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< Uda Equation of Dirac Field >

The Hamiltonian H of the Dirac Field is as follows.
H = ∫d3x ψ†(x0(-iγkk + m)ψ(x)
Here (γ0γk)† = γk†γ0† = (-γk)(γ0) = -γkγ0 = γ0γk that is to say hermitian.
The quantization condition is as follows.
α(x), ψβ†(y)]+ = δαβδ3(x - y),
α(x), ψβ(y)]+ = 0.
The Schrodinger equation is as follows.
ih(d/dt)|ΨS(t)> = H|ΨS(t)>.

As a preparation, let's consider a quantum mechanics of two real degrees x, y of freedom.
H = c(x - iy)(x + iy),
[x + iy, x - iy]+ ≡ (x + iy)(x - iy) + (x - iy)(x + iy) = 1,
[x + iy, x + iy]+ ≡ (x + iy)(x + iy) + (x + iy)(x + iy) = 0.
ih(d/dt)|ΨS(t)> = H|ΨS(t)>
where c is real.

(x + iy)(x - iy) + (x - iy)(x + iy) = 2x2 + 2y2
∴ x2 + y2 = 1/2

(x + iy)(x + iy) = x2 - y2 + i(xy + yx)
∴ [x, y]+ = i(x2 - y2) = i(2x2 - 1/2) = i(1/2 - 2y2)

a ≡ x + iy, [a, a†]+ = 1, [a, a]+ = 0.

[a†a, a]- = a†aa - aa†a = -aa†a = -(1 - a†a)a = -a,
[a†a, a†]- = a†aa† - a†a†a = a†aa† = (1 - aa†)a† = a†.

a†a|n> = n|n>, <n|n> = 1.

a†aa|n> = 0 = 0a|n> ∵ aa = 0
(a|n>, a|n>) = <n|a†a|n> = n<n|n> = n
∴ a|n> = (√n)|0>

a†aa†|n> = (a†a†a + [a†a, a†]-)|n> = [a†a, a†]-|n> = a†|n> = 1a†|n>,
(a†|n>, a†|n>) = <n|aa†|n> = <n|(1 - a†a)|n> = (1 - n)<n|n> = 1 - n
∴ a†|n> = [√(1 - n)]|1>

n2<n|n> = (a†a|n>, a†a|n>) = <n|a†aa†a|n> = <n|a†(1 - a†a)a|n> = <n|a†a|n> = n<n|n>
∴ n2 = n
∴ n(n - 1) = 0
∴ n = 0 or 1.

a|0> = 0, a|1> = |0>, a†|0> = |1>, a†|1> = 0.

This is possible because there exsists the following concrete representation.
 

|0> and |1> are the following functions.
φ0(0) = 1 and φ0(1) = 0 ・・・ |0>
φ1(0) = 0 and φ1(1) = 1 ・・・ |1>

a and a† are the following linear operators.
0 = 0 and aφ1 = φ0,
a†φ0 = φ1 and a†φ1 = 0.

An arbitrary function φ is expressed as
φ = φ(0)φ0 + φ(1)φ1
and so
aφ = φ(0)aφ0 + φ(1)aφ1 = φ(1)φ0,
a†φ = φ(0)a†φ0 + φ(1)a†φ1 = φ(0)φ1,
a†aφ = [aφ(0)]φ1 = φ(1)φ0(0)φ1 = φ(1)φ1
∴ (aφ)(n) = φ(1)φ0(n) = δn0Σk=01δk1φ(k),
  (a†φ)(n) = φ(0)φ1(n) = δn1Σk=01δk0φ(k),
  (a†aφ)(n) = φ(1)φ1(n) = δn1φ(1) = δn1φ(n).

A quantum history is represented by a functional Φ whose variable is a function n and whose value is a complex number.
Φ[n] ∈ C,
n(t) = 0 or 1,
t ∈ R.

ih limε→+0(1/ε){Φ[n(□ -ε)] - Φ[n]} = cα∫-∞dt δn(t),1Φ[n].

tΦ[n] ≡ limε→+0(1/ε){Φ[n'] - Φ[n'']},
 t - ε≦ t' < t ⇒ n'(t') = 1 and n''(t') = 0,
 t' < t - ε or t' ≧ t ⇒ n'(t') = n''(t') = n(t').

Φ[n(□ -ε)] - Φ[n] ~ -ε∫-∞dt [dn(t)/dt]⊿tΦ[n]
because dn(t)/dt is a sum of delta functions.

-ih-∞dt [dn(t)/dt]⊿tΦ[n] = cα∫-∞dt δn(t),1Φ[n].
∴∫-∞dt {ih[dn(t)/dt]⊿t + cαδn(t),1]Φ[n]
  witten at 2019/04/18/14:57.


On the other hand, the Schrodinger equation of the same system in the old grammar is derived as follows.

ih(∂/∂t)[φ(0, t)|0> + φ(1, t)|1>] = ca†a[φ(0, t)|0> + φ(1, t)|1>]
∴ ih(∂/∂t)φ(0, t)|0> + ih(∂/∂t)φ(1, t)|1> = cφ(1, t)|1>
∴ ih(∂/∂t)φ(0, t) = 0 and ih(∂/∂t)φ(1, t) = cφ(1, t).

A solution of this set of equations must be as follows.
φ(0, t) = c1, φ(1, t) = c2 exp[-(i/h)ct].
The normalization condition fort this solution is as follows.
|φ(0, t)|2 + |φ(1, t)|2 = 1
∴ |c1|2 + |c2|2 = 1.
I am glad of how plausible this solution is.













Author Yuichi Uda, Write start at 2019/04/15/19:09JST, Last edit at 2019/04/18/16:58JST