Discrete Variable Representation

A discrete variable representation (DVR) converts a finite basis expansion into values associated with quadrature nodes. It combines a transformed kinetic operator with an approximately diagonal local potential. Its efficiency comes from a specific basis–quadrature relationship; an arbitrary grid with a diagonal potential is not automatically the DVR of a chosen variational basis.

Required background. Basis representations explains unitary changes of coordinates and overlap matrices. Helpful background. Spectral methods distinguishes projection from collocation, and Rayleigh–Ritz explains finite-basis energy estimates.

Quadrature defines the discrete transformation

Consider $N$ orthonormal functions $\phi_n(x)$ in the physical measure $dx$, with quadrature nodes $x_i$ and positive weights $w_i$ such that quadrature integrates every product $\phi_m^*\phi_n$ exactly:

$$\sum_{i=1}^N w_i\phi_m(x_i)^*\phi_n(x_i)=\delta_{mn}.$$

Define $U_{in}=\sqrt{w_i}\phi_n(x_i)$. Then $U^\dagger U=I$; since $U$ is square, $UU^\dagger=I$ as well. For a basis coefficient vector $c$, its DVR coordinate vector is

$$d=Uc,\qquad d_i=\sqrt{w_i}\,\psi_N(x_i),\qquad \|d\|^2=\|c\|^2.$$

The cardinal functions

$$\chi_i(x)=\sum_n\phi_n(x)U_{in}^*$$

satisfy $\chi_i(x_j)=\delta_{ij}/\sqrt{w_j}$ and are orthonormal in the exact finite-basis inner product. These identities are useful implementation tests. For an orthogonal-polynomial construction whose quadrature uses a weight $\rho(x)$, account for that weight in the functions or physical measure explicitly; do not silently reinterpret weighted quadrature as integration against $dx$.

Why a local potential becomes diagonal

The exact transformed potential is $UV^{\mathrm{basis}}U^\dagger$. Quadrature instead gives

$$\langle\chi_i|V|\chi_j\rangle \approx\sum_k w_k\chi_i(x_k)^*V(x_k)\chi_j(x_k) =V(x_i)\delta_{ij}.$$

Thus the common DVR Hamiltonian is

$$H_{\mathrm{DVR}}=UT^{\mathrm{basis}}U^\dagger +\operatorname{diag}(V(x_i)).$$

The kinetic transformation is exact within the selected basis if its original matrix elements are exact. The potential approximation introduces a separate quadrature error. Light, Hamilton, and Lill formalize this relationship between DVR and variational basis representations.

For real $V$, the displayed Hamiltonian remains Hermitian. This alone does not make its eigenvalues variational upper bounds: the quadrature-modified potential need not equal the projection of the physical operator. Restore exact potential matrix elements to recover ordinary Rayleigh–Ritz assumptions, or study quadrature convergence independently.

A two-node example that exposes the approximation

On $[-1,1]$, take $\phi_0=1/\sqrt2$ and $\phi_1=\sqrt{3/2}\,x$. Two-point Gauss–Legendre quadrature has $x_\pm=\pm1/\sqrt3$, $w_\pm=1$. Ordering the negative node first,

$$U=\frac1{\sqrt2} \begin{pmatrix}1&-1\\1&1\end{pmatrix}.$$

For $V=x$, the basis matrix is

$$V^{\mathrm{basis}}= \begin{pmatrix}0&1/\sqrt3\\1/\sqrt3&0\end{pmatrix}.$$

Its transform is exactly $\operatorname{diag}(-1/\sqrt3,1/\sqrt3)$. For $V=x^2$, however, the exact basis matrix is $\operatorname{diag}(1/3,3/5)$, whereas node sampling yields $(1/3)I$. The $\phi_1^*x^2\phi_1$ integrand has degree four, beyond the rule’s exactness through degree three. The discrepancy is not roundoff or an eigensolver defect.

This example concerns multiplication operators. The two polynomials alone do not specify a self-adjoint Schrödinger boundary-value problem. Choose kinetic basis functions and boundary conditions together when constructing an actual Hamiltonian.

Convergence and multidimensional cost

Increase basis size and compare physical observables, coefficient tails, and selected exact or higher-order potential matrix elements. Singular potentials and nonsmooth wavefunctions can converge slowly even when the cardinal identities hold to machine precision. Coordinate mappings require transformed kinetic energy and Jacobian factors, as described under coordinate transformations.

A direct product of $d$ one-dimensional DVRs with $N$ nodes each has $N^d$ coefficients. A diagonal potential reduces application cost but does not remove exponential storage growth. Exploit separability, symmetry, sparse representations, or low-rank states only when their approximation errors are controlled. The tensor-network chapter discusses one possible compression strategy.

Exercises

For $V=x^2$ in the two-node example, calculate the exact potential matrix in DVR coordinates and compare it with the diagonal approximation.

Solution

Multiplying $U\operatorname{diag}(1/3,3/5)U^\dagger$ gives diagonal entries $7/15$ and off-diagonal entries $-2/15$. The sampled potential is $(1/3)I$. Their difference has eigenvalues $0$ and $4/15$, displaying a finite quadrature error despite exact orthogonality of $U$.

Why is using $d_i=\psi(x_i)$ instead of $\sqrt{w_i}\psi(x_i)$ generally incorrect?

Solution

An unweighted Euclidean sum of samples does not represent the physical norm when quadrature weights vary. It also changes the transformation’s orthogonality and matrix adjoint. One can retain raw samples only by consistently retaining their weight metric in all inner products and operator relations.

References

• Light, J. C., I. P. Hamilton, and J. V. Lill, “Generalized Discrete Variable Approximation in Quantum Mechanics,” Journal of Chemical Physics 82, 1400–1409 (1985). DOI.
• Tannor, David J. Introduction to Quantum Mechanics: A Time-Dependent Perspective. University Science Books, 2007.