Quantization,

• Be covariant quantization of electromagnetic field (3) (Nakanishi - Lautrap theory)
  http://maldoror-ducasse.cocolog-nifty.com/blog/2009/06/3-lautrap-2191.html
  But, because □a Μ =0 is not, a Μ (x) = (2π) -3/2 ∫d 3 p (2| p |) -1/2 {a Μ ^ (p) exp (- i| p |t+i px) +a Μ ^ + (p) exp (i| p |t-i px)}} = (2π) -3/2 ∫d 4. Θ (p 0) Δ (p 2) {a Μ ^ (p) exp (- ipx) +a Μ ^ + (p) exp (ipx) with what it cannot express, is being the neck of this problem

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  Aber, weil □a Î œ =0 nicht ist, ein Î œ (X) = (2π) -3/2 ∫d 3 p (2| p |) -1/2 {ein Î œ ^ (P) exp (- i| p |t+i px) +a Î œ ^ + (P) exp (i| p |Ti px)}} = (2π) -3/2 ∫d 4. Î ˜ (p 0) Î ” (p 2) {ein Î œ ^ ist (P) exp (- IPX) +a Î œ ^ + ein (P) exp (IPX) mit, was es nicht ausdrücken kann, der Ansatz dieses Probleme
  
  
• Japanese weblog
  http://maldoror-ducasse.cocolog-nifty.com/blog/2010/04/propagator-theo.html
  Namely, it is simple, but rough approximation: u + (t) =u - (t) =1 s'= u + (XXINF) - ∫ - XXINF XXINF u - + (t) {∂/∂t+ (i/h c) h 1 (t)}u + (t) dt, and s^=1- (i/h c) ∫ - XXINF XXINF [u - + (t) h 1 (t) + h 1 (t) u + (t)]dt+ (i/h c) ∫ - XXINF XXINF u - + (t) h 1 (t) u + (t)]dt+ (i/h c) 2 ∫ - XXINF XXINF the ∫ - XXINF XXINF u - + (t) h 1 (t) Θ (t-t') h 1 (t') u + (t') dtdt'[ni] it substitutes

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  Nämlich ist es einfacher, aber rauer Näherungswert: u + (T) =u - (T) =1 s'= u + (XXINF) - ∫ - XXINF XXINF u - + (T) {∂/∂t+ (i/h c) h 1 (T)} u + (T) Papierlösekorotron und s^=1- (i/h c) ∫ - XXINF XXINF [u - + (T) h 1 (T) + h 1 (T) u + (T)] dt+ (i/h c) ∫ - XXINF XXINF u - + (T) h 1 (T) u + (T)] dt+ (i/h c) 2 ∫ - XXINF XXINF das ∫ - XXINF XXINF u - + (T) h 1 (T) Î ˜ (t-t') h 1 (t') u + (t') dtdt'[Ni] ersetzt es
  
  
• weblog title
  http://maldoror-ducasse.cocolog-nifty.com/blog/2009/06/2-3e4f.html
  Then assistant. Because place b (x) d'Alembert equation □b (x) fills up =0, Kirchhoff's integral calculus indication b (x) =∫d 3 z [{∂d (x-z)/∂x 0} b (z) +d (x-z) ∂b (z) was given with/∂z 0

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  Dann Assistent. Weil Platz b (X) d'Alembert Gleichung □b (X) =0 auffüllt, Kirchhoffs wurde Anzeige des integralen Kalküls b (X) =∫d mit 3 z [{∂d (x-z)/∂x 0} b (Z) +d (x-z) ∂b (z)/∂z 0 gegeben