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When φ'(x, t) = exp[φ(x, t)] is an wave function in the old quantum mechanics and
Φ[χ] = exp[α∫dtφ(χ(t), t)],
measure of a functional integration
∫Dχ Φ[χ]*Φ[χ]
is not trivial.

Because
∫dx1[φ'(x1, t)1/n]*φ'(x1, t)1/n ・・・ ∫dxn[φ'(xn, t)1/n]*φ'(xn, t)1/n ≠ ∫dx φ'(x, t)*φ'(x, t).

Let χε(・・・, x-2, x-1, x0, x1, x2, ・・・) be a function such that
if kε≦ t ＜ (k + 1)ε, [χε(x)](t) = xk.

∫dx Φ[χε(x)]*Φ[χε(x)]
= ∫dx exp{α∫dt[φ([χε(x)](t), t)* + φ([χε(x)](t), t)]}
= ∫dx exp{αΣk(k+1)εdt[φ([χε(x)](t), t)* + φ([χε(x)](t), t)]}
= ∫dx exp{αΣk(k+1)εdt[φ(xk, t)* + φ(xk, t)]}
≒ ∫dx exp{αΣk(k+1)εdt[φ(xk, kε)* + φ(xk, kε)]}
= ∫dx exp{αεΣk[φ(xk, kε)* + φ(xk, kε)]}
= ∫dx Πk exp{αε[φ(xk, kε)* + φ(xk, kε)]}
= ∫dx Πk {exp[φ(xk, kε)* + φ(xk, kε)]}αε
= ∫dx Πk {exp[φ(xk, kε)*]exp[φ(xk, kε)]}αε
= ∫dx Πk {[φ'(xk, kε)]*[φ'(xk, kε)]}αε
≠ 1 (ε≠1/α)

∫dx {Φ[χε(x)]1/(αε)}*Φ[χε(x)]1/(αε)
= ∫dx exp{(1/ε)∫dt[φ([χε(x)](t), t)* + φ([χε(x)](t), t)]}
= ∫dx exp{(1/ε)Σk(k+1)εdt[φ([χε(x)](t), t)* + φ([χε(x)](t), t)]}
= ∫dx exp{(1/ε)Σk(k+1)εdt[φ(xk, t)* + φ(xk, t)]}
≒ ∫dx exp{(1/ε)Σk(k+1)εdt[φ(xk, kε)* + φ(xk, kε)]}
= ∫dx exp{Σk[φ(xk, kε)* + φ(xk, kε)]}
= ∫dx Πk exp{[φ(xk, kε)* + φ(xk, kε)]}
= ∫dx Πk exp[φ(xk, kε)*]exp[φ(xk, kε)]
= ∫dx Πk [φ'(xk, kε)]*[φ'(xk, kε)]
= 1

limε→0∫dx {Φ[χε(x)]1/(αε)}*Φ[χε(x)]1/(αε) = 1

Author Yuichi Uda, Write start at 2016/04/05/20:48JST, Last edit at 2016/04/09/15:48JST