since 2006
Help Sitemap
< Forum >
< Problems >
< Quantum History Theory >
< Application of the Stacked Daruma Game Formula >

In this page, I try applying
the stacked Daruma game formula to the solution derived in Discretized Quantum Mechanical Model.

Φ[x] ≒ [Πs=-∞Σi(s)=12-∞des] {A0[e,i] + βA1[e,i]}Πk=-∞ue(k)i(k)(xk),
A0[e,i] = Πk=-∞δ(ek),
A1[e,i] = 2mαΣk=-∞(ek + ek+1)-1[・・・δ(ek-13-i(k)(0,ek3-i(k+1)(0,ek+1)δ(ek+2)・・・].

This solution is of the following form.
Φ[x] = Σk=-∞ ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・
That is to say, this is a sum of functions, in each of which only adjoining two degrees of freedom are entangled.

Φ[・・・, xp-1, a, xp, ・・・]
= Σk≦p-3 ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・φ(xp-1)φ(a)φ(xp)・・・
 + ・・・φ(xp-3)ψ(xp-2, xp-1)φ(a)φ(xp)・・・
 + ・・・φ(xp-2)ψ(xp-1, a)φ(xp)・・・

 + ・・・φ(xp-1)ψ(a, xp)φ(xp+1)・・・
 + ・・・φ(xp-1)φ(a)ψ(xp, xp+1)φ(xp+2)・・・
 + Σk≧p+1 ・・・φ(xp-1)φ(a)φ(xp)・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・.

Φ[・・・, xq-1, b, xq, ・・・]
= Σk≦q-3 ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・φ(xq-1)φ(b)φ(xq)・・・
 + ・・・φ(xq-3)ψ(xq-2, xq-1)φ(b)φ(xq)・・・
 + ・・・φ(xq-2)ψ(xq-1, b)φ(xq)・・・

 + ・・・φ(xq-1)ψ(b, xq)φ(xq+1)・・・
 + ・・・φ(xq-1)φ(b)ψ(xq, xq+1)φ(xq+2)・・・
 + Σk≧q+1 ・・・φ(xq-1)φ(b)φ(xq)・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・.

Φ[・・・, xq-1, b, xq, ・・・]* Φ[・・・, xp-1, a, xp, ・・・]

= [Σk≦q-3 ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・φ(xq-1)φ(b)φ(xq)・・・]*i≦p-3 ・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・φ(xp-1)φ(a)φ(xp)・・・]
 + [Σk≦q-3 ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・φ(xq-1)φ(b)φ(xq)・・・]* [・・・φ(xp-3)ψ(xp-2, xp-1)φ(a)φ(xp)・・・]
 + [Σk≦q-3 ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・φ(xq-1)φ(b)φ(xq)・・・]* [・・・φ(xp-2)ψ(xp-1, a)φ(xp)・・・]

 + [Σk≦q-3 ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・φ(xq-1)φ(b)φ(xq)・・・]* [・・・φ(xp-1)ψ(a, xp)φ(xp+1)・・・]
 + [Σk≦q-3 ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・φ(xq-1)φ(b)φ(xq)・・・]* [・・・φ(xp-1)φ(a)ψ(xp, xp+1)φ(xp+2)・・・]
 + [Σk≦q-3 ・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・φ(xq-1)φ(b)φ(xq)・・・]*i≧p+1 ・・・φ(xp-1)φ(a)φ(xp)・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・]

 + [・・・φ(xq-3)ψ(xq-2, xq-1)φ(b)φ(xq)・・・]*i≦p-3 ・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・φ(xp-1)φ(a)φ(xp)・・・]
 + [・・・φ(xq-3)ψ(xq-2, xq-1)φ(b)φ(xq)・・・]* [・・・φ(xp-3)ψ(xp-2, xp-1)φ(a)φ(xp)・・・]
 + [・・・φ(xq-3)ψ(xq-2, xq-1)φ(b)φ(xq)・・・]* [・・・φ(xp-2)ψ(xp-1, a)φ(xp)・・・]

 + [・・・φ(xq-3)ψ(xq-2, xq-1)φ(b)φ(xq)・・・]* [・・・φ(xp-1)ψ(a, xp)φ(xp+1)・・・]
 + [・・・φ(xq-3)ψ(xq-2, xq-1)φ(b)φ(xq)・・・]* [・・・φ(xp-1)φ(a)ψ(xp, xp+1)φ(xp+2)・・・]
 + [・・・φ(xq-3)ψ(xq-2, xq-1)φ(b)φ(xq)・・・]*i≧p+1 ・・・φ(xp-1)φ(a)φ(xp)・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・]

