Quantum Field Theory on the Time-Axis

This topic is dedicated to discussion of operator formulation of the new grammar version of quantum mechanics.

It can be called quantum field theory on the time-axis.

It will be used to imitate the derivation of Feynman's path integral in the new grammar version of quantum mechanics.
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Last edited at 2012/06/22/15:13JST

|χ> is a functional such that |χ>[χ'] = δ[χ'-χ].
<χ| is a linear functional such that <χ|Φ = Φ[χ], and so <χ|χ'> = |χ'>[χ] = δ[χ-χ'].

[∫Dχ|χ><χ|Φ][χ']=∫Dχ(<χ|Φ)|χ>[χ']=∫Dχ Φ[χ]δ[χ'-χ]=Φ[χ']
∴ ∫Dχ|χ><χ|Φ=Φ
∴ ∫Dχ|χ><χ|=1
where
∫Dχ=lim_n→∞Π_k=1ⁿ∫dχ(kT/n).

[X(t)Φ][χ] = χ(t)Φ[χ]

[X(t)|χ>][χ'] = χ'(t)|χ>[χ'] = χ'(t)δ[χ-χ'] = χ(t)δ[χ-χ'] = χ(t){|χ>[χ']} = [χ(t)|χ>][χ']
∴ X(t)|χ> = χ(t)|χ>

[P(t)Φ][χ] = -(ih/α)[δ/δχ(t)]Φ[χ]

{[X(t),P(t')]_-Φ}[χ]
={[X(t)P(t')-P(t')X(t)]Φ}[χ]
={X(t)P(t')Φ-P(t')X(t)Φ}[χ]
=[X(t)P(t')Φ][χ]-[P(t')X(t)Φ][χ]
={X(t)[P(t')Φ]}[χ]-{P(t')[X(t)Φ]}[χ]
=χ(t)[P(t')Φ][χ]-(-ih/α)[δ/δχ(t')][X(t)Φ]}[χ]
=χ(t)(-ih/α)[δ/δχ(t')]Φ[χ]-(-ih/α)[δ/δχ(t')]{χ(t)Φ[χ]}
=[χ(t),-(ih/α)[δ/δχ(t')]]_-Φ[χ]
=(ih/α)δ(t-t')Φ[χ]
=[(ih/α)δ(t-t')Φ][χ]
∴ [X(t),P(t')]_- = (ih/α)δ(t-t')

|π> is defined as a functional such that |π>[χ] = exp[(iα/h)∫dt π(t)χ(t)].

<χ|π>=|π>[χ]=exp[(iα/h)∫dt π(t)χ(t)]

[P(t)|π>][χ] = -(ih/α)[δ/δχ(t)]|π>[χ]
= -(ih/α)[δ/δχ(t)]exp[(iα/h)∫dt' π(t')χ(t')]
= π(t)exp[(iα/h)∫dt' π(t')χ(t')]
= π(t)|π>[χ]
= [π(t)|π>][χ]
∴ P(t)|π> = π(t)|π>

<π| is defined as a linear functional such that <π|Φ = ∫Dχ exp[-(iα/h)∫dt π(t)χ(t)]Φ[χ].

<π|χ>=∫Dχ' exp[-(iα/h)∫dt π(t)χ'(t)]|χ>[χ']
=∫Dχ' exp[-(iα/h)∫dt π(t)χ'(t)]δ[χ'-χ]
=exp[-(iα/h)∫dt π(t)χ(t)]

[∫Dπ|π><π|Φ][χ]
=∫Dπ(<π|Φ)|π>[χ]
=∫Dπ{∫Dχ' exp[-(iα/h)∫dt π(t)χ'(t)]Φ[χ']}exp[(iα/h)∫dt' π(t')χ(t')]
=∫Dχ'∫Dπ exp{(iα/h)∫dt π(t)[χ(t)-χ'(t)]}Φ[χ']
=∫Dχ' lim_n→∞[(α/h)(T/n)/(2×3.14・・・)]ⁿ[Π_k=1ⁿ∫dπ(kT/n)]exp{(iα/h)(T/n)Σ_j=1ⁿ π(jT/n)[χ(jT/n)-χ'(jT/n)]}Φ[χ']
=∫Dχ' Φ[χ']lim_n→∞[(α/h)(T/n)/(2π)]ⁿΠ_k=1ⁿ(2π)[(α/h)(T/n)]^-1δ(χ(jT/n)-χ'(jT/n))
=∫Dχ' Φ[χ']lim_n→∞Π_k=1ⁿδ(χ(jT/n)-χ'(jT/n))
=∫Dχ' Φ[χ']δ[χ-χ']
=Φ[χ]
∴ ∫Dπ|π><π|Φ = Φ
∴ ∫Dπ|π><π| = 1
where
∫Dπ=lim_n→∞[(α/h)(T/n)/(2π)]ⁿΠ_k=1ⁿ∫dπ(kT/n).

