Kaku M.- Quantum Field Theory

  • June 2020
  • PDF

This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA


Overview

Download & View Kaku M.- Quantum Field Theory as PDF for free.

More details

  • Words: 1,714
  • Pages: 805
所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

Useful Formulae ® 闻水国 ©

Useful Formulae

= = 6.58 ×10−25 GeV sec = 1 =c = 0.197 GeV F = 1 (1 GeV) −2 = 0.389 mb

α=

e− (− E ) < 0

e+ ≡ E >0

time

e2 1 = 4π 137 ⎛∂ ⎞ p μ = ( E , p) = i ⎜ , −∇ ⎟ = i∂ μ ⎝ ∂t ⎠ 2 2 2 μ p ≡ pμ p = E − p = m 2

x μ = (t , x), xμ = (t , − x) p ⋅ x = Et − p ⋅ x, (,2 + m 2 )φ = 0,

J PC P = −(−1) L C = (−1) L + S

(i γ μ ∂ μ − m)ψ = 0. μ

μ

μ

In an electromagnetic filed, i∂ → i∂ + eA (charge − e)

j μ = −ie(φ * ∂ μφ − φ∂ μφ *),

j μ = −eψ γ μ ψ

γ-Matrices Tra/ 1 " a/ n = 0 (for n odd)

γ μ γν + γν γ μ = 2 g μν , γ μ † = γ 0 γ μ γ 0 . †

γ0 = γ0 ,



γ 0 γ 0 = I ; γ k = −γ k ,

γ 5 = i γ 0 γ1 γ 2 γ 3 ,

γ k γ k = − I , k = 1, 2,3. †

γ μ γ 5 + γ 5 γ μ = 0, γ 5 = γ 5 .

(Trace theorems on pages 104 in relativistic quantum mechanics of Bjorken) Standard representation:

⎛I 0 ⎞ ⎛ 0 γ0 ≡ β = ⎜ ⎟ , γ ≡ βα = ⎜ ⎝ 0 −I ⎠ ⎝ −σ

σ⎞

⎛0 I ⎞ 5 ⎟, γ = ⎜ ⎟ 0⎠ ⎝ I 0⎠

⎡0 1 ⎤ ⎡ 0 −i ⎤ ⎡1 0 ⎤ ⎡1 0 ⎤ , σ2 = ⎢ , σ3 = ⎢ , σ0 = ⎢ ⎥ ⎥ ⎥ ⎥. ⎣1 0 ⎦ ⎣i 0 ⎦ ⎣0 −1⎦ ⎣0 1 ⎦

σ1 = ⎢

Spinors ⎧(p/ − m)u = 0 ; ⎨ ⎩u (p/ − m) = 0 †

u ( r ) u ( s ) = 2 Eδ rs ,

u ( r )u ( s ) = 2mδ rs ,

1 (1 − γ 5 )u ≡ uL , 2

1 (1 + γ 5 )u ≡ uR . 2

∑u

s =1,2

(s)

⎧⎪u ≡ u † γ 0 , Γ ≡ γ 0 Γ + γ 0 ⎨ μ / ≡ γ μ Aμ ⎪⎩ p/ ≡ pμ γ , A

u ( s ) = p/ + m = 2mΛ +

1 1 If m = 0 or E  m, then uL has helicity λ = − , u L has λ = + . 2 2

标量 (φφ * )

J P = 0+

赝标 (φγ 5φ * )

J P = 0−

矢量 (φγ μφ * )

J P = 1−

轴矢 (φγ 5γ μφ * )

J P = 1+

Mandelstam variables

p3

p1

p2

p4

s = ( p1 + p2 ) 2 = ( p3 + p4 ) 2 t = ( p1 − p3 ) 2 = ( p2 − p4 ) 2 u = ( p1 − p4 ) 2 = ( p2 − p3 ) 2

p1

p3

p2

p4 s-channel

p1

p3

p1

p3

p2

p4

p2

p4

t-channel

Time

Feynman parameter integral 1 1 dx =∫ ab 0 [ax + b(1 − x)]2

Trace Theorems μ ν a/ ≡ aμ γ μ , ab / / = a ⋅ b − iσ μν a b

Tra/ 1 " a/ n = 0 for n odd;

p μ pν = g μν p 2 / d ;

Trγ 5 = 0;

Tr1 = 4;

Trab / / = 4a ⋅ b;

Trγ 5 ab / / = 0;

α β γ δ Trγ 5 abcd / / / / = 4iε αβγδ a b c d ;

γ μ a/ γ μ = −2a/ ;

μ μ γ μ ab / / / γ = −2cba / / /; / / γ = 4a ⋅ b; γ μ abc

Tr[γ μ γ ν γ λ γ δ ] = 4( g μν g λδ − g μλ gνδ + g μδ gνλ ); Tr[a/ 1a/ 2 a/ 3a/ 4 ] = 4[a1 ⋅ a2 a3 ⋅ a4 − a1 ⋅ a3a2 ⋅ a4 + a1 ⋅ a4 a2 ⋅ a3 ]

u-channel

所有者:闻水国

you do, you will! take it, easyly!