 + [・・・φ(xq-2)ψ(xq-1, b)φ(xq)・・・]*i≦p-3 ・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・φ(xp-1)φ(a)φ(xp)・・・]
 + [・・・φ(xq-2)ψ(xq-1, b)φ(xq)・・・]* [・・・φ(xp-3)ψ(xp-2, xp-1)φ(a)φ(xp)・・・]
 + [・・・φ(xq-2)ψ(xq-1, b)φ(xq)・・・]* [・・・φ(xp-2)ψ(xp-1, a)φ(xp)・・・]
 + [・・・φ(xq-2)ψ(xq-1, b)φ(xq)・・・]* [・・・φ(xp-1)ψ(a, xp)φ(xp+1)・・・]
 + [・・・φ(xq-2)ψ(xq-1, b)φ(xq)・・・]* [・・・φ(xp-1)φ(a)ψ(xp, xp+1)φ(xp+2)・・・]
 + [・・・φ(xq-2)ψ(xq-1, b)φ(xq)・・・]*i≧p+1 ・・・φ(xp-1)φ(a)φ(xp)・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・]

 + [・・・φ(xq-1)ψ(b, xq)φ(xq+1)・・・]*i≦p-3 ・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・φ(xp-1)φ(a)φ(xp)・・・]
 + [・・・φ(xq-1)ψ(b, xq)φ(xq+1)・・・]* [・・・φ(xp-3)ψ(xp-2, xp-1)φ(a)φ(xp)・・・]
 + [・・・φ(xq-1)ψ(b, xq)φ(xq+1)・・・]* [・・・φ(xp-2)ψ(xp-1, a)φ(xp)・・・]
 + [・・・φ(xq-1)ψ(b, xq)φ(xq+1)・・・]* [・・・φ(xp-1)ψ(a, xp)φ(xp+1)・・・]
 + [・・・φ(xq-1)ψ(b, xq)φ(xq+1)・・・]* [・・・φ(xp-1)φ(a)ψ(xp, xp+1)φ(xp+2)・・・]
 + [・・・φ(xq-1)ψ(b, xq)φ(xq+1)・・・]*i≧p+1 ・・・φ(xp-1)φ(a)φ(xp)・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・]

 + [・・・φ(xq-1)φ(b)ψ(xq, xq+1)φ(xq+2)・・・]*i≦p-3 ・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・φ(xp-1)φ(a)φ(xp)・・・]
 + [・・・φ(xq-1)φ(b)ψ(xq, xq+1)φ(xq+2)・・・]* [・・・φ(xp-3)ψ(xp-2, xp-1)φ(a)φ(xp)・・・]
 + [・・・φ(xq-1)φ(b)ψ(xq, xq+1)φ(xq+2)・・・]* [・・・φ(xp-2)ψ(xp-1, a)φ(xp)・・・]
 + [・・・φ(xq-1)φ(b)ψ(xq, xq+1)φ(xq+2)・・・]* [・・・φ(xp-1)ψ(a, xp)φ(xp+1)・・・]
 + [・・・φ(xq-1)φ(b)ψ(xq, xq+1)φ(xq+2)・・・]* [・・・φ(xp-1)φ(a)ψ(xp, xp+1)φ(xp+2)・・・]
 + [・・・φ(xq-1)φ(b)ψ(xq, xq+1)φ(xq+2)・・・]*i≧p+1 ・・・φ(xp-1)φ(a)φ(xp)・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・]

 + [Σk≧q+1 ・・・φ(xq-1)φ(b)φ(xq)・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]*i≦p-3 ・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・φ(xp-1)φ(a)φ(xp)・・・]
 + [Σk≧q+1 ・・・φ(xq-1)φ(b)φ(xq)・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]* [・・・φ(xp-3)ψ(xp-2, xp-1)φ(a)φ(xp)・・・]
 + [Σk≧q+1 ・・・φ(xq-1)φ(b)φ(xq)・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]* [・・・φ(xp-2)ψ(xp-1, a)φ(xp)・・・]
 + [Σk≧q+1 ・・・φ(xq-1)φ(b)φ(xq)・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]* [・・・φ(xp-1)ψ(a, xp)φ(xp+1)・・・]
 + [Σk≧q+1 ・・・φ(xq-1)φ(b)φ(xq)・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]* [・・・φ(xp-1)φ(a)ψ(xp, xp+1)φ(xp+2)・・・]
 + [Σk≧q+1 ・・・φ(xq-1)φ(b)φ(xq)・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]*i≧p+1 ・・・φ(xp-1)φ(a)φ(xp)・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・].

