What in the world is an action functional?

This problem first appeared in a message on a web forum on 2007/12/26.

This problem was proposed at JPS 2010 Spring Meeting 20pBJ-1.

This problem was discussed in my presentation material for JPS 2011 Spring Meeting 25aGC-1.

A conclusion was shown in JPS 2011 Autumn Meeting.
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Last edited at 2012/05/01/14:51JST

Following the idea on 2012/04/11, I want to decide the potential energy functional so that the integrand of Feynman's path integral is a solution of the new grammar version of Schrödinger equation.

First I will look for a potential energy functional V such that the following functional Φ is a solution of the new grammar version of Schrödinger equation.
Φ[χ]=exp[∫dt₁∫dt₂ χ(t₁)D(t₁-t₂)χ(t₂)]
where D is a even function.

δΦ[χ]/δχ(t)
=[∫dt₂ D(t-t₂)χ(t₂)+∫dt₁ χ(t₁)D(t₁-t)]exp[∫dt₁∫dt₂ χ(t₁)D(t₁-t₂)χ(t₂)]
=[∫dt₂ D(t-t₂)χ(t₂)+∫dt₂ χ(t₂)D(t₂-t)]exp[∫dt₁∫dt₂ χ(t₁)D(t₁-t₂)χ(t₂)]
=2[∫dt₂ D(t-t₂)χ(t₂)]exp[∫dt₁∫dt₂ χ(t₁)D(t₁-t₂)χ(t₂)] ∵D(t₂-t)=D(t-t₂)
=2[∫dt₂ D(t-t₂)χ(t₂)]Φ[χ]

[δ/δχ(t)]²Φ[χ]
=2D(t-t)Φ[χ]+2[∫dt₂ D(t-t₂)χ(t₂)][δ/δχ(t)]Φ[χ]
=2D(0)Φ[χ]+4[∫dt₂ D(t-t₂)χ(t₂)]²Φ[χ]
=[4∫dt₁∫dt₂ χ(t₁)D(t₁-t)D(t-t₂)χ(t₂)+2D(0)]Φ[χ]
∵[∫dt₂ D(t-t₂)χ(t₂)]²
　=[∫dt₂ D(t-t₂)χ(t₂)][∫dt₂ D(t-t₂)χ(t₂)]
　=∫dt₁ D(t-t₁)χ(t₁)∫dt₂ D(t-t₂)χ(t₂)
　=∫dt₁ D(t₁-t)χ(t₁)∫dt₂ D(t-t₂)χ(t₂) ∵D(t-t₁)=D(t₁-t)
　=∫dt₁∫dt₂ χ(t₁)D(t₁-t)D(t-t₂)χ(t₂)

∫dt [1/(2m)][(-ih/α)δ/δχ(t)]²Φ[χ]
　=-[1/(2m)](h/α)²{4∫dt₁∫dt₂ χ(t₁)[∫dt D(t₁-t)D(t-t₂)]χ(t₂)+2∫dt D(0)}Φ[χ]
　=-[1/(2m)](h/α)²{4∫dt₁∫dt₂ χ(t₁)[∫dt D(t₁-t)D(t-t₂)]χ(t₂)+2TD(0)}Φ[χ]
where T is the length of the circular time axis.

Φ[χ(□-ε)]=exp[∫dt₁∫dt₂ χ(t₁-ε)D(t₁-t₂)χ(t₂-ε)]
　=exp[∫dt'₁∫dt'₂ χ(t'₁)D((t'₁+ε)-(t'₂+ε))χ(t'₂)] where t'₁=t₁-ε, t'₂=t₂-ε
　=exp[∫dt'₁∫dt'₂ χ(t'₁)D(t'₁-t'₂)χ(t'₂)]
　=Φ[χ]
∴ (d/dε)Φ[χ(□-ε)]=0

To realize the equation
(ih/α)(d/dε)Φ[χ(□-ε)]|_ε=0 = {∫dt [1/(2m)][(-ih/α)δ/δχ(t)]²+V[χ]}Φ[χ]=0,
the following potential energy functional V is needed.
V[χ]=[1/(2m)](h/α)²{4∫dt₁∫dt₂ χ(t₁)[∫dt D(t₁-t)D(t-t₂)]χ(t₂) + 2TD(0)}

D(t₁-t₂) → (i/h)[-(m/2)δ⁽²⁾(t₁-t₂)-(k/2)δ(t₁-t₂)]