<π|P(t)Φ=∫Dχ exp[-(iα/h)∫dt' π(t')χ(t')][P(t)Φ][χ]
=∫Dχ exp[-(iα/h)∫dt' π(t')χ(t')](-ih/α)[δ/δχ(t)]Φ[χ]
=∫Dχ Φ[χ](ih/α)[δ/δχ(t)]exp[-(iα/h)∫dt' π(t')χ(t')]
=∫Dχ Φ[χ]π(t)exp[-(iα/h)∫dt' π(t')χ(t')]
=π(t)<π|Φ
∴ <π|P(t) = π(t)<π|

The new grammar version of Schrödinger equation is as follows.
(ih/α)(d/dε)<χ(□-ε)|Φ>|_ε=0 = <χ|∫dt{[P(t)]²/(2m)+V(X(t))}|Φ>

Φ_ε[χ] ≡ Φ[χ(□-ε)]
<χ|Φ_ε>=<χ(□-ε)|Φ>
(ih/α)(d/dε)|Φ_ε> = ∫dt{[P(t)]²/(2m)+V(X(t))}|Φ_ε>

H=∫dt{[P(t)]²/(2m)+V(X(t))}
|Φ_ε>=exp[(-iα/h)εH]|Φ₀>
=lim_n→∞{exp[(-iα/h)(ε/n)H]}ⁿ|Φ₀>
=lim_n→∞∫Dχ₁・・・∫Dχ_n exp[(-iα/h)(ε/n)H]|χ_n><χ_n|・・・|χ₂><χ₂|exp[(-iα/h)(ε/n)H]|χ₁><χ₁|Φ₀>

<χ_k+1|exp[(-iα/h)(ε/n)H]|χ_k>
=∫Dπ<χ_k+1|π><π|exp[(-iα/h)(ε/n)H]|χ_k>
=∫Dπ exp[(-iα/h)(ε/n)∫dt{[π(t)]²/(2m)+V(χ_k(t))}]<χ_k+1|π><π|χ_k>
=∫Dπ exp[(-iα/h)(ε/n)∫dt{[π(t)]²/(2m)+V(χ_k(t))}]exp{(iα/h)∫dt π(t)[χ_k+1(t)-χ_k(t)]}
=∫Dπ exp{(iα/h)(ε/n)∫dt[π(t)[χ_k+1(t)-χ_k(t)]/(ε/n)-{[π(t)]²/(2m)+V(χ_k(t))}]}
=lim_n'→∞[(α/h)(T/n')/(2×3.14・・・)]^n'Π_j=1^n'∫dπ(jT/n')exp{(iα/h)(ε/n)(T/n')Σ_s=1^n'[-[1/(2m)]{π(sT/n')-m[χ_k+1(sT/n')-χ_k(sT/n')]/(ε/n)}²-V(χ_k(sT/n'))+(m/2){[χ_k+1(sT/n')-χ_k(sT/n')]/(ε/n)}²]}
=lim_n'→∞[(α/h)(T/n')/(2×3.14・・・)]^n'{(2m×3.14・・・)/[(iα/h)(ε/n)(T/n')]}^n'/2exp[(iα/h)(ε/n)∫dt{(m/2){[χ_k+1(t)-χ_k(t)]/(ε/n)}²-V(χ_k(t))}]
=lim_n'→∞{(α/h)(T/n')m/[2i(ε/n)×3.14・・・]}^n'/2exp[(iα/h)(ε/n)∫dt{(m/2){[χ_k+1(t)-χ_k(t)]/(ε/n)}²-V(χ_k(t))}]
=lim_n'→∞{(α/h)(T/n')m/[2πi(ε/n)]}^n'/2exp[(iα/h)(ε/n)∫dt{(m/2){[χ_k+1(t)-χ_k(t)]/(ε/n)}²-V(χ_k(t))}]

C≡lim_n'→∞{(α/h)(T/n')m/[2πi(ε/n)]}^n'/2=0
This is a difficulty!

ξ(kε/n, t)≡χ_k(t)

<χ_k+1|exp[(-iα/h)(ε/n)H]|χ_k>
=C exp[(iα/h)(ε/n)∫dt{(m/2)[(d/dτ)ξ(τ,t)]²-V(ξ(τ,t))}]|_τ=kε/n (n→∞)

<χ_n+1|Φ_ε>=∫Dξexp[(iα/h)∫₀^εdτ∫₀^Tdt{(m/2)[(d/dτ)ξ(τ,t)]²-V(ξ(τ,t))}]<ξ(ε/n,□)|Φ₀>
where
∫Dξ=lim_n→∞Cⁿlim_n'→∞Π_k=1ⁿΠ_j=1^n'∫dξ(kε/n,jT/n')
　=lim_n→∞lim_n'→∞{(α/h)(T/n')m/[2πi(ε/n)]}^nn'/2Π_k=1ⁿΠ_j=1^n'∫dξ(kε/n,jT/n')
Here the difficulty is absorbed into the functional integral measure.

