Discrete Variable Representation (Fourier DVR)

These notes outline the construction of a discrete variable representation (DVR) for a one-dimensional quantum particle in a box with periodic boundary conditions, using a Fourier (plane-wave) primitive basis. The goal is to obtain a representation in which the potential energy operator is diagonal while the kinetic energy retains a simple, analytic form.

I. The 1D Hamiltonian

Consider a particle of mass $m$ confined to the interval $x \in [0,L]$ with periodic boundary conditions. The Hamiltonian is

\[ \hat{H} = \hat{T} + \hat{V} = \frac{\hat{p}^2}{2m} + V(\hat{x}), \tag{1} \]

with the canonical commutation relation $[\hat{x},\hat{p}] = i\hbar$. The DVR aims to construct a basis in which $V(\hat{x})$ is diagonal, while $\hat{T}$ has a convenient analytic matrix representation.

II. Choice of primitive basis: Fourier (plane-wave) states

We choose $N$ orthonormal plane-wave basis functions

\[ \phi_n(x) = \langle x \vert \phi_n \rangle = \frac{1}{\sqrt{L}} e^{i k_n x}, \qquad n = 1,2,\dots,N, \tag{2} \]

where the wavevectors are

\[ k_n = \frac{2\pi}{L} \left( n - \frac{N+1}{2} \right). \tag{3} \]

The truncated Hilbert space is then

\[ \mathcal{H}_N = \mathrm{span}\{\, \vert \phi_1 \rangle, \dots, \vert \phi_N \rangle \,\}. \]

III. Kinetic energy in the primitive basis

In this plane-wave basis, momentum is diagonal:

\[ \hat{p} \vert \phi_n \rangle = \hbar k_n \vert \phi_n \rangle, \tag{4} \]

so the kinetic energy matrix elements are

\[ T_{nm} = \langle \phi_n \vert \hat{T} \vert \phi_m \rangle = \frac{\hbar^2 k_n^2}{2m} \,\delta_{nm}. \tag{5} \]

Thus the kinetic energy is diagonal in the primitive basis,

\[ T = \mathrm{diag} \left( \frac{\hbar^2 k_1^2}{2m}, \dots, \frac{\hbar^2 k_N^2}{2m} \right). \]

IV. Projected position operator

Define the projector onto the truncated space

\[ \hat{P} = \sum_{n=1}^N \vert \phi_n \rangle \langle \phi_n \vert, \]

and the projected position operator

\[ \hat{X} = \hat{P}\,\hat{x}\,\hat{P}. \]

Its matrix elements in the primitive basis are

\[ \begin{aligned} X_{nm} &= \langle \phi_n \vert \hat{x} \vert \phi_m \rangle = \frac{1}{L} \int_0^L x\, e^{-i k_n x} e^{i k_m x}\,dx \\ &= \frac{1}{L} \int_0^L x\, e^{i (k_m - k_n)x}\,dx, \end{aligned} \tag{6} \]

where we define $\Delta k_{nm} = k_m - k_n$.

a. Diagonal elements ($n = m$)

\[ X_{nn} = \frac{1}{L} \int_0^L x\,dx = \frac{L}{2}. \tag{7} \]

b. Off-diagonal elements ($n \ne m$)

\[ \begin{aligned} X_{nm} &= \frac{1}{L} \int_0^L x\, e^{i \Delta k_{nm} x}\,dx \\ &= \frac{1}{L} \left[ e^{i \Delta k L} \left( \frac{L}{i\Delta k} - \frac{1}{(\Delta k)^2} \right) + \frac{1}{(\Delta k)^2} \right]. \end{aligned} \tag{8–9} \]

This defines the full Hermitian matrix $X$ in the primitive Fourier basis.

V. Diagonalization of the projected position operator

Since $X$ is Hermitian, it has $N$ real eigenvalues and orthonormal eigenvectors:

\[ X U = U\,x_\mathrm{diag}, \tag{10} \]

where $x_\mathrm{diag} = \mathrm{diag}(x_1,x_2,\dots,x_N)$ and the columns of $U$ are the normalized eigenvectors.

For the Fourier basis, the eigenvalues are known analytically:

\[ x_\alpha = \frac{L}{N}(\alpha - 1), \qquad \alpha = 1,\dots,N. \tag{11} \]

These $x_\alpha$ are equally spaced grid points on $[0,L)$.

