since 2006
Help　Sitemap
< Forum >
< Problems >
< Quantum History Theory >
< Discretized Quantum Mechanical Solution >

the new grammar version of Schrödinger equation in the case that V = 0.

To do so, I use a sequence of numbers:
・・・, χ(-2/α), χ(-1/α), χ(0), χ(1/α), χ(2/α), ・・・
instead of a function χ representing a classical history.

Accordingly, I replace the functional Φ in the equation with a function as follows.
Φ[χ] → Φ(・・・, χ-2, χ-1, χ0, χ1, χ2, ・・・)
where χk ≡χ(k/α).

Moreover, I rewrite the original equation in the form of an equation for the following function F.
Φ(χ) = F(a1(χ), b1(χ); a2(χ), b2(χ); a3(χ), b3(χ); ・・・)
where
an(χ) = Σk=-∞χk cos(kπ/n),
bn(χ) = Σk=-∞χk sin(kπ/n).

---

I calculate some differentiations as a preparation for a while.

(d/dε)an(χ(□ - ε))|ε=0 ≒ α[an(χ(□ - 1/α)) - an(χ)]
= αΣk=-∞k-1 - χk)cos(kπ/n)
= αΣk=-∞χk[cos((k+1)π/n) - cos(kπ/n)]
= αΣk=-∞χk[cos(kπ/n)cos(π/n) - sin(kπ/n)sin(π/n) - cos(kπ/n)]
= α[cos(π/n) - 1]Σk=-∞χk cos(kπ/n) - αsin(π/n)Σk=-∞χk sin(kπ/n)
= α[cos(π/n) - 1]an(χ) - αsin(π/n)bn(χ)

(d/dε) bn(χ(□ - ε))|ε=0 ≒ α[bn(χ(□ - 1/α)) - bn(χ)]
= αΣk=-∞k-1 - χk)sin(kπ/n)
= αΣk=-∞χk[sin((k+1)π/n) - sin(kπ/n)]
= αΣk=-∞χk[sin(kπ/n)cos(π/n) + cos(kπ/n)sin(π/n) - sin(kπ/n)]
= α[cos(π/n) - 1]Σk=-∞χk sin(kπ/n) + αsin(π/n)Σk=-∞χk cos(kπ/n)
= α[cos(π/n) - 1]bn(χ) + αsin(π/n)an(χ)

(d/dε) Φ[χ(□ - ε)]|ε=0
= Σn=1 [(dan/dε)∂F/∂an + (dbn/dε)∂F/∂bn]|ε=0
= αΣn=1 [{[cos(π/n) - 1]an - sin(π/n)bn}∂F/∂an + {[cos(π/n) - 1]bn + sin(π/n)an}∂F/∂bn] ・・・ (1)

This is a part of the left hand side of the original equation.

---

To rewrite the right-hand side of the original equation, I do some calculations below as a preparation.

∂/∂χk = Σn=1[(∂an/∂χk)∂/∂an + (∂bn/∂χk)∂/∂bn}
= Σn=1[cos(kπ/n)∂/∂an + sin(kπ/n)∂/∂bn]

Σk=-∞(∂/∂χk)2 = Σk=-∞Σn=1Σs=1[cos(kπ/n)∂/∂an + sin(kπ/n)∂/∂bn][cos(kπ/s)∂/∂as + sin(kπ/s)∂/∂bs]
= Σn=1Σs=1Σk=-∞[cos(kπ/n)cos(kπ/s)(∂/∂an)(∂/∂as) + sin(kπ/n)sin(kπ/s)(∂/∂bn)(∂/∂bs) + cos(kπ/n)sin(kπ/s)(∂/∂an)(∂/∂bs) + sin(kπ/n)cos(kπ/s)(∂/∂bn)(∂/∂as)]
= (1/2)Σn=1Σs=1Σk=-∞[cos(kπ(1/n + 1/s)) + cos(kπ(1/n - 1/s))](∂/∂an)(∂/∂as)
+ (1/2)Σn=1Σs=1Σk=-∞[-cos(kπ(1/n + 1/s)) + cos(kπ(1/n - 1/s))](∂/∂bn)(∂/∂bs)
+ (1/2)Σn=1Σs=1Σk=-∞[sin(kπ(1/n + 1/s)) - sin(kπ(1/n - 1/s))](∂/∂an)(∂/∂bs)
+ (1/2)Σn=1Σs=1Σk=-∞[sin(kπ(1/n + 1/s)) + sin(kπ(1/n - 1/s))](∂/∂bn)(∂/∂as)

