# Quantum versus Classical Partition Function

I am confused as to which partition functions are classical and which are quantum. I am interested in the canonical ensemble.

Several places I have seen the "classical" partition function as:

$$Z = \sum_{i=0}^{\infty}e^{{-\beta E_{i}}}$$ (1)

as well as:

$$Z=\frac{1}{h^3}\int_{0}^{\infty}e^{-\beta H(p,q)}dpdq$$ (2)

I have seen the "quantum" partition function as:

$$Z = \sum_{i=0}^{\infty}e^{{-\beta E_{i}}}$$ , contradicting (1)

$$Z = \sum_{i=0}^{\infty}g(E_{i}) e^{{-\beta E_{i}}}$$ (3)

$$Z = \sum_{n=0}^{\infty}e^{{-\beta \sum_{i}^{\infty}n_{i}E_{i}}}$$ (4)

I have probably come across more as well, so it's all a bit confusing.

Thanks

Formula $$(1)$$ can be considered a general, formal expression valid for classical and quantum systems. The critical point is to have clear in mind the meaning of the summation index $$i$$. The partition function $$(1)$$ is a sum over the microstates.

In a classical system with a finite set of microstates, like, for example, the Ising model, $$i$$ labels one of the $$2^N$$ possible microstates.

In a classical system with an infinite (continuous) set of microstates, like, for example, a system of atoms, the index $$i$$ and the symbol of summation must be intended as a symbolic notation for the more formally correct expression $$(2)$$ (however, in that formula $$dpdq$$ should be intended as the $$6N$$-dimensional phase space volume element and the integral should be divided by $$h^{3N}$$). The correspondence is evident in any discretized approximation of the integral in formula $$(2)$$: all the infinite microstates in a small volume $$dpdq$$ of the $$6N$$-dimensional classical phase space are considered as a set of $$dpdq/h^{3N}$$ states all with the same energy $$E_i$$.

In a quantum system, the canonical partition function is the trace of the unnormalized canonical density matrix operator $$e^{-\beta \hat H}$$, where $$\hat H$$ is the Hamiltonian operator. By choosing a basis of eigenstates of the Hamiltonian, this trace reduces to expression $$(1)$$, provided $$i$$ is now intended as the complete set of quantum numbers labeling each eigenvector of the Hamiltonian.

For quantum and classical discrete systems, the expression $$(1)$$ can be rewritten in the form of the expression $$(3)$$, provided the index $$i$$ is intended as the label of the $$i$$-th energy level. In this formula, the factor $$g(E_i)$$ represents the degeneracy of each energy level (i.e., the number of microstates having the same energy $$E_i$$.

Finally, the expression $$(4)$$, with some corrections, would correspond to the particular case of the canonical partition functions for a separable non-interacting system. However, in that case, it should be written as $$\sum_{\{ n_i \}} e^{-\beta \sum_{i} n_i \varepsilon_i}$$ and the meaning of the symbols is as follows

• $${\{ n_i \}}$$ is the set of all possible occupation number configurations of one-particle microstates;
• $$\sum_{\{ n_i \}}$$ has to be intended as $$\sum_{n_1} \sum_{n_1} \dots \sum_{n_1}$$, where each sum goes over the allowed occupation numbers, taking into account every possible constraint on them;
• $$i$$ in these formulas has to be intended as the label of the $$i$$-th one particle microstate.
• $$\varepsilon_i$$ is the energy of the $$i$$-th one-particle state, which is different from the energy of the $$N$$-particles state.
• Just a small mistake here: "In a quantum system the canonical partition function is the trace of the canonical density matrix". The trace of the density matrix is always 1 (normalisation of probabilities). You should instead say that $Z$ is the normalisation factor of the canonical density matrix, so that $\mathrm{Tr} \rho = 1$.
– Cham
May 17, 2022 at 23:31
• @Cham I had in mind the unnormalized canonical density matrix $\hat \rho = e^{-\beta \hat H}$. However, thanks. I am going to clarify this point. May 18, 2022 at 5:08

For a classical system defined on a phase-space $$\{q_1,...,q_n,p_1,....,p_n\}$$ and a Hamiltonian $$H(\{q_n,p_n\})$$ the partition function for a canonical ensemble is given by $$Z_{\text{classical}}= \prod_i^n\iint dq_idp_i \exp[-\beta H(\{q_n,p_n\})].$$ For a quantum system with a discrete spectrum $$\{n_1,n_2,n_3,...\}$$ with energies $$E_i$$ and a continuous spectrum parameterized by $$\mathbf{k}$$ the partition function is given by $$Z_{\text{quantum}}=\sum_{n_i} \exp(-\beta E_i) + \int d^3k\exp(-\beta E(\mathbf{k}))$$ I think we should always be able to paramterize the continuous spectrum such that $$E(\mathbf{k})=E_{ij}k_ik_j$$ but I'm not entirely sure. The quantum case can be formulated more elegantly by introducing the thermal (mixed) state $$\hat{\rho} = \frac{1}{Z}e^{-\beta \hat{H}},$$ so the idea is that a thermalized system is in this mixed state. Normalization requires $$\text{Tr}\hat{\rho} = 1$$ which means that $$Z=\text{Tr}~e^{-\beta \hat{H}}=\sum_{n_i}\langle n_i|e^{-\beta \hat{H}}|n_i\rangle=\sum_{n_i} \exp(-\beta E_i)$$ is just the quantum partition function (with the continuum spectrum implied). The thermal expectation value for an observable $$\Omega$$ also follows from this as $$\langle\Omega\rangle=\text{Tr}(\Omega\hat{\rho})=\frac{1}{Z}\sum_{n_i}\langle n_i|\Omega|n_i \rangle\exp(-\beta E_i).$$ Note that the thermal state is also constant in time since $$[\hat{\rho},\hat{H}]$$ vanishes, which is nice.