Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

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.

share|improve this answer

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 :

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.