ##\mu## can play a role of this Lagrange multiplier. The factor ##\sim\exp(-\mu N)## in the Gibbsian ensambles can be regarded as having fixed ##\mu## and variable ##N## (the grand-canonical ensamble), but it also can be treated as having fixed ##N## and variable ##\mu## - this latter case would correspond to including the constant-particle-number condition as an additional constraint.

Perhaps this is "overkill", but discussions on changing the relevant variables from ##\mu## to ##N## and vice versa can be found, e.g., there:

1) Abrikosov, Gorkov, Dzyaloshinski - "Methods of Quantum Field Theory in Statistical Physics" (Prentice-Hall, Inc., 1963); page 56 & page 65

2) Lifshitz, Pitaevskii - "Statistical Physics. Part 2" (Butterworth-Heinemann, 2002 {reprint}); ##\S##8
 
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  • #3
div_grad said:
The factor ##\sim\exp(-\mu N)## in the Gibbsian ensambles ... also can be treated as having fixed ##N## and variable ##\mu## - this latter case would correspond to including the constant-particle-number condition as an additional constraint.
Do you know a reference where it is done explicitly? (Preferably without field theory and Green function techniques.)
 
  • #4
Demystifier said:
Do you know a reference where it is done explicitly? (Preferably without field theory and Green function techniques.)
I remember doing similar calculations with incorporating different constraints through Lagrange multipliers (certainly with no field theory involved). I'll look into my old notes later, maybe I'll have some references or direct computations relevant precisely to your question.
 
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  • #5
It seems that a very similar problem is solved in Kardar, Statistical Physics of Particles, Sec. 7. 2.
 
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  • #6
It also occurred to me how the technical problem above can be avoided (not solved) by a simple mental trick. If ##V## is the volume of the ampule, the number ##N## of atoms in it does not fluctuate, so I cannot use the chemical potential ##\mu##, which makes things complicated. However, I can consider the volume ##V'=V/2##, i.e. half the ampule, in which the number ##N'## does fluctuate. Hence I can use ##\mu'## to express the average ##\langle N'\rangle## with a grand-canonical ensemble. Then the fixed number of particles in the full ampule is
$$N=2 \langle N'\rangle$$
I think this is a nice and instructive illustration of the general principle that "all kinds of ensembles are equivalent". Perhaps the most surprising and counterintuitive moral of this, at least to me, is that it is simpler to compute with a fluctuating number of particles than with a fixed one.
 
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  • #7
div_grad said:
particle-number operator
Note that my original problem considers a fixed number ##N## of atoms, so ##N## is not an operator. Of course, formally one can always introduce a second quantization of atoms and express everything in this higher language, but I'm not convinced that this makes the problem simpler. Physically, the change of the number of atoms in a subsystem due to thermal motions is not the same as creation/destruction of atoms in this subsystem, even if it can be treated so formally.
 
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  • #8
Demystifier said:
It seems that a very similar problem is solved in Kardar, Statistical Physics of Particles, Sec. 7. 2.
Now I have read it more carefully, so I can confirm that this indeed solves my problem. It introduces an additional complication by working with wave functions in the position space, but it has a value by giving an additional physical insight. The position space representation shows that quantum partition function differs from the semi-classical one through the quantum-exchange terms that can be interpreted as effective nonlocal potentials responsible for the quantum corrections to the classical pressure.

Even though my initial motivation was purely technical, this insight is also interesting from a quantum foundations point of view. Since the quantum pressure is measurable, this can also be interpreted as a measurable manifestation of nonlocal forces in Bohmian mechanics. The catch is that it is only measurable on the average macroscopic level of many particles, not at the level of a few particles, so one cannot take it as a direct evidence that the Bohmian interpretation of individual particles is "true". But at least it makes the Bohmian interpretation more plausible. This is also somewhat similar to my result that Bohmian trajectories can be seen experimentally at the macroscopic many-particle level in superconductors https://arxiv.org/abs/2003.14049
 
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  • #9
Demystifier said:
Note that my original problem considers a fixed number ##N## of atoms, so ##N## is not an operator. Of course, formally one can always introduce a second quantization of atoms (...). Physically, the change of the number of atoms in a subsystem due to thermal motions is not the same as creation/destruction of atoms in this subsystem (...).
What I had in mind when speaking about ##\hat{N}## was the mental picture that the particle-number operator "by default" counts the occupation numbers in states expressed in the Fock representation. So if I have a fixed number of (true, physical) particles equal ##N##, and there are ##s## possible states which can be occupied by these particles, I can write the total state of the system in the Fock space as $$\sim |n_1, n_2, \cdot\cdot\cdot, n_s \rangle$$ which is to be understood as describing a situation in which ##n_1## of those ##N## particles are in some state "1"; ##n_2## particles are in the state "2"; etc. Then the family of occupation-number operators ##\hat{n}_i## defined by their action on the states as $$\hat{n}_i |n_1, n_2, \cdot\cdot\cdot, n_s \rangle = n_i |n_1, n_2, \cdot\cdot\cdot, n_s \rangle$$ can be combined into ##\hat{N}## in an obvious way: ##\hat{N} = \sum_{i=1}^s \hat{n}_i##. The corresponding constraint on the fixed number of particles $$N = \sum_{i=1}^s n_i \equiv \text{const}$$ is expressed as a sum over the ##s## available states, and note that while ##N## is fixed once you specify the number of physical particles in some gas container the same is not true for the individual occupation numbers ##n_i##, as there are several ways in which particles can be distributed among these ##s## "bins" which lead to the same value of ##N##.

Then the appropriate "creation" and "annihilation" operators acting in such established Fock space will create and annihilate the occupations within the kets ## |n_1, n_2, \cdot\cdot\cdot, n_s \rangle## and ##\textbf{not}## the physical particles inside the container. With this picture in mind, I suspect that in order to have a fixed number of particles ##N## one must set up the formalism in such way that when any of the occupation numbers inside the kets is, e.g., reduced by one then any other occupation number within the same ket must increase by one so that the constraint condition on ##N## is always satisfied. And if one relaxes this condition, the physical ##N## could indeed change and one introduces the characteristic "grand-canonical" fluctuations in ##N##. But this is just a comment, not a solution to the problem :P
 
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  • #10
Demystifier said:
Do you know a reference where it is done explicitly? (Preferably without field theory and Green function techniques.)
I've looked into my old notes, and it turns out I got confused when I said that I considered similar problems in the past. What I actually did then was to determine distributions which maximize the logarithms of the number of microstates of given systems (standard task in stat mech), and despite the fact that I included additional constraints using Lagrange multipliers there, it has nothing to do with your original question; sorry for the confusion.
 
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