 since 2006 Help　Sitemap < Forum > < Problems > < Quantum History Theory > < Uda Equation of QED > At first I extract the definition of the quantum electrodynamics from a summarization note written by me in 1995. Now again I admire the note for its describing important necessary details omitted in any other literature. Readers can not grasp exact definition of QED if such details are not explicitly written. However, all textbooks except for my note omit such details. I suppose that some of even authors of such textbooks do not grasp those details. Expressions of my note are more complex than the expressions of the same things in other literatures. My note is difficult to read proportionally to the complexity of its expressions. However, the complexity is not mine but of QED itself, and so my note is kinder than other textbooks. The Hamiltonian H of QED is as follows.(Q-51-23～28) H = ∫d3x ψ†(x)γ0(-iγk∂k + m)ψ(x) 　　　+ (1/2)∫d3x [Ak(β; x)Ak(β; x) + ∂kAj(α; x)・∂kAj(α; x)] 　　　+ e∫d3x ψ†(x)γ0[(1/2)γ0A0(ψ; x) + γkAk(α; x)]ψ(x) Ak and A0 are defined as follows. Q-45-11～13 Ak(α; x) ≡ (2π)-3/2∫d3p Στ=12 ε(τ)k(p)[ατ(p)exp(ip・x) + ατ†(p)exp(-ip・x)] The integration is taken over p ∈ R3+. R3+ is half of R3. ε(τ)(p) is a real normal spacial vector perpendicular to p. ε(1)(p) is perpendicular to ε(2)(p). 　p・ε(τ)(p) ≡ Σk=13pkε(τ)k(p) = 0, 　ε(λ)(p)・ε(τ)(p) ≡ Σk=13ε(λ)k(p)ε(τ)k(p) = δλτ. Q-45-18,19 A0(ψ; x) = [e/(4π)]∫d3x' ψ†(x')ψ(x')/|x' - x| To fix gauge, I use α: R3 → C2 instead of the vector potential A: R3 → R4 as fundamental fields representing classical electromagnetic field. It fixes gauge to the Coulomb gauge. βτ(x) is conjugate momentum of ατ(x) and is defined by the following algebra. The quantization condition is as follows.(Q-54-2～14) [ψα(x), ψβ†(y)]+ = δαβδ3(x - y), [αλ(p), βτ†(q)]- = iδλτδ3(p - q), [αλ(p), ατ†(q)]- = [βλ(p), βτ†(q)]- = 0, [αλ(p), ατ(q)]- = [βλ(p), βτ(q)]- = 0, [αλ(p), βτ(q)]- = [ψα(x), ψβ(y)]+ = 0, [ατ(p), ψα(x)]- = [βτ(p), ψα(x)]- = 0, [ατ(p), ψα†(x)]- = [βτ(p), ψα†(x)]- = 0. Please notice that α and β in ψα(x) and ψβ(x) are different from α and β in ατ(p) and βτ(p). I can not avoid using same letters because usable letters are fewer than my needs. The Schrodinger equation is as follows. ih(d/dt)|ΨS(t)> = H|ΨS(t)>. As a preparation, let's consider a quantum mechanics of a complex degree X of freedom and a real degree Q of freedom. H = aX†X + b2P2 + c2Q2 + eX†QX, [X, X†]+ = 1, [X, X]+ = 0, [Q, P]- = ih, [Q, Q]- = 0, [P, P]- = 0, [Q, X]- = 0, [P, X]- = 0. X|0, q> = 0, X|1, q> = |0, q>, X†|0, q> = |1, q>, X†|1, q> = 0, X†X|n, q> = n|n, q>, Q|n, q> = q|n, q>, P|n, q> = ih(∂/∂q)|n, q>. ih(d/dt)Σn=01∫-∞∞dq φ(n, q; t)|n, q> = HΣn=01∫-∞∞dq φ(n, q; t)|n, q>. P is the conjugate momentum of Q. a, b, c and e are positive real numbers. Σn=01∫-∞∞dq ih(∂/∂t)φ(n, q; t)|n, q> 　= Σn=01∫-∞∞dq φ(n, q; t)(an - b2h2∂2/∂q2 + c2q2 + enq)|n, q> ∴ ih(∂/∂t)φ(n, q; t) = (an - b2h2∂2/∂q2 + c2q2 + enq)φ(n, q; t) 　　written at 2019/04/20/15:49. Y ≡ {√[c/(2hb)]}Q + i{√[b/(2hc)]}P, Y† ≡ {√[c/(2hb)]}Q - i{√[b/(2hc)]}P. Y†Y = [c/(2hb)]Q2 + [b/(2hc)]P2 + i[1/(2h)][Q, P]- 　　　= [c/(2hb)]Q2 + [b/(2hc)]P2 - (1/2). H = aX†X + 2hbcY†Y + hbc + e{√[hb/(2c)]}X†(Y + Y†)X. X|0, s> = 0, X|1, s> = |0, s>, X†|0, s> = |1, s>, X†|1, s> = 0. Y|n, s> = (√s)|n, s - 1> if s≧1. Y|n, 0> = 0. Y†|n, s> = [√(s + 1)]|n, s + 1>. n = 0, 1; s = 0, 1, 2, 3, ・・・. X†(Y + Y†)X|0, s> = 0. X†(Y + Y†)X|1, 0> = X†(Y + Y†)|0, 0> = X†Y†|0, 0> = X†|0, 1> = |1, 1>. X†(Y + Y†)X|1, s> = X†(Y + Y†)|0, s> = (√s)X†|0, s - 1> + [√(s + 1)]X†|0, s + 1> 　　　= (√s)|1, s - 1> + [√(s + 1)]|1, s + 1> if s≧1. H|0, s> = (2hbcs + hbc)|0, s> = hbc(2s + 1)|0, s>, H|1, 0> = (a + hbc)|1, 0> + e{√[hb/(2c)]}|1, 1>, H|1, s> = [a + hbc(2s + 1)]|1, s> + e{√[hb/(2c)]}{(√s)|1, s - 1> + [√(s + 1)]|1, s + 1>} if s≧1. HΣn=01Σs=0∞φ(n, s)|n, s> = Σs=0∞φ(0, s)H|0, s> + φ(1, 0)H|1, 0> + Σs=1∞φ(1, s)H|1, s> = hbcΣs=0∞(2s + 1)φ(0, s)|0, s> 　　　+ (a + hbc)φ(1, 0)|1, 0> + e{√[hb/(2c)]}φ(1, 0)|1, 1> 　　　+ Σs=1∞[a + hbc(2s + 1)]φ(1, s)|1, s> 　　　+ e{√[hb/(2c)]}Σs=1∞(√s)φ(1, s)|1, s - 1> 　　　+ e{√[hb/(2c)]}Σs=1∞[√(s + 1)]φ(1, s)|1, s + 1> = hbcΣs=0∞(2s + 1)φ(0, s)|0, s> 　　　+ (a + hbc)φ(1, 0)|1, 0> 　　　+ e{√[hb/(2c)]}φ(1, 0)|1, 1> 　　　+ Σs=1∞[a + hbc(2s + 1)]φ(1, s)|1, s> 　　　+ e{√[hb/(2c)]}Σs=0∞[√(s + 1)]φ(1, s + 1)|1, s> 　　　+ e{√[hb/(2c)]}Σs=2∞(√s)φ(1, s - 1)|1, s> ih(d/dt)Σn=01Σs=0∞φ(n, s; t)|n, s> = HΣn=01Σs=0∞φ(n, s; t)|n, s>. ----------------------- ih(∂/∂t)φ(0, s; t) = hbc(2s + 1)φ(0, s; t) for all s. ih(∂/∂t)φ(1, 0; t) = (a + hbc)φ(1, 0; t) + e{√[hb/(2c)]}φ(1, 1; t). ih(∂/∂t)φ(1, 1; t) = e{√[hb/(2c)]}φ(1, 0; t) + (a + 3hbc)φ(1, 1; t) + e{√[hb/(2c)]}(√2)φ(1, 2; t). ih(∂/∂t)φ(1, s; t) 　= [a + hbc(2s + 1)]φ(1, s; t) + e{√[hb/(2c)]}[√(s + 1)]φ(1, s + 1; t) + e{√[hb/(2c)]}(√s)φ(1, s - 