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t' = t + a, x' = x + b,
χ'(t') = χ(t) + b,
Φ'[χ'] = Φ[χ].

χ'(t') = χ(t' - a) + b
∴χ' = χ(□ - a) + b
where b(t') = b.

Φ'[χ(□ - a) + b] = Φ[χ]

[δ/δχ(τ)]Φ[χ]
= (∂/∂ε)Φ[χ+εδ(□ - τ)]|ε=0
= (∂/∂ε)Φ'[[χ+εδ(□ - τ)](□ - a) + b]|ε=0

[[χ+εδ(□ - τ)](□ - a) + b](t')
= [χ+εδ(□ - τ)](t' - a) + b
= χ(t' - a) +εδ((t' - a) - τ) + b
= [χ(□ - a) +εδ(□ - a - τ) + b](t')
= [χ' +εδ(□ - a - τ)](t')
∴ [χ+εδ(□ - τ)](□ - a) + b = χ' +εδ(□ - a - τ)

[δ/δχ(τ)]Φ[χ]
= (∂/∂ε)Φ'[χ' +εδ(□ - a - τ)]|ε=0
= [δ/δχ'(τ+ a)]Φ'[χ']

[δ/δχ(τ)]2Φ[χ]
= [δ/δχ(τ)][δ/δχ'(τ+ a)]Φ'[χ']
= [δ/δχ(τ)]Ψ'[χ']
= [δ/δχ(τ)]Ψ[χ]
= [δ/δχ'(τ+ a)]Ψ'[χ']
= [δ/δχ'(τ+ a)]2Φ'[χ']
where Ψ'[χ'] ≡ [δ/δχ'(τ+ a)]Φ'[χ'].

∫dt [1/(2m)][(-ih/α)δ/δχ(t)]2Φ[χ]
= ∫dt [1/(2m)][(-ih/α)δ/δχ'(t + a)]2Φ'[χ']
= ∫dt' [1/(2m)][(-ih/α)δ/δχ'(t')]2Φ'[χ']

[χ(□-ε)](t) + b
= χ(t - ε) + b
= χ'((t - ε)')
= χ'(t -ε+ a)
= χ'(t' -ε)
= [χ'(□-ε)](t')
∴ [χ(□-ε)]' = χ'(□-ε)

(ih/α)(∂/∂ε)Φ[χ(□-ε)]|ε=0
= (ih/α)(∂/∂ε)Φ'[χ'(□-ε)]|ε=0

(ih/α)(∂/∂ε)Φ[χ(□-ε)]|ε=0 = ∫dt [1/(2m)][(-ih/α)δ/δχ(t)]2Φ[χ]

(ih/α)(∂/∂ε)Φ'[χ'(□-ε)]|ε=0 = ∫dt' [1/(2m)][(-ih/α)δ/δχ'(t')]2Φ'[χ']

The Uda equation is invariant under any translation when V = 0.

Author Yuichi Uda, Write start at 2016/04/09/16:05JST, Last edit at 2016/04/10/15:26JST