Φ_ε[ξ(ε,□)] = ∫Dξ exp[(iα/h)∫₀^εdτ∫₀^Tdt{(m/2)[(d/dτ)ξ(τ,t)]²-V(ξ(τ,t))}]Φ₀[ξ(0,□)]

Φ_ε[ξ(ε,□)] = Φ[ξ(ε,□-ε)]

Φ[ξ(ε,□-ε)] = ∫Dξ exp[(iα/h)∫₀^εdτ∫₀^Tdt{(m/2)[(d/dτ)ξ(τ,t)]²-V(ξ(τ,t))}]Φ[ξ(0,□)]

This message follows the idea on 2012/05/01.
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Last edited at 2012/07/18/14:53JST

Though it is impossible because entanglement is inevitable, if
Φ[χ] = lim_n→∞ Π_k=1ⁿ φ(χ(kT/n),kT/n),
the conclusion of the previous message is understood as follows.

φ(ξ(ε,kT/n),kT/n+ε)
= ∫Dξ(□,kT/n) exp[(iα/h)(T/n)∫₀^εdτ{(m/2)[(d/dτ)ξ(τ,kT/n)]²-V(ξ(τ,kT/n))}]φ(ξ(0,kT/n),kT/n)
∵
Φ[ξ(ε,□-ε)] = lim_n→∞ Π_k=1ⁿ φ(ξ(ε,kT/n-ε),kT/n) = lim_n→∞ Π_k=1ⁿ φ(ξ(ε,kT/n),kT/n+ε),
Φ[ξ(0,□)] = lim_n→∞ Π_k=1ⁿ φ(ξ(0,kT/n),kT/n).

Therefore,
φ(χ(ε),kT/n+ε) = ∫Dχ exp[(iα/h)(T/n)∫₀^εdτ{(m/2)[(d/dτ)χ(τ)]²-V(χ(τ)}]φ(χ(0),kT/n).

This means that φ obeys the ordinary Schrodinger equation when T/n=1/α, and so the new theory is natural extension of the old one.
That is to say, the length of an instant which defines a quantum state is 1/α.

Then we can write an approximate solution as follows.
G(χ(ε),ε;χ(0),0) = ∫Dχ exp[(i/h)∫₀^εdτ{(m/2)[(d/dτ)χ(τ)]²-V(χ(τ)}],
φ(χ(ε),k/α+ε) = ∫dχ(0) G(χ(ε),ε;χ(0),0) φ(χ(0),k/α)　(k=1,2,・・・,αT),
Φ[χ] = Π_k=1^αT φ(χ(k/α),k/α)
where G is the Green's function.
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Last edited at 2012/07/17/15:43JST

When
Φ[χ] = ∫dx Π_k=1^αT G(χ(k/α),k/α;x,t0),
∫_{χ(a)=A,χ(b)=B}Dχ Φ[χ] ≠ C G(A,a;B,b)


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Last edited at 2012/07/21/14:42JST

Consider the case that T/n = 1/α and the functional:
Φ[χ] = ∫dx Π_k=1^αT G(χ(k/α),k/α;x,t₀)
where G is the Green's function.

This functional Φ is a sum of quantum histories which give x for measurement at time t₀.

∫Dξ exp[(iα/h)∫₀^εdτ∫₀^Tdt{(m/2)[(d/dτ)ξ(τ,t)]²-V(ξ(τ,t))}]Φ[ξ(0,□)]
≒ ∫Dξ exp[(iα/h)(T/n)Σ_k=1ⁿ∫₀^εdτ{(m/2)[(d/dτ)ξ(τ,kT/n)]²-V(ξ(τ,kT/n))}]Φ[ξ(0,□)]
= ∫Dξ Π_k=1^αTexp[(i/h)∫₀^εdτ{(m/2)[(d/dτ)ξ(τ,k/α)]²-V(ξ(τ,k/α))}]Φ[ξ(0,□)]
= [Π_k=1^αT∫dξ(0,k/α) G(ξ(ε,k/α),ε;ξ(0,k/α),0)]Φ[ξ(0,□)]
= [Π_k=1^αT∫dξ(0,k/α) G(ξ(ε,k/α),ε;ξ(0,k/α),0)]∫dx Π_k=1^αT G(ξ(0,k/α),k/α;x,t₀)
= ∫dx Π_k=1^αT∫dξ(0,k/α) G(ξ(ε,k/α),ε;ξ(0,k/α),0) G(ξ(0,k/α),k/α;x,t₀)
= ∫dx Π_k=1^αT G(ξ(ε,k/α),k/α+ε;x,t₀)
= ∫dx Π_k=1^αT G(ξ(ε,k/α-ε),k/α;x,t₀)
= Φ[ξ(ε,□-ε)]

Therefore,
Φ[χ] = ∫dx Π_k=1^αT G(χ(k/α),k/α;x,t₀)
is an approximate solution.

∫dx is not necessary, but I expect that the entanglement caused by it rationalizes the probability interpretation.

This message follows the idea on 2012/07/02.
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Last edited at 2012/07/22/11:04JST