Σk=-NN cos(kπ(1/n + 1/s)) = (Σk=-N-1 + Σk=00 + Σk=1N)cos(kπ(1/n + 1/s))
= Σk=1N cos(-kπ(1/n + 1/s)) + 1 + Σk=1N cos(kπ(1/n + 1/s))
= (1/2)sin[-(N + 1/2)π(1/n + 1/s)]/sin[-π(1/n + 1/s)/2] - 1/2
+ 1
+ (1/2)sin[(N + 1/2)π(1/n + 1/s)]/sin[π(1/n + 1/s)/2] - 1/2
= sin[(N + 1/2)π(1/n + 1/s)]/sin[π(1/n + 1/s)/2]
→ 0 (N → ∞)

Σk=-NN sin(kπ(1/n + 1/s)) = (Σk=-N-1 + Σk=00 + Σk=1N)sin(kπ(1/n + 1/s))
= Σk=1Nsin(-kπ(1/n + 1/s)) + 0 + Σk=1Nsin(kπ(1/n + 1/s))
= 0

Σk=-NN cos(kπ(1/n - 1/s)) = (Σk=-N-1 + Σk=00 + Σk=1N)cos(kπ(1/n - 1/s))
= Σk=1N cos(-kπ(1/n - 1/s)) + 1 + Σk=1N cos(kπ(1/n - 1/s))
= (1/2)sin[-(N + 1/2)π(1/n - 1/s)]/sin[-π(1/n - 1/s)/2] - 1/2
+ 1
+ (1/2)sin[(N + 1/2)π(1/n - 1/s)]/sin[π(1/n - 1/s)/2] - 1/2
= sin[(N + 1/2)π(1/n - 1/s)]/sin[π(1/n - 1/s)/2]
→ 0 (N → ∞, n≠s)

Σk=-NN cos(kπ(1/n - 1/s)) = (Σk=-N-1 + Σk=00 + Σk=1N)cos(kπ(1/n - 1/s))
= 2N + 1 (n = s)

Σk=-NN sin(kπ(1/n - 1/s)) = (Σk=-N-1 + Σk=00 + Σk=1N)sin(kπ(1/n - 1/s))
= Σk=1Nsin(-kπ(1/n - 1/s)) + 0 + Σk=1Nsin(kπ(1/n - 1/s))
= 0

∴ Σk=-∞(∂/∂χk)2 = limN→∞ [(2N + 1)/2]Σn=1[(∂/∂an)2 + (∂/∂bn)2] ・・・ (2)

This is a part of the right hand side of the original equation.

---

The original equation is rewritten as follows.

(ih/α)(d/dε)Φ[χ(□ - ε)]|ε=0 = [1/(2m)](1/α)Σk=-∞[(-ih/α)α∂/∂χk]2Φ[χ]

(d/dε)Φ[χ(□ - ε)]|ε=0 = [ih/(2m)]Σk=-∞(∂/∂χk)2Φ[χ] ・・・ (3)

---

Substituting (1) and (2) into (3), the following equation is obtained.
αΣn=1 [{[cos(π/n) - 1]an - sin(π/n)bn}∂F/∂an + {[cos(π/n) - 1]bn + sin(π/n)an}∂F/∂bn]
= [ih/(2m)][(2N + 1)/2]Σn=1[(∂/∂an)2 + (∂/∂bn)2]F

∴ Σn=1{(∂/∂an)2 + (∂/∂bn)2
- {4mα/[(ih)(2N + 1)]}[{[cos(π/n) - 1]an - sin(π/n)bn}∂/∂an + {[cos(π/n) - 1]bn + sin(π/n)an}∂/∂bn]}F = 0 ・・・ (4)

---

Here I specialize my investigation as
F(a1, b1; a2, b2; a3, b3; ・・・) = f(an, bn).

