Backflow transformation for Gaussian orbitals
Posted: Tue Aug 26, 2025 7:51 am
Hello QMC experts.
In his article "Fermion Nodes" "Journal o f Statistical Physics, Vol. 63, Nos. 5/6, 1991", DOI :10.1007/BF01030009 D. M. Ceperley make a crude approximation of correcting the trial function by the local energy itself:
ψ₂(R) = Aψ₁(R)exp(-𝜏E₁(R)) (eq. 19)
This leads to a new Slater matrix with elements:
Mᵢⱼ = φᵢ(rⱼ)exp[∇ⱼln(φᵢ(rⱼ))•∇ⱼU(R)] (eq. 20)
Ceperley notes that ψ₂(R) contains two new terms:
1. A backflow term, which is the correction inside the determinant and affects the nodes.
2. A three-body boson-like correction, which does not affect the nodes.
Considering the case of free particles under periodic boundary conditions, where the single-particle orbitals are plane waves exp(ik•r), the result implies that instead of the bare coordinates rᵢ, one should use dressed "quasiparticle" coordinates:
xᵢ = rᵢ + ∇ᵢU(R) (eq. 21)
Introducing the notation ∇ᵢU(R) = ξᵢ where U(R) is an effective potential (symmetric with respect to allowed permutations) to be optimized, this can be viewed as a general mapping from real coordinates to quasiparticle coordinates. A simple one-body wave function is then constructed from these quasiparticle coordinates. However, atomic calculations typically employ Gaussian orbitals rather than plane waves. The question is, how does this formulation change in that case?
Consider a primitive Gaussian orbital:
φ(r) = x^l * y^m * z^n * exp(-αr²) = exp(l*ln(x) + m*ln(y) + n*ln(z) - αr²)
The matrix element then becomes:
Mᵢⱼ = φᵢ(rⱼ) * exp[∇ⱼln(φᵢ(rⱼ))•ξⱼ] = φᵢ(rⱼ)*exp[(-2αrⱼ + l/xⱼ + m/yⱼ + n/zⱼ)•ξⱼ]
We can incorporate φᵢ(rⱼ) into the exponent:
Mᵢⱼ = exp[l*ln(xⱼ) + m*ln(yⱼ) + n*ln(xⱼ) + (-αrⱼ²-2αrⱼ + l/xⱼ + m/yⱼ + n/zⱼ)•ξⱼ]
Using the approximations:
ln(x + ξ) ≈ ln(x) + ξ/x
-αrⱼ²-2αrⱼξⱼ = -α|rⱼ + ξⱼ| + α|ξⱼ|²
the expression can be rewritten as:
Mᵢⱼ = (xⱼ + ξⱼ(x))^l(yⱼ + ξⱼ(y))^m(zⱼ + ξⱼ(z))^n * exp(-α|rⱼ + ξⱼ|²) * exp(α|ξⱼ|²)
Or, more compactly:
Mᵢⱼ = φᵢ(rⱼ + ξⱼ) * exp(α|ξⱼ|²)
This result is not equivalent to the simple coordinate shift obtained in the plane-wave case due to the presence of the extra exponential factor exp(α|ξⱼ|²). Perhaps it is worth considering the case of Slater orbitals?
Best Vladimir.
In his article "Fermion Nodes" "Journal o f Statistical Physics, Vol. 63, Nos. 5/6, 1991", DOI :10.1007/BF01030009 D. M. Ceperley make a crude approximation of correcting the trial function by the local energy itself:
ψ₂(R) = Aψ₁(R)exp(-𝜏E₁(R)) (eq. 19)
This leads to a new Slater matrix with elements:
Mᵢⱼ = φᵢ(rⱼ)exp[∇ⱼln(φᵢ(rⱼ))•∇ⱼU(R)] (eq. 20)
Ceperley notes that ψ₂(R) contains two new terms:
1. A backflow term, which is the correction inside the determinant and affects the nodes.
2. A three-body boson-like correction, which does not affect the nodes.
Considering the case of free particles under periodic boundary conditions, where the single-particle orbitals are plane waves exp(ik•r), the result implies that instead of the bare coordinates rᵢ, one should use dressed "quasiparticle" coordinates:
xᵢ = rᵢ + ∇ᵢU(R) (eq. 21)
Introducing the notation ∇ᵢU(R) = ξᵢ where U(R) is an effective potential (symmetric with respect to allowed permutations) to be optimized, this can be viewed as a general mapping from real coordinates to quasiparticle coordinates. A simple one-body wave function is then constructed from these quasiparticle coordinates. However, atomic calculations typically employ Gaussian orbitals rather than plane waves. The question is, how does this formulation change in that case?
Consider a primitive Gaussian orbital:
φ(r) = x^l * y^m * z^n * exp(-αr²) = exp(l*ln(x) + m*ln(y) + n*ln(z) - αr²)
The matrix element then becomes:
Mᵢⱼ = φᵢ(rⱼ) * exp[∇ⱼln(φᵢ(rⱼ))•ξⱼ] = φᵢ(rⱼ)*exp[(-2αrⱼ + l/xⱼ + m/yⱼ + n/zⱼ)•ξⱼ]
We can incorporate φᵢ(rⱼ) into the exponent:
Mᵢⱼ = exp[l*ln(xⱼ) + m*ln(yⱼ) + n*ln(xⱼ) + (-αrⱼ²-2αrⱼ + l/xⱼ + m/yⱼ + n/zⱼ)•ξⱼ]
Using the approximations:
ln(x + ξ) ≈ ln(x) + ξ/x
-αrⱼ²-2αrⱼξⱼ = -α|rⱼ + ξⱼ| + α|ξⱼ|²
the expression can be rewritten as:
Mᵢⱼ = (xⱼ + ξⱼ(x))^l(yⱼ + ξⱼ(y))^m(zⱼ + ξⱼ(z))^n * exp(-α|rⱼ + ξⱼ|²) * exp(α|ξⱼ|²)
Or, more compactly:
Mᵢⱼ = φᵢ(rⱼ + ξⱼ) * exp(α|ξⱼ|²)
This result is not equivalent to the simple coordinate shift obtained in the plane-wave case due to the presence of the extra exponential factor exp(α|ξⱼ|²). Perhaps it is worth considering the case of Slater orbitals?
Best Vladimir.