Uda Equation of QED @ Quantum History Theory @ Problems @ Forum @ www.GrammaticalPhysics.ac

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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(-iγ^k∂_k + m)ψ(x)
　　　+ (1/2)∫d³x [A_k(β; x)A_k(β; x) + ∂_kA_j(α; x)・∂_kA_j(α; x)]
　　　+ e∫d³x ψ†(x)γ⁰[(1/2)γ⁰A₀(ψ; x) + γ^kA_k(α; x)]ψ(x)

A_k and A₀ are defined as follows.

Q-45-11～13
A_k(α; x) ≡ (2π)^-3/2∫d³p Σ_τ=1² ε^(τ)_k(p)[α_τ(p)exp(ip・x) + α_τ†(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) ≡ Σ_k=1³p^kε^(τ)_k(p) = 0,
　ε^(λ)(p)・ε^(τ)(p) ≡ Σ_k=1³ε^(λ)_k(p)ε^(τ)_k(p) = δ_λτ.

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.
β_τ(x) is conjugate momentum of α_τ(x) and is defined by the following algebra.

The quantization condition is as follows.(Q-54-2～14)
[ψ_α(x), ψ_β†(y)]₊ = δ_αβδ³(x - y),
[α_λ(p), β_τ†(q)]_- = iδ_λτδ³(p - q),
[α_λ(p), α_τ†(q)]_- = [β_λ(p), β_τ†(q)]_- = 0,
[α_λ(p), α_τ(q)]_- = [β_λ(p), β_τ(q)]_- = 0,
[α_λ(p), β_τ(q)]_- = [ψ_α(x), ψ_β(y)]₊ = 0,
[α_τ(p), ψ_α(x)]_- = [β_τ(p), ψ_α(x)]_- = 0,
[α_τ(p), ψ_α†(x)]_- = [β_τ(p), ψ_α†(x)]_- = 0.
Please notice that α and β in ψ_α(x) and ψ_β(x) are different from α and β in α_τ(p) and β_τ(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)|Ψ_S(t)> = H|Ψ_S(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)Σ_n=0¹∫_-∞^∞dq φ(n, q; t)|n, q> = HΣ_n=0¹∫_-∞^∞dq φ(n, q; t)|n, q>.
P is the conjugate momentum of Q.
a, b, c and e are positive real numbers.

Σ_n=0¹∫_-∞^∞dq ih(∂/∂t)φ(n, q; t)|n, q>
　= Σ_n=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.

HΣ_n=0¹Σ_s=0^∞φ(n, s)|n, s>
= Σ_s=0^∞φ(0, s)H|0, s> + φ(1, 0)H|1, 0> + Σ_s=1^∞φ(1, s)H|1, s>
= hbcΣ_s=0^∞(2s + 1)φ(0, s)|0, s>
　　　+ (a + hbc)φ(1, 0)|1, 0> + e{√[hb/(2c)]}φ(1, 0)|1, 1>
　　　+ Σ_s=1^∞[a + hbc(2s + 1)]φ(1, s)|1, s>
　　　+ e{√[hb/(2c)]}Σ_s=1^∞(√s)φ(1, s)|1, s - 1>
　　　+ e{√[hb/(2c)]}Σ_s=1^∞[√(s + 1)]φ(1, s)|1, s + 1>
= hbcΣ_s=0^∞(2s + 1)φ(0, s)|0, s>
　　　+ (a + hbc)φ(1, 0)|1, 0>
　　　+ e{√[hb/(2c)]}φ(1, 0)|1, 1>
　　　+ Σ_s=1^∞[a + hbc(2s + 1)]φ(1, s)|1, s>
　　　+ e{√[hb/(2c)]}Σ_s=0^∞[√(s + 1)]φ(1, s + 1)|1, s>
　　　+ e{√[hb/(2c)]}Σ_s=2^∞(√s)φ(1, s - 1)|1, s>

ih(d/dt)Σ_n=0¹Σ_s=0^∞φ(n, s; t)|n, s> = HΣ_n=0¹Σ_s=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

