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Instantiation Cost Functional — Coordinate Computations

Jean-Paul Niko · Sole Author

Purpose

The CS Operator Theory defines \(\mathcal{E}(\psi) = \langle \psi, (I - C^*C)\psi \rangle\) abstractly. This page computes it explicitly for four systems: harmonic oscillator, hydrogen atom, free scalar field, and Yang-Mills vacuum. The goal: turn the abstract cost into numbers that connect to experiment or to the Millennium Problems.

1. The Cost Functional (Review)

For a state \(\psi \in \mathcal{H}_Q\), the instantiation cost is:

\[\boxed{\mathcal{E}(\psi) = \|\psi\|_Q^2 - \|C\psi\|_P^2 = \langle \psi, (I - C^*C)\psi \rangle}\]

In the SVD basis \(C\psi_n = \sigma_n \phi_n\):

\[\mathcal{E}(\psi) = \sum_n (1 - \sigma_n^2) |\langle \psi, \psi_n \rangle|^2\]
  • \(\sigma_n = 1\): zero cost (perfect instantiation)
  • \(\sigma_n = 0\): maximum cost (mode is in \(\ker(C)\) — dark sector)
  • \(0 < \sigma_n < 1\): partial cost (lossy projection)

2. Harmonic Oscillator

2.1 Setup

The quantum harmonic oscillator with Hamiltonian \(H = \frac{1}{2}(p^2 + \omega^2 x^2)\) in natural units (\(\hbar = m = 1\)).

Energy eigenstates: \(|n\rangle\) with \(E_n = \omega(n + \frac{1}{2})\).

2.2 RTSG Identification

The QS states are the full Hilbert space \(\mathcal{H}_Q = L^2(\mathbb{R})\). The PS states are classical phase-space distributions — Wigner functions that are non-negative (classical limit).

The instantiation operator \(C\) maps quantum states to their classical shadows. The natural candidate: the Husimi projection (coherent-state representation):

\[C|n\rangle = Q_n(x, p) = |\langle \alpha | n \rangle|^2, \qquad \alpha = \frac{x + ip}{\sqrt{2}}\]

The Husimi function \(Q_n\) is a smoothed Wigner function — always non-negative, always classical. The smoothing loses quantum information.

2.3 Singular Values

The overlap between the \(n\)-th energy eigenstate and a coherent state \(|\alpha\rangle\) is:

\[\langle \alpha | n \rangle = \frac{\alpha^n}{\sqrt{n!}} e^{-|\alpha|^2/2}\]

The norm of the Husimi function:

\[\|Q_n\|^2 = \int |Q_n(x,p)|^2 \, \frac{dx\,dp}{2\pi} = \frac{1}{2^n \cdot n! \cdot \sqrt{\pi}} \cdot \frac{(2n)!}{2^n \cdot n!}\]

For the ground state (\(n=0\)): \(\|Q_0\|^2 = \frac{1}{\sqrt{\pi}} \int e^{-2|\alpha|^2} d^2\alpha = \frac{1}{2}\).

Since \(\||0\rangle\|^2 = 1\), the singular value for the ground state is:

\[\sigma_0^2 = \|C|0\rangle\|^2 / \||0\rangle\|^2 = \frac{1}{2}\]
\[\boxed{\sigma_0 = \frac{1}{\sqrt{2}} \approx 0.707}\]

2.4 Cost for Each Level

For the \(n\)-th level, the singular value (from the Husimi \(L^2\) norm) scales as:

\[\sigma_n^2 \sim \frac{1}{\sqrt{4\pi n}} \quad \text{for large } n \quad (\text{Stirling})\]

So the instantiation cost grows with excitation:

\[\mathcal{E}(|n\rangle) = 1 - \sigma_n^2 \approx 1 - \frac{1}{\sqrt{4\pi n}}\]
State \(\sigma_n^2\) \(\mathcal{E}\) Interpretation
\(\|0\rangle\) (vacuum) 0.500 0.500 Half the quantum info survives classical projection
\(\|1\rangle\) 0.375 0.625 First excitation: more quantum, higher cost
\(\|10\rangle\) \(\sim 0.089\) \(\sim 0.911\) Highly excited: mostly quantum, expensive to instantiate
\(\|n \to \infty\rangle\) \(\to 0\) \(\to 1\) Fully quantum: instantiation cost approaches maximum

Physical meaning: Excited states are "more quantum" and cost more to instantiate into the classical world. The ground state is the cheapest to instantiate — it has the highest singular value. This is the harmonic oscillator's version of "gravity is Stage 0" — the ground state (vacuum, minimal complexity) instantiates with maximum efficiency.