1; t)　if s ≧ 2. ------------------------written at 2019/04/21/14:18 = hbcδn'0(2s' + 1)φ(0, s') + (a + hbc)δn'1δs'0φ(1, 0) + e{√[hb/(2c)]}δn'1δs'1φ(1, 0) 　+ Σs=1∞[a + hbc(2s + 1)]δn'1δs'sφ(1, s) 　+ e{√[hb/(2c)]}[√(s' + 1)]δn'1φ(1, s' + 1) 　+ e{√[hb/(2c)]}Σs=2∞(√s)δn'1δs'sφ(1, s - 1) = [ hbcδn'0(2s' + 1)Σs=0∞δs's 　　+ (a + hbc)δn'1δs'0Σs=0∞δ0s 　　+ e{√[hb/(2c)]}δn'1δs'1Σs=0∞δ0s 　　+ δn'1[a + hbc(2s' + 1)]Σs=1∞δs's 　　+ e{√[hb/(2c)]}δn'1[√(s' + 1)]Σs=0∞δs'+1,s 　　+ e{√[hb/(2c)]}δn'1(√s')Σs=1∞δs',s+1 ]φ(n', s) -ih∫-∞∞dt {[dn(t)/dt]⊿1t + [ds(t)/dt]⊿2t}Φ[n, s] = cαεlimε→0Σk=-∞∞ [ hbcδn(kε),0(2s(kε) + 1)Φ[n, s] 　　　　　　　　　　　+ (a + hbc)δn(kε),1δs(kε),0 Φk[n, s; 0] 　　　　　　　　　　　+ e{√[hb/(2c)]}δn(kε),1δs(kε),1Φk[n, s; 0] 　　　　　　　　　　　+ δn(kε),1[a + hbc(2s(kε) + 1)]{Φ[n, s] - δs(kε),0Φk[n, s; 0]} 　　　　　　　　　　　+ e{√[hb/(2c)]}δn(kε),1[√(s(kε) + 1)] Φk[n, s; s + 1] 　　　　　　　　　　　+ e{√[hb/(2c)]}δn(kε),1(√s(kε)){Φk[n, s; s - 1] - δs(kε),1Φk[n, s; 0]}. 　　　　　　　　　　　　　　　written at 2019/04/22/19:38. Φk[n, s; s'] ≡ Φ[n, s''] s''(t) = s'(t)　if　 kε ≦ t ＜ (k + 1)ε. s''(t) = s(t)　if　 t ＜ kε or t ≧ (k + 1)ε. (s + 1)(t) ≡ s(t) + 1. written at 2019/04/22/19:38. I write the above things again for not touching the above writings. ⊿1tΦ[n, s] ≡ limε→+0(1/ε){Φ[n', s] - Φ[n'', s]}, 　t - ε≦ t' ＜ t ⇒ n'(t') = 1 and n''(t') = 0, 　t' ＜ t - ε or t' ≧ t ⇒ n'(t') = n''(t') = n(t'). ⊿2tΦ[n, s] ≡ limε→+0(1/ε){Φ[n, s'] - Φ[n, s]}, 　t - ε≦ t' ＜ t ⇒ s'(t') = s(t') + 1, 　t' ＜ t - ε or t' ≧ t ⇒ s'(t') = s(t'). Φε,k[n, s; n', s'] ≡ Φ[n'', s''], 　kε ≦ t ＜ (k + 1)ε ⇒ n''(t) = n'(t) and s''(t) = s'(t), 　t ＜ kε or t ≧ (k + 1)ε ⇒ n''(t) = n(t) and s''(t) = s(t). -ih∫-∞∞dt {[dn(t)/dt]⊿1t + [ds(t)/dt]⊿2t}Φ[n, s] 　= αεlimε→+0Σk=-∞∞ [ hbcδn(kε),0[2s(kε) + 1]Φε,k[n, s; 0, s] 　　　　　　　　　　　　　　　　　+ (a + hbc)δn(kε),1δs(kε),0Φε,k[n, s; 1, 0] 　　　　　　　　　　　　　　　　+ e{√[hb/(2c)]}δn(kε),1δs(kε),1Φε,k[n, s; 1, 0] 　　　　　　　　　　　　　　　+ Σj=1∞[a + hbc(2j + 1)]δn(kε),1δs(kε),jΦε,k[n, s; 1, j] 　　　　　　　　　　　　　　+ e{√[hb/(2c)]}{√[s(kε) + 1]}δn(kε),1Φε,k[n, s; 1, s + 1] 　　　　　　　　　　　　　+ e{√[hb/(2c)]}Σj=2∞(√j)δn(kε),1δs(kε),jΦε,k[n, s; 1, s - 1] ]. 　　　　　　　　　　　　　　　　written at 2019/04/23/19:14. --- ⊿/⊿n(t) ≡ ⊿1t. -ih∫-∞∞dt {[dn(t)/dt][⊿/⊿n(t)] + [dq(t)/dt][δ/δq(t)]}Φ[n, q] 　= α∫-∞∞dt {an(t) + b2[(-ih/α)δ/δq(t)]2 + c2q(t)2 + en(t)q(t)}Φ[n, q]. written at 2019/05/02/16:55 Author Yuichi Uda, Write start at 2019/04/19/15:25JST, Last edit at 2019/05/02/16:55JST