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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