Below, I investigate the case that q is much larger than p.
I assume that
∫dx φ(x)*φ(x) = 1.

∫Dx Φ[・・・, xq-1, b, xq, ・・・]* Φ[・・・, xp-1, a, xp, ・・・]

= Aφ(b)*φ(a)
 + φ(b)*∫dxp-1 f1(xp-1)ψ(xp-1, a)
 + φ(b)*∫dxp f2(xp)ψ(a, xp)
 + ∫dxq-1 f3(xq-1)ψ(xq-1, b)*φ(a)
 + ∫dxq f4(xq)ψ(b, xq)*φ(a)
 + B1∫dxq-1dxp-1φ(xq-1)ψ(xq-1, b)*ψ(xp-1, a)φ(xp-1)*
 + B2∫dxq-1dxpφ(xq-1)ψ(xq-1, b)*ψ(a, xp)φ(xp)*
 + B3∫dxqdxp-1φ(xq)ψ(b, xq)*ψ(xp-1, a)φ(xp-1)*
 + B4∫dxqdxpφ(xq)ψ(b, xq)*ψ(a, xp)φ(xp)*

= Aφ(b)*φ(a)
 + φ(b)*∫dx [f1(x)ψ(x, a) + f2(x)ψ(a, x)]

 + ∫dx [f3(x)ψ(x, b)*+ f4(x)ψ(b, x)*]φ(a)
 + ∫dx dyφ(x)ψ(x, b)*[B1ψ(y, a) + B2ψ(a, y)]φ(y)*
 + ∫dx dyφ(x)ψ(b, x)*[B3ψ(y, a) + B4ψ(a, y)]φ(y)*.

A = Σk≠q-1Σi≠p-1∫Dx [・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]* [・・・φ(xi-1)ψ(xi, xi+1)φ(xi+2)・・・].

B1 = B2 = B3 = B4 = 1.

f1(xp-1) = Σk≠q-1(・・・∫dxp-2∫dxp・・・)[・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]* [・・・φ(xp-2)φ(xp)・・・],
f2(xp) = Σk≠q-1(・・・∫dxp-1∫dxp+1・・・)[・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・]* [・・・φ(xp-1)φ(xp+1)・・・],
f3(xq-1) = Σk≠p-1(・・・∫dxq-2∫dxq・・・)[・・・φ(xq-2)φ(xq)・・・]* [・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・],
f4(xq) = Σk≠p-1(・・・∫dxq-1∫dxq+1・・・)[・・・φ(xq-1)φ(xq+1)・・・]* [・・・φ(xk-1)ψ(xk, xk+1)φ(xk+2)・・・],

∴ f1(x) = f2(x) = f3(x)* = f4(x)*.

∫Dx Φ[・・・, xq-1, b, xq, ・・・]* Φ[・・・, xp-1, a, xp, ・・・]

= Aφ(b)*φ(a)
 + φ(b)*∫dx [f1(x)ψ(x, a) + f1(x)ψ(a, x)]

 + ∫dx [f1(x)*ψ(x, b)*+ f1(x)*ψ(b, x)*]φ(a)
 + ∫dx dyφ(x)ψ(x, b)*[ψ(y, a) + ψ(a, y)]φ(y)*
 + ∫dx dyφ(x)ψ(b, x)*[ψ(y, a) + ψ(a, y)]φ(y)*

= Aφ(b)*φ(a)
 + φ(b)*∫dx f1(x)[ψ(x, a) + ψ(a, x)]
 + ∫dx f1(x)*[ψ(x, b)* + ψ(b, x)*]φ(a)
 + ∫dx dyφ(x)[ψ(x, b)* + ψ(b, x)*] [ψ(y, a) + ψ(a, y)]φ(y)*.

If A-1/2f1(x) = φ(x)*, this expression reduces to the following one.

{A1/2φ(b)* + A-1/2∫dx f1(y)*[ψ(y, b)* + ψ(b, y)*] } {A1/2φ(a) + A-1/2∫dx f1(x)[ψ(x, a) + ψ(a, x)] }.

However, it is not so.

Therefore, the p-th degree of freedom and the q-th degree of freedom are entangled with each other.
Though entanglement exists, ∫Dx Φ[・・・, xq-1, b, xq, ・・・]* Φ[・・・, xp-1, a, xp, ・・・] has no dependence on q - p at all, and so this result can not reproduce the result of the old quantum mechanics.

Author Yuichi Uda, Write start at 2015/05/04/22:22JST, Last edit at 2015/06/25/21:02JST