Is there a classical limit that connects classical volume/phase space with the scalar product of momentum and position eigenstates?

Question

The book Understanding Molecular Simulation: From Algorithms to Applications, 2002, Daan Frenkel and Berend Smit, states the following

$$ \langle r|k\rangle \langle k|r\rangle = 1/V^N $$

where $|r\rangle$ are position states, $|k\rangle$ are momentum states, $V$ is volume and $N$ is number of particles. This is partof chapter 2.2 Classical Statistical Mechanics(pages 13-15), in a "derivation" connecting quantum partition function with the classical partition function. It seems to imply that the given expression somehow holds in the classical limit $\hbar \rightarrow 0$ but I am not seeing it. Is this a valid expression or some handwavy argument or some adhoc definition ?

The expression occurs in a derivation that connects the quantum expression for the partition function with a classical expression.

The derivation in the book goes as follows, $$\begin{aligned} \text{Tr}\exp(-\beta \hat H) &= \text{Tr}\left[\exp(-\beta \hat U)\exp(-\beta \hat T)\exp(-\beta \mathcal O([U,T]))\right]\\ \lim_{\hbar \rightarrow 0} \text{Tr}\exp(-\beta \hat H) &= \text{Tr}\left[\exp(-\beta \hat U)\exp(-\beta \hat T)\right]\\ \end{aligned}$$ This is called the classical limit and allows to pull kinetic and potential energy apart since all commutators, that are bundled in the term $\mathcal O([U,T])$, are in orders of $\hbar$ and vanish in the limit.

The next step is given as $$ \text{Tr}\exp(-\beta \hat H) =\sum_{r,k} \langle r|\exp(-\beta \hat U)|r\rangle \langle r|k\rangle \langle k|\exp(-\beta \hat T)|k\rangle\langle k|r\rangle \\ $$

Now we replace the quantum expressions by their classical analogs, including the expression that confuses me $\langle r|k\rangle \langle k|r\rangle = 1/V^N$ to obtain, $$\begin{aligned} \text{Tr}\exp(-\beta \hat H) &\approx \frac{1}{h^{dN}N!} \int dp^Ndr^N\ \exp(-\beta(U(r^N)+T(p^N)))\\ &\equiv Q_{\text{classical}} \end{aligned}$$

$U(r^N)$ is the classical potential energy as function of the N particle positions and $T(p^N)$ is the classical kinetic energy function of the N particle momenta.

Is the expression

$$ \langle r|k\rangle \langle k|r\rangle = 1/V^N $$

meaningful and if not, how does the proper derivation work ?

Answer 1

One way to justify that factor is to consider non-interacting particles in a cubic box with side length $L$. For simplicity let's assume periodic boundary conditions.

The energy eigenstates are also momentum eigenstates. Up to a normalization constant that in general could depend on $p$ (although it will turn out not to in this case), which we will call $\mathcal{N}_p$, the single-particle momentum eigenstates in the position basis are given by \begin{equation} \langle \vec{r} | \vec{p} \rangle = \mathcal{N}_{p} e^{i \vec{p} \cdot \vec{x}/\hbar} \end{equation} Because we are dealing with a box, only certain discrete values of the momentum are allowed, in order that the wavefunction is single-valued when any of the three Cartesian coordinates lying along edges of the box is increased by $L$. For instance, in the $x$ direction, we have \begin{equation} p_x = \frac{2\pi \hbar n_x}{L}, \ \ n_x=1, 2, 3, \cdots \end{equation} To fix the normalization constant, we normalize the states using a Kronecker delta normalization (the advantage of using a box is that we don't have to use a delta function normalization, which resolves some singular behavior). If we label the states by $n_x, n_y, n_z$ (in a hopefully obvious way), then we impose \begin{equation} \langle n_x, n_y, n_z | n_x', n_y', n_z' \rangle = \delta_{n_x, n_x'} \delta_{n_y, n_y'} \delta_{n_z, n_z'} \end{equation} Working through the math leads to the result \begin{equation} \mathcal{N}_p = \sqrt{\frac{1}{L^3}} = \sqrt{\frac{1}{V}} \end{equation} where $V=L^3$ is the volume of the cubic box.

Since the particles are non-interacting, an energy eigenbasis for an $N$ particle state is simply given by a product of $N$ one particle states, thus \begin{equation} \langle \vec{r}_1, \vec{r}_2 \cdots \vec{r}_N | \vec{p}_1 \vec{p}_2 \cdots \vec{p}_N \rangle = \left(\frac{1}{V}\right)^{N/2} \exp\left(i \sum_{a=1}^N \frac{\vec{p}_a \cdot \vec{r}_a}{\hbar}\right) \end{equation} The norm squared of this state is then \begin{equation} |\langle \vec{r}_1, \vec{r}_2 \cdots \vec{r}_N | \vec{p}_1 \vec{p}_2 \cdots \vec{p}_N \rangle|^2 = \frac{1}{V^N} \end{equation} which is the result your book quotes (with slightly more explicit notation on the left hand side).

More generally, this normalization (at least up to a dimensionless constant) follows on dimensional grounds. The normalization of the phase space distribution $p(\vec{r}_1, \cdots, \vec{r}_N, \vec{p}_1, \cdots, \vec{p}_N)$ has to have units of $1/V^N$. For non-interacting particles, physical volume the particles occupy is the only relevant volume scale.