Soit E 2 ∈ R n {\displaystyle E^{2}\in \mathbb {R} ^{n}} une sous-variété différentielle munie d'une métrique sur R n {\displaystyle \mathbb {R} ^{n}}
E = ∮ Ω 2 ∂ 2 Ψ ∂ ω 2 μ 0 G ν μ 3 4 π ϵ 0 d ω {\displaystyle E=\oint _{\Omega ^{2}}{\frac {\partial ^{2}\Psi }{\partial \omega ^{2}}}{\frac {\mu _{0}G_{\nu \mu }^{3}}{4\pi \epsilon _{0}}}d\omega }
i ℏ E ( t ) = ∑ k = 1 ∞ k ∇ 2 Ψ ( x , t + k ) {\displaystyle i\hbar E(t)=\sum _{k=1}^{\infty }k\nabla ^{2}\Psi (x,t+k)}
− ∇ p 2 d ˙ l ∂ G 3 = | Ψ ∗ ⟩ {\displaystyle -\nabla {\frac {p^{2}{\dot {d}}l}{\partial G^{3}}}=|\Psi *\rangle }
Il vient trivialement que :
∮ Ω 2 i ℏ E ( t ) ∂ 2 Ψ ∂ ω 2 μ 0 G ν μ 3 4 π ϵ 0 d ω + | Ψ ∗ ⟩ = − ∇ p 2 d ˙ l ∂ G 3 Λ ν {\displaystyle \oint _{\Omega ^{2}}i\hbar E(t){\frac {\partial ^{2}\Psi }{\partial \omega ^{2}}}{\frac {\mu _{0}G_{\nu \mu }^{3}}{4\pi \epsilon _{0}}}d\omega +|\Psi *\rangle =-\nabla {\frac {p^{2}{\dot {d}}l}{\partial G^{3}}}\Lambda _{\nu }}
Ce qui conclut la preuve. (Quelle PREUVE ?)