Useful Formulae ® 闻水国 ©

Useful Formulae

= = 6.58 ×10−25 GeV sec = 1 =c = 0.197 GeV F = 1 (1 GeV) −2 = 0.389 mb

α=

e− (− E ) < 0

e+ ≡ E >0

time

e2 1 = 4π 137 ⎛∂ ⎞ p μ = ( E , p) = i ⎜ , −∇ ⎟ = i∂ μ ⎝ ∂t ⎠ 2 2 2 μ p ≡ pμ p = E − p = m 2

x μ = (t , x), xμ = (t , − x) p ⋅ x = Et − p ⋅ x, (,2 + m 2 )φ = 0,

J PC P = −(−1) L C = (−1) L + S

(i γ μ ∂ μ − m)ψ = 0. μ

μ

μ

In an electromagnetic filed, i∂ → i∂ + eA (charge − e)

j μ = −ie(φ * ∂ μφ − φ∂ μφ *),

j μ = −eψ γ μ ψ

γ-Matrices Tra/ 1 " a/ n = 0 (for n odd)

γ μ γν + γν γ μ = 2 g μν , γ μ † = γ 0 γ μ γ 0 . †

γ0 = γ0 ,



γ 0 γ 0 = I ; γ k = −γ k ,

γ 5 = i γ 0 γ1 γ 2 γ 3 ,

γ k γ k = − I , k = 1, 2,3. †

γ μ γ 5 + γ 5 γ μ = 0, γ 5 = γ 5 .

(Trace theorems on pages 104 in relativistic quantum mechanics of Bjorken) Standard representation:

⎛I 0 ⎞ ⎛ 0 γ0 ≡ β = ⎜ ⎟ , γ ≡ βα = ⎜ ⎝ 0 −I ⎠ ⎝ −σ

σ⎞

⎛0 I ⎞ 5 ⎟, γ = ⎜ ⎟ 0⎠ ⎝ I 0⎠

⎡0 1 ⎤ ⎡ 0 −i ⎤ ⎡1 0 ⎤ ⎡1 0 ⎤ , σ2 = ⎢ , σ3 = ⎢ , σ0 = ⎢ ⎥ ⎥ ⎥ ⎥. ⎣1 0 ⎦ ⎣i 0 ⎦ ⎣0 −1⎦ ⎣0 1 ⎦

σ1 = ⎢

Spinors ⎧(p/ − m)u = 0 ; ⎨ ⎩u (p/ − m) = 0 †

u ( r ) u ( s ) = 2 Eδ rs ,

u ( r )u ( s ) = 2mδ rs ,

1 (1 − γ 5 )u ≡ uL , 2

1 (1 + γ 5 )u ≡ uR . 2

∑u

s =1,2

(s)

⎧⎪u ≡ u † γ 0 , Γ ≡ γ 0 Γ + γ 0 ⎨ μ / ≡ γ μ Aμ ⎪⎩ p/ ≡ pμ γ , A

u ( s ) = p/ + m = 2mΛ +

1 1 If m = 0 or E  m, then uL has helicity λ = − , u L has λ = + . 2 2

标量 (φφ * )

J P = 0+

赝标 (φγ 5φ * )

J P = 0−

矢量 (φγ μφ * )

J P = 1−

轴矢 (φγ 5γ μφ * )

J P = 1+

Mandelstam variables

p3

p1

p2

p4

s = ( p1 + p2 ) 2 = ( p3 + p4 ) 2 t = ( p1 − p3 ) 2 = ( p2 − p4 ) 2 u = ( p1 − p4 ) 2 = ( p2 − p3 ) 2

p1

p3

p2

p4 s-channel

p1

p3

p1

p3

p2

p4

p2

p4

t-channel

Time

Feynman parameter integral 1 1 dx =∫ ab 0 [ax + b(1 − x)]2

Trace Theorems μ ν a/ ≡ aμ γ μ , ab / / = a ⋅ b − iσ μν a b

Tra/ 1 " a/ n = 0 for n odd;

p μ pν = g μν p 2 / d ;

Trγ 5 = 0;

Tr1 = 4;

Trab / / = 4a ⋅ b;

Trγ 5 ab / / = 0;

α β γ δ Trγ 5 abcd / / / / = 4iε αβγδ a b c d ;

γ μ a/ γ μ = −2a/ ;

μ μ γ μ ab / / / γ = −2cba / / /; / / γ = 4a ⋅ b; γ μ abc

Tr[γ μ γ ν γ λ γ δ ] = 4( g μν g λδ − g μλ gνδ + g μδ gνλ ); Tr[a/ 1a/ 2 a/ 3a/ 4 ] = 4[a1 ⋅ a2 a3 ⋅ a4 − a1 ⋅ a3a2 ⋅ a4 + a1 ⋅ a4 a2 ⋅ a3 ]

u-channel

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

所有者:闻水国

you do, you will! take it, easyly!

Related Documents