3. Hydrogen Atom

3.1 Setup

Bound states \(|n, \ell, m\rangle\) with \(E_n = -13.6/n^2\) eV.

3.2 RTSG Identification

\(C\) maps the quantum wave function \(\psi_{n\ell m}(\mathbf{r})\) to its classical counterpart — the probability density \(|\psi_{n\ell m}|^2\) interpreted as a classical charge distribution.

The loss: quantum phase information \(\arg(\psi)\) is destroyed. Coherences between different \((\ell, m)\) components are lost.

3.3 Singular Values by Quantum Numbers

The singular value depends on how "classical" the orbital is:

\[\sigma_{n\ell m}^2 \propto \frac{\text{classical content of } |n\ell m\rangle}{\text{total content}}\]

For circular orbits (\(\ell = n-1\), maximum angular momentum): the Bohr correspondence says these are the most classical. The wave function concentrates on a thin shell at \(r = n^2 a_0\), resembling a classical orbit.

\[\sigma_{n, n-1, m}^2 \sim 1 - O(1/n) \quad \text{(approaches 1 for large } n\text{)}\]

For \(s\)-orbitals (\(\ell = 0\)): the wave function has a node at the origin, penetrates the nucleus, has no classical analogue.

\[\sigma_{n, 0, 0}^2 \sim 1/n \quad \text{(drops to 0 for large } n\text{)}\]
State Classical analogue \(\sigma^2\) (estimate) \(\mathcal{E}\)
\(1s\) None (fully quantum) \(\sim 0.3\) \(\sim 0.7\)
\(2p\) (\(\ell=1\)) Elliptical orbit \(\sim 0.5\) \(\sim 0.5\)
\(n, n-1\) (circular) Bohr orbit \(\to 1\) \(\to 0\)
Rydberg (\(n \gg 1\), \(\ell = 0\)) None \(\to 0\) \(\to 1\)

The Bohr correspondence as an instantiation statement: Circular Rydberg orbits have \(\sigma \to 1\) — they instantiate almost perfectly. The \(1s\) ground state, despite being lowest energy, is deeply quantum (\(\sigma \sim 0.5\)). Energy and instantiation cost are not the same — the most quantum state is not the most energetic.

4. Free Scalar Field (QFT)

4.1 Setup

Free massive scalar field \(\phi\) with Hamiltonian \(H = \int d^3k \, \omega_k (a_k^\dagger a_k + \frac{1}{2})\), \(\omega_k = \sqrt{k^2 + m^2}\).

4.2 RTSG Identification

\(\mathcal{H}_Q\) = Fock space. \(\mathcal{H}_P\) = classical field configurations. \(C\) maps quantum field states to their classical expectation values:

\[C |\psi\rangle = \langle \psi | \hat{\phi}(x) | \psi \rangle = \phi_{\text{cl}}(x)\]

The vacuum \(|0\rangle\) maps to \(\phi_{\text{cl}} = 0\). A coherent state \(|\alpha_k\rangle\) maps to \(\phi_{\text{cl}}(x) = \int \alpha_k e^{ikx} d^3k\) — a classical wave.

4.3 Cost per Mode

Each mode \(k\) is an independent harmonic oscillator. From §2:

\[\sigma_k^2(\text{vacuum}) = \frac{1}{2} \quad \text{for each mode}\]

The total vacuum cost (summed over modes) diverges:

\[\mathcal{E}(\text{vacuum}) = \sum_k (1 - \sigma_k^2) = \sum_k \frac{1}{2} = \infty\]

This is the UV divergence of vacuum instantiation cost. In QFT language: the vacuum has infinite zero-point energy. In RTSG language: it costs infinite energy to instantiate the full vacuum because there are infinitely many quantum modes that each lose half their content upon classical projection.

4.4 Renormalization as Instantiation Regularization

Renormalization subtracts the vacuum cost: \(\mathcal{E}_{\text{ren}}(\psi) = \mathcal{E}(\psi) - \mathcal{E}(|0\rangle)\).

\[\mathcal{E}_{\text{ren}}(|1_k\rangle) = (1 - \sigma_{k,1}^2) - (1 - \sigma_{k,0}^2) = \sigma_{k,0}^2 - \sigma_{k,1}^2 = \frac{1}{2} - \frac{3}{8} = \frac{1}{8}\]

RTSG reading of renormalization: We don't measure absolute instantiation cost; we measure relative cost against the vacuum. Renormalization is not a trick — it's the statement that PS only detects differences in instantiation cost, not the absolute cost.

⚠ Connection to Hilbert-Schmidt conjecture (CS Operator Theory §5.1): the divergence of \(\sum_k (1 - \sigma_k^2)\) suggests \(C\) is not Hilbert-Schmidt for a free field. The Hilbert-Schmidt conjecture may need a renormalized version: \((I - C^*C)_{\text{ren}}\) is trace-class after vacuum subtraction.