∫dt₁∫dt₂ χ(t₁)D(t₁-t₂)χ(t₂)]
　→ -(i/h)(m/2)∫dt₁∫dt₂ χ(t₁)δ⁽²⁾(t₁-t₂)χ(t₂)-(i/h)(k/2)∫dt₁∫dt₂ χ(t₁)δ(t₁-t₂)χ(t₂)
　=(i/h)(m/2)∫dt₁∫dt₂ χ(t₁)[(∂/∂t₁)(∂/∂t₂)δ(t₁-t₂)]χ(t₂)-(i/h)(k/2)∫dt₁∫dt₂ χ(t₁)δ(t₁-t₂)χ(t₂)
　=(i/h)(m/2)∫dt₁∫dt₂ χ⁽¹⁾(t₁)δ(t₁-t₂)χ⁽¹⁾(t₂)-(i/h)(k/2)∫dt₁∫dt₂ χ(t₁)δ(t₁-t₂)χ(t₂)
　=(i/h)(m/2)∫dt χ⁽¹⁾(t)χ⁽¹⁾(t)-(i/h)(k/2)∫dt χ(t)χ(t)
　=(i/h)∫dt{(m/2)[(d/dt)χ(t)]²-(k/2)[χ(t)]²}

Φ[χ] → exp{(i/h)∫dt[(m/2)(dχ(t)/dt)² - (k/2)(χ(t))²]} = exp{(i/h)S[χ]}

∫dt D(t₁-t)D(t-t₂)
　→ (i/h)²∫dt{-(m/2)δ⁽²⁾(t₁-t)-(k/2)δ(t₁-t)}{-(m/2)δ⁽²⁾(t-t₂)-(k/2)δ(t-t₂)}
　=(i/h)²∫dt{(m/2)²δ⁽²⁾(t₁-t)δ⁽²⁾(t-t₂)+(k/2)²δ(t₁-t)δ(t-t₂)
　　　　　　　　+(m/2)(k/2)δ⁽²⁾(t₁-t)δ(t-t₂)+(k/2)(m/2)δ(t₁-t)δ⁽²⁾(t-t₂)}
　=(i/h)²∫dt{(m/2)²δ⁽⁴⁾(t₁-t)δ(t-t₂)+(k/2)²δ(t₁-t)δ(t-t₂)
　　　　　　　　+(m/2)(k/2)δ⁽²⁾(t₁-t)δ(t-t₂)+(k/2)(m/2)δ(t₁-t)δ⁽²⁾(t-t₂)}
　=(i/h)²{(m/2)²δ⁽⁴⁾(t₁-t₂)+(k/2)²δ(t₁-t₂)+(m/2)(k/2)δ⁽²⁾(t₁-t₂)+(k/2)(m/2)δ⁽²⁾(t₁-t₂)}
　=(1/4)(i/h)²{m²δ⁽⁴⁾(t₁-t₂)+k²δ(t₁-t₂)+2mkδ⁽²⁾(t₁-t₂)}

V[χ]
→ [1/(2m)](h/α)²{∫dt₁∫dt₂ χ(t₁)(i/h)²[m²δ⁽⁴⁾(t₁-t₂)+k²δ(t₁-t₂)
　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　+2mkδ⁽²⁾(t₁-t₂)]χ(t₂)+2TD(0)}
　=(h/α)²(T/m)D(0)
　　　+[1/(2m)]{-(m/α)²∫dt₁∫dt₂ χ(t₁)δ⁽⁴⁾(t₁-t₂)χ(t₂)-(k/α)²∫dt₁∫dt₂ χ(t₁)δ(t₁-t₂)χ(t₂)
　　　　　　　　　　-2(mk/α²)∫dt₁∫dt₂ χ(t₁)δ⁽²⁾(t₁-t₂)χ(t₂)}
　=(h/α)²(T/m)D(0)
　　　+[1/(2m)]{-(m/α)²∫dt[(d/dt)²χ(t)]²-(k/α)²∫dt[χ(t)]²+2(mk/α²)∫dt[(d/dt)χ(t)]²}
　=(h/α)²(T/m)D(0)
　　　-[m/(2α²)]{∫dt[(d/dt)²χ(t)]²+(k/m)²∫dt[χ(t)]²-2(k/m)∫dt[(d/dt)χ(t)]²}
　=(h/α)²(T/m)D(0)
　　　-[m/(2α²)]{∫dt[(d/dt)²χ(t)]²-2(k/m)∫dt[(d/dt)χ(t)]²+(k/m)²∫dt[χ(t)]²}
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Last edited at 2012/05/01/15:05JST

Reduced form of the new grammar version of Schrödinger equation tells us that an action functional is a meta potential energy.
On the other hand, as shown in 物理学正典CAN-5-1-14(in Japanese language), time developement is caused by a potential energy in the interaction picture in the old quantum mechanics.
So, I want to think meta time developement caused by an action functional.

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Last edited at 2012/10/02/14:42JST