The corresponding eigenvectors are given by the discrete Fourier transform (DFT):

\[ U_{n\alpha} = \frac{1}{\sqrt{N}} \exp(i k_n x_\alpha) = \frac{1}{\sqrt{N}} \exp\left[ i\frac{2\pi}{L} \left( n - \frac{N+1}{2} \right) x_\alpha \right]. \tag{12} \]

VI. Definition of DVR basis states

The DVR basis $\{\vert \chi_\alpha \rangle\}$ is defined by the unitary transformation

\[ \vert \chi_\alpha \rangle = \sum_{n=1}^N \vert \phi_n \rangle U_{n\alpha}, \qquad \alpha = 1,\dots,N. \tag{13} \]

Within the truncated space,

\[ \hat{X} \vert \chi_\alpha \rangle = x_\alpha \vert \chi_\alpha \rangle, \]

so the DVR basis diagonalizes the projected position operator.

VII. Potential operator in DVR (diagonal form)

The potential operator is a function of $\hat{x}$:

\[ \hat{V} = V(\hat{x}). \]

In the DVR basis, its matrix elements are

\[ V^{(\mathrm{DVR})}_{\alpha\beta} = \langle \chi_\alpha \vert \hat{V} \vert \chi_\beta \rangle \approx V(x_\alpha)\,\delta_{\alpha\beta}. \tag{14} \]

Thus the potential energy is simply

\[ V^{(\mathrm{DVR})} = \mathrm{diag}\big( V(x_1),\dots,V(x_N) \big). \tag{15} \]

In other words, the potential is diagonal and obtained by sampling $V(x)$ at the DVR grid points $\{x_\alpha\}$.

VIII. Kinetic energy operator in DVR

The kinetic operator in the DVR basis is obtained via the unitary transform

\[ \begin{aligned} T^{(\mathrm{DVR})}_{\alpha\beta} &= \sum_{n,m} U^\ast_{n\alpha} T_{nm} U_{m\beta} = (U^\dagger T U)_{\alpha\beta}. \end{aligned} \tag{16} \]

Since $T$ is diagonal in the plane-wave basis, $T_{nm} = \frac{\hbar^2 k_n^2}{2m}\delta_{nm}$, one can derive a closed-form expression for the Fourier DVR kinetic operator:

\[ T^{(\mathrm{DVR})}_{\alpha\beta} = \begin{cases} \displaystyle \frac{\hbar^2}{2m} \frac{\pi^2}{3L^2} \big(N^2 - 1\big), & \alpha = \beta, \\[1.2em] \displaystyle \frac{\hbar^2}{2m} \frac{(-1)^{\alpha-\beta}}{L^2} \frac{\pi^2}{\sin^2\!\left(\dfrac{\pi}{N}(\alpha-\beta)\right)}, & \alpha \neq \beta. \end{cases} \tag{17} \]

IX. Full Hamiltonian in the DVR basis

Combining kinetic and potential contributions, the DVR Hamiltonian matrix is

\[ H^{(\mathrm{DVR})}_{\alpha\beta} = T^{(\mathrm{DVR})}_{\alpha\beta} + V(x_\alpha)\,\delta_{\alpha\beta}. \tag{18} \]

The time-independent Schrödinger equation becomes the matrix eigenvalue problem

\[ \sum_{\beta=1}^N H^{(\mathrm{DVR})}_{\alpha\beta} C^{(k)}_\beta = E_k C^{(k)}_\alpha, \tag{19} \]

where the $E_k$ approximate the eigenenergies of the Hamiltonian, and

\[ \psi_k(x_\alpha) \approx C^{(k)}_\alpha \tag{20} \]

is the approximate wavefunction evaluated at the DVR grid points.

X. Summary of Fourier DVR procedure

The Fourier DVR construction can be summarized as follows:

  1. Choose the plane-wave basis $\phi_n(x)$ and compute the wavevectors $k_n$ for $n=1,\dots,N$.
  2. Construct the projected position matrix $X_{nm} = \langle \phi_n \vert \hat{x} \vert \phi_m \rangle$ and diagonalize it.
  3. The eigenvalues give the DVR grid points $x_\alpha = \dfrac{L}{N}(\alpha - 1)$.
  4. Form the DVR kinetic energy matrix using the analytic formula in Eq. (17).
  5. Form the DVR potential energy by evaluating $V(x_\alpha)$ at each grid point.
  6. Diagonalize the resulting DVR Hamiltonian $H^{(\mathrm{DVR})}$ to obtain eigenvalues $E_k$ and eigenvectors $C^{(k)}_\alpha$, which approximate the energies and wavefunctions of the 1D Hamiltonian.

This yields an efficient and systematically convergent numerical method for solving one-dimensional quantum Hamiltonians with arbitrary potentials using a Fourier-based DVR.

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