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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?

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  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 $w_n = e^{-E_n/\theta}/z$.

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  • $\begingroup$ I have two questions for what you wrote. You have no boundaries in the 2nd integral. Does that mean the boundaries are $-\infty$ and $+\infty$? . "Thermal states of a quantum system" since they are described by a density matrix, that implies that these thermal states can be either microstates or macrostates, correct? $\endgroup$
    – imbAF
    Jan 10, 2022 at 18:06
  • $\begingroup$ @imbAF Yes, boundaries in the second integral are $-\infty$ and $\infty$ for each $p_i$. I am confused if we can say "macrostate" for one particle. For a system of many particles, $\hat{\rho}$ describes a macrostate, and $|n>$ are microstates. $\endgroup$
    – Gec
    Jan 10, 2022 at 19:40
  • $\begingroup$ I am a bit confused with all the different names for states. If the system is, in classical terms a statistical ensemble, a canonical one. In QM this translates to a mixed state, which is represented via the density matrix. So the mixed state of the system=macrostate (in classical thermodynamics). The microstates of the system (classical thermodynamics)=pure states (QM), and at the same time the microstates/pure states are eigenstates of the hamiltonian. Am I correct ? Or no? $\endgroup$
    – imbAF
    Jan 10, 2022 at 19:58
  • $\begingroup$ @imbAF I would say, yes, you are correct. A set of microstates (=eigenstates of hamiltonian) and corresponding probabilities define a macrostate, which, from the other side, can be described by macroscopic quantities like temperature, pressure, and others. $\endgroup$
    – Gec
    Jan 10, 2022 at 20:08
  • $\begingroup$ first of all thx. I needed a confirmation because things get a bit messy sometimes. I do have one more additional question, if you have time. $\endgroup$
    – imbAF
    Jan 10, 2022 at 20:10

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