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- 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?
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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