Then, (4) reduces as follows.
{(∂/∂ξ)2 + (∂/∂η)2 - [(βξ - γη)∂/∂ξ + (βη + γξ)∂/∂η]}f(ξ,η) = 0 ・・・ (5)
where
β = {4mα/[(ih)(2N + 1)]}[cos(π/n) - 1],
γ = {4mα/[(ih)(2N + 1)]}sin(π/n),
N is a number to tend to infinity.

---

To solve Equation (5), I rewrite it in the form of an equation for the following function g.
f(ξ,η) = ∫dp∫dq g(p, q)exp(ipξ + iqη).

ξ(∂/∂ξ)f(ξ,η) = ∫dp∫dq g(p, q)ipξexp(ipξ + iqη)
= ∫dp∫dq g(p, q)ip(-i)(∂/∂p)exp(ipξ + iqη)
= -∫dp∫dq exp(ipξ + iqη)(∂/∂p)[p g(p, q)]

η(∂/∂ξ)f(ξ,η) = ∫dp∫dq g(p, q)ipηexp(ipξ + iqη)
= ∫dp∫dq g(p, q)ip(-i)(∂/∂q)exp(ipξ + iqη)
= -∫dp∫dq exp(ipξ + iqη)(∂/∂q)[p g(p, q)]

η(∂/∂η)f(ξ,η) = ∫dp∫dq g(p, q)iqηexp(ipξ + iqη)
= ∫dp∫dq g(p, q)iq(-i)(∂/∂q)exp(ipξ + iqη)
= -∫dp∫dq exp(ipξ + iqη)(∂/∂q)[q g(p, q)]

ξ(∂/∂η)f(ξ,η) = ∫dp∫dq g(p, q)iqξexp(ipξ + iqη)
= ∫dp∫dq g(p, q)iq(-i)(∂/∂p)exp(ipξ + iqη)
= -∫dp∫dq exp(ipξ + iqη)(∂/∂p)[q g(p, q)]

The equation ia rewritten as follows.
[-p2 -q2 + (β∂/∂p - γ∂/∂q)p + (β∂/∂q + γ∂/∂p)q]g(p, q) = 0.
∴ [-p2 -q2 + p(β∂/∂p - γ∂/∂q) + q(β∂/∂q + γ∂/∂p) +2β]g(p, q) = 0 ・・・ (6)

---

To solve Equation (6), I rewrite it in the form of an equation for the following function h.
g(p, q) = h(r, θ)
where p = r cosθ, q = r sinθ.

∂/∂r = (cosθ)∂/∂p + (sinθ)∂/∂q,
∂/∂θ = -r(sinθ)∂/∂p + r(cosθ)∂/∂q

∂/∂p = (cosθ)∂/∂r - (1/r)(sinθ)∂/∂θ,
∂/∂q = (sinθ)∂/∂r + (1/r)(cosθ)∂/∂θ

p∂/∂p + q∂/∂q = r∂/∂r,
q∂/∂p - p∂/∂q = -∂/∂θ

The equation ia rewritten as follows.
(-r2 + βr∂/∂r - γ∂/∂θ + 2β)h(r, θ) = 0 ・・・ (7)

---

To solve Equation (7), I rewrite it in the form of an equation for the following function R.
h(r, θ) = Σk=-∞ Rk(r)exp(ikθ).

The equation is rewritten as follows.
(-r2 + βr∂/∂r - iγk + 2β)Rk(r) = 0 ・・・ (8).
The solution of this equation is as follows.
Rk(r) = λk[r^(-2 + iγk/β)]exp[r2/(2β)] ・・・ (9).

Author Yuichi Uda, Write start at 2015/10/03/01:48JST, Last edit at 2016/02/07/19:52JST