Backflow transformation for Gaussian orbitals

[Vladimir_Konjkov]

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.

[Vladimir_Konjkov]

ChatGPT-5 told me that this is a consequence of the type of function defined in eq 20 In article "Fermion Nodes" "Journal o f Statistical Physics, Vol. 63, Nos. 5/6, 1991", DOI :10.1007/BF01030009

Screenshot_20250826_184834.png (37.91 KiB) Viewed 73 times

a plane wave is a complex exponential, so the formula also works

[Vladimir_Konjkov]

I can't understand what Ceperly wanted to say. So if I start with the unsymmetrical version of the pair product trial function.

I get a slightly different formula.

1.png (42.33 KiB) Viewed 44 times

Substitute the value of the expressions into the discrete step of the generalized Feynman-Kac / DMC for the trial function with corrections
Ψ(R, t + dt) = Ψ(R, t) * exp[-dt * (EL(R) - EL_ΨT)]

2.png (77.39 KiB) Viewed 44 times

3.png (95.71 KiB) Viewed 44 times

this is a very interesting result, it shows that the whole function should be backflow transformed.
It also became clear why the claim that the inverse transformation should not be included in the Jastrow factor is false.
viewtopic.php?f=3&t=227

I could give a link to the GPT chat, but the problem statement there is in russian.

Since ξi, or more precisely dξi depends on t, it is possible to integrate.

Best, Vladimir.

[Vladimir_Konjkov]

and the equation

4.png (4.89 KiB) Viewed 35 times

has a stationary point at which the optimization of ξi will converge and this is obvious flattened (corrected) trial function.