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< Many-worlds Solution of Uda Equation > | ||||||||||||
In this page, I note down what I thought about a possibility that many-worlds interpretation is expressed and supported by my new quantum grammar. At present(2018/06/26), I have got to think that this approach is wrong. Below I use energy representation. In my new quantum grammar, Fe represents an unentangled quantum history represented by a wave-function πe in the old grammar. πe(e'', t) = πe'(e'', t) if t <0. πe(e', t) = exp[(-i/ In position representation, a general wave-function ψ corresponding to a wave-function π in energy representation is as follows. ψ(x, t) =∫de π(e, t)ue(x) where ue is an eigenfunction of Hamiltonian with an eigenvalue e. {[- As for πe, position representation wave function ψe corresponding to it is as follows. ψe(x, t) =∫de' πe(e', t)ue'(x). I wondered if each Fe is not a solution of the Uda equation and F = ΣeCeFe is a solution of the Uda equation for some complex coefficients Ce. This idea was presented at JPS 2018 Spring Meeting and is faced with deadlock in this page. F[E] = Σe Ce Fe[E]. As a sketch, Fe[E] ~ exp{α∫-∞0dt[(-i/ The first factor of the right hand side of this equation was derived in Energy Representation of Uda Equation @ Quantum History Theory. Precise expression may be the following one. Fe[E] = δ(limt→-∞[tE(t) + i ∫-∞0dt [dE(t)/dt]δFe[E]/δE(t) = α∫-∞0dt [dE(t)/dt][(-i/ = α{[(-i/ ∴ (-i = [∫-∞0dt E(t)]Fe[E] + {[-tE(t)]t=-∞0 - i ∫T∞dt [dE(t)/dt]δFe[E]/δE(t) = exp{α∫-∞0dτ[(-i/ = 0 ∵ [E(t') - e][δ/δE(t)]∫DW exp{iα∫T∞dτ W(τ)[E(τ) - e]} ~ [E(t') - e]δ'(E(t) - e)Πτ≠tδ(E(τ) - e) ~ -δ(t' - t)δ(E(t) - e)Πτ≠tδ(E(τ) - e) = -δ(t' - t)Πτδ(E(τ) - e) = -δ(t' - t)∫DW exp{iα∫T∞dτ W(τ)[E(τ) - e]} ∴ (d/dt')[E(t') - e][δ/δE(t)]∫DW exp{iα∫T∞dτ W(τ)[E(τ) - e]} ~ -δ'(t' - t)∫DW exp{iα∫T∞dτ W(τ)[E(τ) - e]} ∴ [dE(t')/dt'][δ/δE(t)]∫DW exp{iα∫T∞dτ W(τ)[E(τ) - e]} ~ -δ'(t' - t)∫DW exp{iα∫T∞dτ W(τ)[E(τ) - e]} ∴ [dE(t)/dt][δ/δE(t)]∫DW exp{iα∫T∞dτ W(τ)[E(τ) - e]} = 0. ∫T∞dt E(t) Fe[E] = ∫T∞dt e Fe[E] ・・・ ※. (-i (-i Because of these two equations, energy representation of Uda equation reduces to the following equation. (-i Because of this equation and ※, (-i ∴ ΣeCe∫0Tdt {(-i = ΣeCe[∫T∞dt e + i I can not go further from here. I assumed that tE(t) + i Below I investigated how functional differentiation is by using discretized sketch. ・・・・(a-2 + ε-2)(a-1 + ε-1)(a0 + Δ)(a1 + ε1)(a2 + ε2)・・・ = ・・・・ a-2 a-1 a0 a1 a2 ・・・ + ・・・・ a-2 a-1 Δ a1 a2 ・・・ + Σk≠0(εk/ak)・・・・ a-2 a-1(a0 + Δ)a1 a2 ・・・ ∴ ・・・・(a-2 + ε-2)(a-1 + ε-1)(a0 + Δ)(a1 + ε1)(a2 + ε2)・・・ - ・・・・ a-2 a-1 a0 a1 a2 ・・・ = ・・・・ a-2 a-1 Δ a1 a2 ・・・ + Σk≠0 εk(∂/∂ak)・・・・ a-2 a-1(a0 + Δ)a1 a2 ・・・ Fe[E] = Πk=-∞0f(E(k/α), k/α)Πk=1∞δ(E(k/α) - e) Fe[E(□ - 1/α)] = Πk=-∞0f(E(k/α - 1/α), k/α)Πk=1∞δ(E(k/α - 1/α) - e) = Πk'=-∞-1f(E(k'/α), (k' + 1)/α)Πk'=0∞δ(E(k'/α) - e) = Πk=-∞-1f(E(k/α), (k + 1)/α)Πk=0∞δ(E(k/α) - e) Fe[E(□ - 1/α)] - Fe[E] = Πk=-∞-1f(E(k/α), k/α)[δ(E(0/α) - e) - f(E(0/α), 0/α)]Πk=1∞δ(E(k/α) - e) + Σk=-∞-1 (1/α)[∂/∂(k/α)]Πk'=-∞-1f(E(k'/α), k'/α)δ(E(0/α) - e)Πk'=1∞δ(E(k'/α) - e) I can not go further from here. I tried to calculate the shift of wave-function at t = 0. ∫DW exp{iα∫0εdτ W(τ)[E(τ) - e]} - exp{α∫0εdt[(-i/ = δ(αε [E(0) - e]} - exp{αε[g(E(0))+h(0)]} if ε → +0. This also does not give any hint to me. Below I looked for Hamiltonian which causes wave contraction in measurement along the old grammar. U(t1, t2) = exp[(-i/ U(t1, t2) = exp[(-i/ U(t1, t2) = exp[(-i/ i = exp[(-i/ = HU(t1, t2) + δ(t1)exp[(-i/ ≠ H'U(t1, t2). It seems to be impossible for wave contraction to be caused by Hamiltonian. |
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Author Yuichi Uda, Write start at 2018/05/21/14:06JST, Last edit at 2018/06/26/17:57JST | ||||||||||||
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