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The Hamiltonian H of the Dirac Field is as follows. H = ∫d^{3}x ψ†(x)γ^{0}(iγ^{k}∂_{k} + m)ψ(x) Here (γ^{0}γ^{k})† = γ^{k}†γ^{0}† = (γ^{k})(γ^{0}) = γ^{k}γ^{0} = γ^{0}γ^{k} that is to say hermitian. The quantization condition is as follows. [ψ_{α}(x), ψ_{β}†(y)]_{+} = δ_{αβ}δ^{3}(x  y), [ψ_{α}(x), ψ_{β}(y)]_{+} = 0. The Schrodinger equation is as follows. i 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. i where c is real. (x + iy)(x  iy) + (x  iy)(x + iy) = 2x^{2} + 2y^{2} ∴ x^{2} + y^{2} = 1/2 (x + iy)(x + iy) = x^{2}  y^{2} + i(xy + yx) ∴ [x, y]_{+} = i(x^{2}  y^{2}) = i(2x^{2}  1/2) = i(1/2  2y^{2}) 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†an> = nn>, <nn> = 1. a†aan> = 0 = 0an> ∵ aa = 0 (an>, an>) = <na†an> = n<nn> = n ∴ an> = (√n)0> a†aa†n> = (a†a†a + [a†a, a†]_{})n> = [a†a, a†]_{}n> = a†n> = 1a†n>, (a†n>, a†n>) = <naa†n> = <n(1  a†a)n> = (1  n)<nn> = 1  n ∴ a†n> = [√(1  n)]1> n^{2}<nn> = (a†an>, a†an>) = <na†aa†an> = <na†(1  a†a)an> = <na†an> = n<nn> ∴ n^{2} = n ∴ n(n  1) = 0 ∴ n = 0 or 1. a0> = 0, a1> = 0>, a†0> = 1>, a†1> = 0. This is possible because there exsists the following concrete representation. 0> and 1> are the following functions. φ_{0}(0) = 1 and φ_{0}(1) = 0 ・・・ 0> φ_{1}(0) = 0 and φ_{1}(1) = 1 ・・・ 1> a and a† are the following linear operators. aφ_{0} = 0 and aφ_{1} = φ_{0}, a†φ_{0} = φ_{1} and a†φ_{1} = 0. An arbitrary function φ is expressed as φ = φ(0)φ_{0} + φ(1)φ_{1} and so aφ = φ(0)aφ_{0} + φ(1)aφ_{1} = φ(1)φ_{0}, a†φ = φ(0)a†φ_{0} + φ(1)a†φ_{1} = φ(0)φ_{1}, a†aφ = [aφ(0)]φ_{1} = φ(1)φ_{0}(0)φ_{1} = φ(1)φ_{1} ∴ (aφ)(n) = φ(1)φ_{0}(n) = δ_{n0}Σ_{k=0}^{1}δ_{k1}φ(k), (a†φ)(n) = φ(0)φ_{1}(n) = δ_{n1}Σ_{k=0}^{1}δ_{k0}φ(k), (a†aφ)(n) = φ(1)φ_{1}(n) = δ_{n1}φ(1) = δ_{n1}φ(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. i ⊿_{t}Φ[n] ≡ lim_{ε→+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]⊿_{t}Φ[n] because dn(t)/dt is a sum of delta functions. i ∴∫_{∞}^{∞}dt {i 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. i ∴ i ∴ i A solution of this set of equations must be as follows. φ(0, t) = c_{1}, φ(1, t) = c_{2} exp[(i/ The normalization condition fort this solution is as follows. φ(0, t)^{2} + φ(1, t)^{2} = 1 ∴ c_{1}^{2} + c_{2}^{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  






