The Modern Mathematics of Deep Learning

Hamiltonian potential terms

Penguin learning project

The project uses 342 complete Palmer penguin records: 240 training-pool and 102 held-out rows. The selected training rows determine feature scaling and model parameters. Source CSV

This project trains weights with softmax, unlike the homepage lab, which uses nearest species averages. Same penguin measurements; different classification methods. Predictions can differ.

Quantum models · Describe wavefunctions, particle coordinates and energy relations.

PDE and quantum models are not operations on these penguin measurements.

Consider this concept alongside the project; its source assumptions may differ. These observations do not prove a theorem.

Input: 12 × 2; Held-out evaluation: 102 · #58 · 40.6 mm, 193 mm. Known answer: Adélie; Prediction: Adélie · [≈0.7383, ≈0.2444, ≈0.01737].

Used in these source sections: 8.2 Equations as guides

At nonzero separations, electron–nucleus attraction is negative and electron–electron and nucleus–nucleus terms are positive.

The Hamiltonian combines kinetic and Coulomb interaction contributions.

Source formula example

(\mathcal{H}_{R}\psi)(r)=-\sum_{i=1}^{n}\frac{1}{2}(\Delta_{r_{i}}\psi)(r)-\left(\sum_{i=1}^{n}\sum_{j=1}^{p}\frac{Z_{j}}{\|{r}_{i}-{R}_{j}\|_{2}}-\sum_{i=1}^{p-1}\sum_{j=i+1}^{p}\frac{Z_{i}Z_{j}}{\|{R}_{i}-{R}_{j}\|_{2}}-\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}\frac{1}{\|{r}_{i}-{r}_{j}\|_{2}}\right)\psi(r)

illustration · This evaluates all terms of the operator on one invented wavefunction and configuration; it is not a quantum eigenvalue solver.

Symbols

$rᵢ$: Position of electron i in ℝ³.
$Rⱼ$: Position of nucleus j in ℝ³.
$Zⱼ$: Nuclear charge.
$Δᵣᵢ$: Laplacian in electron i coordinates.
$ψ$: Wavefunction.
$n,p$: Electron and nucleus counts.

Source formula example

This source example uses its own inputs and assumptions. Editing it does not change the project.

Edit a value or the code, then run the example. This is a finite demonstration, not a proof of the paper’s theorem.

Both nuclear charges

This example’s data flow

You choose

input.charge = 1

Initial values in code

electronElectron = 1
laplacianRatio = [-2, 10]

Calculated here

Result expression: { psi, electronNucleus, nucleusNucleus, electronElectron, kineticRatio, potentialRatio, hPsi: (kineticRatio + potentialRatio) * psi, }

q = input.charge Uses: input.charge
psi = Math.exp(-5)
electronNucleus = q * (1 + 1 / 3 + 1 / 2 + 1 / 2) Uses: q
nucleusNucleus = (q * q) / 4 Uses: q
kineticRatio = -0.5 * laplacianRatio.reduce((a, b) => a + b, 0) Uses: laplacianRatio
potentialRatio = -electronNucleus + nucleusNucleus + electronElectron Uses: electronNucleus, nucleusNucleus, electronElectron

Returns

Example result

psi: ≈0.006738
electronNucleus: ≈2.333
nucleusNucleus: 0.25
electronElectron: 1
kineticRatio: -4
potentialRatio: ≈-1.083
hPsi: ≈-0.03425

JavaScript

Read numbers from input and return the result. JavaScript basics.

Result

Enable JavaScript to run this example.

Technical statement

Use two electrons at (1,0,0) and (2,0,0), two nuclei at (0,0,0) and (4,0,0), and ψ(r) = exp(−|r₁|² − |r₂|²). In three dimensions, Δexp(−|r|²)/exp(−|r|²) = 4|r|² − 6. Every electron–nucleus, nucleus–nucleus, and electron–electron distance here is nonzero. The source operator is singular at collisions.

Source: Source expression