since 2006
Help　Sitemap
< Forum >
< Problems >
< Measurement Problem >
< 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/h)t]δ(e' - e) if t ＞ T.
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.
{[-h2/(2m)](d/dx)2 + V(x)}ue(x) ＝eue(x).
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/h)tE(t)+g(E(t))+h(t)]}F'e[E(0～T)]Πτ＞Tδ(E(τ) - e).
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) + ihg(E(t))])exp{α∫-∞0dt[(-i/h)tE(t)+g(E(t))+h(t)]}F'e[E(0～T)]∫DW exp{iα∫Tdt W(t)[E(t) -e ]}.

-∞0dt [dE(t)/dt]δFe[E]/δE(t)
＝ α∫-∞0dt [dE(t)/dt][(-i/h)t + ∂g(E(t))/∂E(t)] Fe[E]
＝ α{[(-i/h)tE(t)]t=-∞0 + (i/h)∫-∞0dt E(t) + [g(E(t))]t=-∞0} Fe[E].
∴ (-ih/α)∫-∞0dt [dE(t)/dt]δFe[E]/δE(t)
＝ [∫-∞0dt E(t)]Fe[E] + {[-tE(t)]t=-∞0 - ih[g(E(t))]t=-∞0}Fe[E]

Tdt [dE(t)/dt]δFe[E]/δE(t)
＝ exp{α∫-∞0dτ[(-i/h)τE(τ)+g(E(τ))+h(τ)]}F'e[E(0～T)]∫Tdt [dE(t)/dt][δ/δE(t)]∫DW exp{iα∫Tdτ' W(τ')[E(τ') -e ]}
＝ 0
∵ [E(t') - e][δ/δE(t)]∫DW exp{iα∫Tdτ 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∫Tdτ W(τ)[E(τ) - e]}
∴ (d/dt')[E(t') - e][δ/δE(t)]∫DW exp{iα∫Tdτ W(τ)[E(τ) - e]}
～ -δ'(t' - t)∫DW exp{iα∫Tdτ W(τ)[E(τ) - e]}
∴ [dE(t')/dt'][δ/δE(t)]∫DW exp{iα∫Tdτ W(τ)[E(τ) - e]}
～ -δ'(t' - t)∫DW exp{iα∫Tdτ W(τ)[E(τ) - e]}
∴ [dE(t)/dt][δ/δE(t)]∫DW exp{iα∫Tdτ W(τ)[E(τ) - e]} ＝ 0.

Tdt E(t) Fe[E] ＝ ∫Tdt e Fe[E] ・・・ ※.

(-ih/α)∫-∞0dt [dE(t)/dt]δF[E]/δE(t) ＝ ∫-∞0dt E(t) F[E] + {[-tE(t)]t=-∞0 - ih[g(E(t))]t=-∞0}F[E].
(-ih/α)∫Tdt [dE(t)/dt]δF[E]/δE(t) ＝ 0.
Because of these two equations,
energy representation of Uda equation reduces to the following equation.
(-ih/α)∫0Tdt [dE(t)/dt]δF[E]/δE(t) ＝ ∫0dt E(t) F[E] - {[-tE(t)]t=-∞0 - ih[g(E(t))]t=-∞0}F[E].

Because of this equation and ※,
(-ih/α)∫0Tdt [dE(t)/dt]δF[E]/δE(t) ＝ ∫0Tdt E(t) F[E] + ΣeCeTdt e Fe[E] - {[-tE(t)]t=-∞0 - ih[g(E(t))]t=-∞0}F[E]
∴ ΣeCe0Tdt {(-ih/α)[dE(t)/dt]δ/δE(t) - E(t)}F'e[E(0～T)]∫DW exp{iα∫Tdτ W(τ)[E(τ) - e]}
＝ ΣeCe[∫Tdt e + ihg(E(0))]F'e[E(0～T)]∫DW exp{iα∫Tdτ W(τ)[E(τ) - e]}.
I can not go further from here.
I assumed that tE(t) + ihg(E(t)) → 0 as t → -∞.

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≠0k/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=1δ(E(k/α) - e)
Fe[E(□ - 1/α)] ＝ Πk=-∞0f(E(k/α - 1/α), k/α)Πk=1δ(E(k/α - 1/α) - e)
＝ Πk'=-∞-1f(E(k'/α), (k' + 1)/α)Πk'=0δ(E(k'/α) - e)
＝ Πk=-∞-1f(E(k/α), (k + 1)/α)Πk=0δ(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=1δ(E(k/α) - e)
+ Σk=-∞-1 (1/α)[∂/∂(k/α)]Πk'=-∞-1f(E(k'/α), k'/α)δ(E(0/α) - e)Πk'=1δ(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/h)tE(t)+g(E(t))+h(t)]}
＝ 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/h)t1H][θ(t1)|e><e| + θ(-t1)]exp[(i/h)t2H], t2 ＜ 0.
U(t1, t2) ＝ exp[(-i/h)(t1 - t2)H] if t1 ＜ 0.
U(t1, t2) ＝ exp[(-i/h)(t1 - 0)H]|e><e|exp[(-i/h)(0 - t2)H] if t1 ＞ 0.
ih(∂/∂t1)U(t1, t2)
＝ exp[(-i/h)t1H]{H[θ(t1)|e><e| + θ(-t1)] + δ(t1)|e><e| - δ(t1)}exp[(i/h)t2H]
＝ HU(t1, t2) + δ(t1)exp[(-i/h)t1H][|e><e| - 1]exp[(i/h)t2H]
≠ H'U(t1, t2).
It seems to be impossible for wave contraction to be caused by Hamiltonian.

Author Yuichi Uda, Write start at 2018/05/21/14:06JST, Last edit at 2018/06/26/17:57JST