Canonical partition function of a system composed of 1 particle in a box. Transition from quantum mechanical expression to classical

Question

While the expression for the classical canonical partition function is derived in my notes, there is a small detail that goes unexplained:

Assuming the particle in the box represents an isolated system and that the potential energy in the box is zero. Therefore:

$$E=\frac{\vec p^2}{2m}$$

Because I assumed that the system is isolated, the microstates of the system are eigenstates of the Hamiltonian.

In QM the expression for the canonical partition function of the 1-Particle in the box system is: $$z_1=\Sigma_n \langle n| e^{-\beta \hat H}|n\rangle$$

$|n\rangle$ is a microstate of the system, and because we assumed that it's also an eigenstate of the Hamiltonian, we get:

$$z_1=\Sigma_n e^{-\beta E_n}$$

where $$E_n=\frac{\pi^2 \hbar^2}{2m}(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2})$$ $$E=\frac{\vec p^2}{2m}$$

From here we get that:

$$\vec p=\pi\hbar(\frac{n_x}{L_x},\frac{n_y}{L_y},\frac{n_z}{L_z})^T$$

which gives us i.e $n_x=\frac{p_x}{\pi\hbar}L_x$. Analog to this we get $n_y$ and $n_z$.

Then : $$\Sigma_n \rightarrow \int_{\vec n}d^3n$$

Substituting the expression we found for each $n_i$ ($i=x,y,z$) component in the above eq. we get:

$$\int_{\vec p}\frac{L_1L_2L_3}{(2\pi\hbar)^3}d^3p$$ In order to keep it short from $\langle n| e^{-\beta \hat H}|n\rangle$ we get $\frac 1 V \int_V e^{-\beta H(\vec r, \vec p)}d^3r $

And combining the two we find for the classical canonical partition function of a 1 particle in a box system: $$z_1=\frac{1}{(2\pi\hbar)^3}\int \int e^{-\beta H(\vec r, \vec p)}d\vec r d\vec p$$

I have two questions:

1. Where does the two in $2\pi\hbar$ comes from? If you look at my derivation of the first term, I have not a 2.

2. This derivation was made possible assuming that the microstates of the system (which are present in the qm expression of the canonical partition function) are also eigenstates of the Hamiltonian. Should this be the case always?

Answer 1

1. $n_i$ are positive integers, so $$ \sum_{n_i} \rightarrow \frac{L_i}{\pi\hbar}\int\limits_0^\infty\ dp_i $$ For an even function of $\vec{p}$ we have $$ \sum_n f(\vec{p}_n) \approx \frac{L_1 L_2 L_3}{(\pi\hbar)^3} \int\limits_0^\infty \int\limits_0^\infty \int\limits_0^\infty f(\vec{p})\ dp_1 dp_2 dp_3 = \frac{V}{(2\pi\hbar)^3}\int f(\vec{p}) d\vec{p} $$

2. Thermal states of a quantum system are described by density matrices. Here, we have two expressions for the density matrix: a canonical distribution $$ \hat{\rho} = \frac1z e^{-\hat{H}/\theta}, \quad z = \mbox{Tr} e^{-\hat{H}/\theta} $$ and a general form $$ \hat{\rho} = \sum_{n} w_n |n><n|. $$ So it is natural to think that $|n>$ are eigenvectors of $\hat{H}$ and $w_n = e^{-E_n/\theta}/z$.