Consider a canonical ensemble of $N$ ideal gas atoms, which could have spin up or spin down. Why is it that the probability of finding the particle in a spin up state generally only involves the single partition function?

In a canonical ensemble, $P_i = \frac{e^{-\frac{E_i}{kT}}}{Z}$.

$$P_{up} = \frac{e^{-\frac{h}{kT}}}{e^{-\frac{h}{kT}}+e^{\frac{h}{kT}}} $$

Why do we only use the one-particle partition function here and not the $N$ particle partition function?

I tried this using a sample Hamiltonian:

$$H = \sum_{i=1}^N \frac{p_i^2}{2m} - h s_i$$

I got:

$$Z = \frac{1}{N!} \left(\frac{V}{\lambda^3} (e^{\beta h} + e^{-\beta h}) \right)^N$$, where $\lambda = \frac{h}{\sqrt{2 \pi m k_B T}}$

Is it possible to reproduce $P_{up}$ from the $N$-particle partition function. Can one do this using the binomial theorem, $(1+x)^n = \sum_{k=0}^n {n \choose k} x^k$?

I'm trying to understand how the single particle and many particle partition functions are related.


Let's say one is after the N-body partition-function of a classical system. $$ Z_N \propto \int d\Omega_N e^{-\beta H} $$ The integral is over the N-body phase-space and $H$ is the full Hamiltonian. If the system is non-interacting, $H$ is the sum of $N$ one-particle Hamiltonians $$ H = H_1+H_2+\cdots+H_N $$ each of which only depends on the coordinates and momentum of a single particle.

The phase-space integral then factorizes into one-body integrals $$ Z_N \propto \left(\int d\Omega_1 e^{-\beta H_1}\right)\left(\int d\Omega_2 e^{-\beta H_2}\right)\cdots\left(\int d\Omega_N e^{-\beta H_N}\right) = \prod_i Z(i)$$ which is just the product of the single-particle partion functions. If the Hamiltonians are all the same, e.g $H_i=p_i^2/2m$, all s.p partion-functions coincide and you indeed have $$ Z_N \propto Z_1^N $$ Now, I write proportional, because as you have noticed, there's a factor of $1/N!$ missing. This comes about when you're dealing with indistinguishable particles. If one does not include this factor, the entropy becomes for example non-additive and you're essentially facing Gibb's paradox.

Note that the notion of indistinguishablity is fundamentally a quantum one! Classically such a thing would not be expected, nevertheless this quantum property prevails in the classical limit.

As to the first part of your question: You speak of the probability of 'the particle' having spin up. I'm not sure that's a sensible question in the context of many bodies. Do you mean the probability of finding exactly one particle in the up-state? If so, think about the energy of such a state and use the definition of $P_i$ you gave.


Writing : $Z = \sum\limits_{\{N_{\lambda_i}\}}e^{-\beta \sum\limits_{\lambda_i} N_{\lambda_i}\epsilon_{\lambda_i}}$, we have :

$\langle N_{\lambda_j}\rangle = \dfrac{1}{Z}\sum\limits_{\{N_{\lambda_i}\}}N_{\lambda_j}e^{-\beta \sum\limits_{\lambda_i} N_{\lambda_i}\epsilon_{\lambda_i}} = \dfrac{1}{Z} \dfrac{-1}{\beta} \dfrac{\partial}{\partial \epsilon_{\lambda_j}} Z = \dfrac{-1}{\beta} \dfrac{\partial \ln Z}{\partial \epsilon_{\lambda_j}}$

In the case where $Z = z^N$ or $Z = \dfrac {z^N}{N!}$, we get : $\ln Z = N \ln z + A$, where $A$ is a constant which does not depend on the $\epsilon_{\lambda_j}$. So we could write :

$\langle N_{\lambda_j}\rangle = \dfrac{-N}{\beta} \dfrac{\partial \ln z}{\partial \epsilon_{\lambda_j}}$

In your case, we have $z = e^{-\beta \epsilon_+} + e^{-\beta \epsilon_-}$, so it is very easy to get $N_+$ and $N_-$ from the above relation :

$\langle N_\pm \rangle = N \dfrac{e^{-\beta \epsilon_\pm}}{e^{-\beta \epsilon_+} + e^{-\beta \epsilon_-}}$


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