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