5. Yang-Mills Vacuum

5.1 Setup

\(SU(N)\) Yang-Mills theory on a lattice \(\Lambda\) with spacing \(a\), coupling \(g\), and Wilson action \(S_W = \frac{2N}{g^2} \sum_{\square} (1 - \frac{1}{N} \text{Re Tr } U_\square)\).

5.2 RTSG Identification

\(\mathcal{H}_Q\) = gauge-invariant wave functionals \(\Psi[A]\). \(\mathcal{H}_P\) = classical gauge configurations (solutions to the YM equations of motion). \(C\) projects quantum gauge field onto classical background:

\[C|\Psi\rangle = A_{\text{cl}}^{\mu} = \langle \Psi | \hat{A}^{\mu} | \Psi \rangle\]

5.3 The Mass Gap from the Cost Functional

The YM vacuum \(|0_{\text{YM}}\rangle\) has \(C|0_{\text{YM}}\rangle = 0\) (the classical vacuum is zero field). The first excited state \(|1_{\text{YM}}\rangle\) is a glueball with mass \(\Delta_{\text{YM}}\).

The cost difference between first excited state and vacuum:

\[\mathcal{E}_{\text{ren}}(|1_{\text{YM}}\rangle) = \sigma_{0}^2 - \sigma_{1}^2\]

In the GL framework, the action near the minimum is:

\[S[W] \approx S[W_0] + \int \left[|\partial \delta W|^2 + 2\alpha |\delta W|^2\right] d\mu\]

The linearized operator \(L = -\nabla^2 + 2\alpha\) has spectral gap \(2\alpha\). If \(L\) is the linearization of \(I - C^*C\) around the vacuum:

\[\sigma_0^2 - \sigma_1^2 = 2\alpha\]
\[\boxed{\Delta_{\text{YM}} = \sqrt{2\alpha} = \sqrt{\sigma_0^2 - \sigma_1^2}}\]

The mass gap is the square root of the gap in the instantiation cost spectrum.

5.4 Numerical Estimate

From the engine's lattice data (L=12, β=2.5):

  • Mass gap \(\Delta \approx 0.37\) (lattice units)
  • String tension \(\sigma_{\text{str}} \approx 0.045\)

If \(\Delta = \sqrt{2\alpha}\), then \(\alpha \approx 0.068\) in lattice units.

The GL parameter \(\alpha\) is the curvature of the instantiation potential at its minimum. It is positive whenever the vacuum is stable — which is the existence of the mass gap. The magnitude of \(\alpha\) is determined by the non-perturbative dynamics of confinement.

⚠ Conjecture. The identification $L = $ linearization of \(I - C^*C\) requires proving that the GL action is the effective action for the CS operator restricted to the gauge sector. This is YM Task 2 assigned to @D_GPT.

6. Summary Table

System Ground state \(\sigma_0^2\) First excited \(\sigma_1^2\) Renorm. gap Physical meaning
Harmonic oscillator \(1/2\) \(3/8\) \(1/8\) Quantum → classical smoothing
Hydrogen (\(1s\)) \(\sim 0.3\) \(\sim 0.5\) (\(2p\)) \(\sim -0.2\) (!) \(s\)-orbitals more quantum than \(p\)
Free scalar (per mode) \(1/2\) \(3/8\) \(1/8\) Same as HO; UV divergence in total
Yang-Mills vacuum \(\sigma_0^2\) \(\sigma_0^2 - 2\alpha\) \(2\alpha\) Mass gap = cost gap

Note on hydrogen: \(\sigma_{2p} > \sigma_{1s}\) — the first excited state is more classical than the ground state! The \(2p\) orbital has a classical analogue (elliptical orbit); the \(1s\) does not. This shows that energy ordering and instantiation ordering are different. The cheapest-to-instantiate state is not always the lowest-energy state.

7. Connection to Thermodynamics

The total instantiation cost for a thermal ensemble at temperature \(T\):

\[\mathcal{E}(T) = \sum_n (1 - \sigma_n^2) \, p_n(T), \qquad p_n = \frac{e^{-E_n/T}}{Z(T)}\]

At high \(T\): all modes populated equally, \(\mathcal{E} \to 1 - \overline{\sigma^2}\) (average cost). At low \(T\): only ground state, \(\mathcal{E} \to 1 - \sigma_0^2\) (minimum cost).

The third law of instantiation: As \(T \to 0\), the system approaches the state of minimum instantiation cost. This is the RTSG reading of the third law of thermodynamics — the ground state is the most efficiently instantiable state.