<n', s'| HΣ_n=0¹Σ_s=0^∞φ(n, s)|n, s>
= hbcδ_n'0(2s' + 1)φ(0, s') + (a + hbc)δ_n'1δ_s'0φ(1, 0) + e{√[hb/(2c)]}δ_n'1δ_s'1φ(1, 0)
　+ Σ_s=1^∞[a + hbc(2s + 1)]δ_n'1δ_s'sφ(1, s)
　+ e{√[hb/(2c)]}[√(s' + 1)]δ_n'1φ(1, s' + 1)
　+ e{√[hb/(2c)]}Σ_s=2^∞(√s)δ_n'1δ_s'sφ(1, s - 1)
= [ hbcδ_n'0(2s' + 1)Σ_s=0^∞δ_s's
　　+ (a + hbc)δ_n'1δ_s'0Σ_s=0^∞δ_0s
　　+ e{√[hb/(2c)]}δ_n'1δ_s'1Σ_s=0^∞δ_0s
　　+ δ_n'1[a + hbc(2s' + 1)]Σ_s=1^∞δ_s's
　　+ e{√[hb/(2c)]}δ_n'1[√(s' + 1)]Σ_s=0^∞δ_s'+1,s
　　+ e{√[hb/(2c)]}δ_n'1(√s')Σ_s=1^∞δ_s',s+1 ]φ(n', s)

-ih∫_-∞^∞dt {[dn(t)/dt]⊿¹_t + [ds(t)/dt]⊿²_t}Φ[n, s]
= cαεlim_ε→0Σ_k=-∞^∞ [ hbcδ_n(kε),0(2s(kε) + 1)Φ[n, s]
　　　　　　　　　　　+ (a + hbc)δ_n(kε),1δ_s(kε),0 Φ_k[n, s; 0]
　　　　　　　　　　　+ e{√[hb/(2c)]}δ_n(kε),1δ_s(kε),1Φ_k[n, s; 0]
　　　　　　　　　　　+ δ_n(kε),1[a + hbc(2s(kε) + 1)]{Φ[n, s] - δ_s(kε),0Φ_k[n, s; 0]}
　　　　　　　　　　　+ e{√[hb/(2c)]}δ_n(kε),1[√(s(kε) + 1)] Φ_k[n, s; s + 1]
　　　　　　　　　　　+ e{√[hb/(2c)]}δ_n(kε),1(√s(kε)){Φ_k[n, s; s - 1] - δ_s(kε),1Φ_k[n, s; 0]}.
　　　　　　　　　　　　　　　written at 2019/04/22/19:38.

Φ_k[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.

⊿¹_tΦ[n, s] ≡ lim_ε→+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').

⊿²_tΦ[n, s] ≡ lim_ε→+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').

Φ_ε,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]⊿¹_t + [ds(t)/dt]⊿²_t}Φ[n, s]
　= αεlim_ε→+0Σ_k=-∞^∞ [ hbcδ_n(kε),0[2s(kε) + 1]Φ_ε,k[n, s; 0, s]
　　　　　　　　　　　　　　　　　+ (a + hbc)δ_n(kε),1δ_s(kε),0Φ_ε,k[n, s; 1, 0]
　　　　　　　　　　　　　　　　+ e{√[hb/(2c)]}δ_n(kε),1δ_s(kε),1Φ_ε,k[n, s; 1, 0]
　　　　　　　　　　　　　　　+ Σ_j=1^∞[a + hbc(2j + 1)]δ_n(kε),1δ_s(kε),jΦ_ε,k[n, s; 1, j]
　　　　　　　　　　　　　　+ e{√[hb/(2c)]}{√[s(kε) + 1]}δ_n(kε),1Φ_ε,k[n, s; 1, s + 1]
　　　　　　　　　　　　　+ e{√[hb/(2c)]}Σ_j=2^∞(√j)δ_n(kε),1δ_s(kε),jΦ_ε,k[n, s; 1, s - 1] ].
　　　　　　　　　　　　　　　　written at 2019/04/23/19:14.

---

⊿/⊿n(t) ≡ ⊿¹_t.

-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