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Quantum Electrodynamcs

November 8, 2017

QED

November 8, 2017

1 / 125

Table of Contents I 1 2 3 4 5 6 7 8 9

10 11 12

Free Dirac equation Properties of γ matrices Calculation of sums over electrons spins Example: Klein’s paradox The norelativistic propagator The Feynman propagator The Klein-Gordon equation The Klein-Gordon propagator The Gordon-Volkov solutions Derivation and properties of Gordon-Volkov solutions The transition amplitude for a particle in an intense laser pulse under the influence of an external perturbation The principle of least action Canonical quantization Conservation laws Charge conjugation QED

November 8, 2017

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Table of Contents II Rotational and translational invariance 13

The Klein Gordon field The real Klein-Gordon field The complex Klein-Gordon field Covariant commutation relations The meson propagator

14

The Dirac field The Number Representation for Fermions The Dirac equation Second quantization The spin statistics theorem The Fermion propagator The Electromagnetic Interaction and Gauge Invariance

15

The electromagnetic field The classical electromagnetic field Covariant quantization QED

November 8, 2017

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Table of Contents III The photon propagator

QED

November 8, 2017

4 / 125

The plane wave solutions of the free Dirac equation I The Dirac equation h

i ∂Ψ cα · p + βmc 2 Ψ(x) = i~ ∂t

(1)

Dirac matrices: γ 0 = β,

γ i = βαi

(2)

four momenta Pµ = i~

∂ ≡ i~∂µ ∂x µ

(3)

Covariant form: h

i Pˆ − mc Ψ(x) = 0,

aˆ = γµ aµ

(4)

Plane wave solutions 1 − ~i p·x Ψ(+) ur (p), r (p; x) = √ e V QED

r = 1, 2

(5)

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The plane wave solutions of the free Dirac equation II i 1 Ψ(−) (p; x) = √ e + ~ p·x vr (p), r V

r = 1, 2

(6)

Properties Electron rest frame:  ur (0) =

χr 0



 ,

  u¯r (0) = χTr , 0 , χ1 = ξ,

u¯r (0)us (0) = δrs ,

X

vr (0) =

0 χr

 (7)

  v¯r (0) = 0, −χTr

(8)

χ2 = η

(9)

v¯r (0)vs (0) = −δrs ,

ur (0) ⊗ u¯r (0) =

r =1,2 QED

u¯r (0)vs (0) = 0

1 + γ0 2

(10)

(11) November 8, 2017

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The plane wave solutions of the free Dirac equation III X

vr (0) ⊗ v¯r (0) = −

r =1,2

1 − γ0 2

(12)

ur (p) = p

1 (mc + p) ˆ ur (0) 2(mc)(p 0 + mc)

(13)

vr (p) = p

1 (mc − p) ˆ vr (0) 2(mc)(p 0 + mc)

(14)

(mc − p)u ˆ r (p) = 0, u¯r (p)us (p) = δrs , X r =1,2

(mc + p)v ˆ r (p) = 0

v¯r (p)vs (p) = −δrs ,

ur (p) ⊗ u¯r (p) =

(15)

u¯r (p)vs (p) = 0

(mc) + pˆ ≡ Λ+ (p) 2mc

QED

(16)

(17)

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The plane wave solutions of the free Dirac equation IV

X

vr (p) ⊗ v¯r (p) = −

r =1,2

mc − pˆ ≡ −Λ− (p) 2mc

(18)

Eigenvectors of S ≡ Σ · pˆ Su1 (p) =

~ u1 (p), 2

~ Su2 (p) = − u2 (p), 2

QED

Sv1 (p) =

~ v1 (p) 2

(19)

~ Sv2 (p) = − v2 (p) 2

(20)

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Properties of γ matrices I We define aˆ = γµ aµ with γµ the Dirac γ matrices and γ5 = iγ 0 γ 1 γ 2 γ 3 with the properties γ52 = I , {γ5 , γµ } = 0. Theorem 1 Tr {ˆ a1 aˆ2 . . . aˆ2n+1 } = 0

(21)

n o Tr {ˆ a1 aˆ2 . . . aˆ2n+1 } = Tr aˆ1 aˆ2 . . . aˆ2n+1 γ52 = Tr {γ5 aˆ1 aˆ2 . . . aˆ2n+1 γ5 }

(22)

Proof:

We use the anticommutation property of γ5 with all γ matrices, γ5 aˆ = −ˆ aγ5 , i.e. Tr {γ5 aˆ1 aˆ2 . . . aˆ2n+1 γ5 } =

(−1)2n+1 Tr {ˆ a1 aˆ2 . . . aˆ2n+1 γ5 γ5 }

=

a1 aˆ2 . . . aˆ2n+1 } (−1)2n+1 Tr {ˆ

(23)

or Tr {ˆ a1 aˆ2 . . . aˆ2n+1 } = (−1)2n+1 Tr {ˆ a1 aˆ2 . . . aˆ2n+1 } = 0

QED

November 8, 2017

(24)

9 / 125

Properties of γ matrices II

Theorem 2 Tr {ˆ a1 aˆ2 } = 4(a1 · a2 )

(25)

Proof: Tr {ˆ a1 aˆ2 } = Tr {ˆ a2 aˆ1 } =

1 1 Tr {ˆ a1 aˆ2 + aˆ2 aˆ1 } = Tr {2I (a1 · a2 )} = 4(a1 · a2 ) 2 2

(26)

Theorem 3 Tr {ˆ a1 aˆ2 aˆ3 aˆ4 } = 4 [(a1 · a2 )(a3 · a4 ) + (a1 · a4 )(a2 · a3 ) − (a1 · a3 )(a2 · a4 )]

(27)

Proof: We use the anticommutation properties; the goal of the calculation is to move the first matrix (ˆ a1 ) at the end of the expression under the Tr (after aˆ4 ) and then to bring it in the original position by a circular permutation which leaves the trace unchanged.

QED

November 8, 2017

10 / 125

Properties of γ matrices III

Tr {ˆ a1 aˆ2 aˆ3 aˆ4 } =

Tr {(2a1 · a2 − aˆ2 aˆ1 )ˆ a3 aˆ4 }

=

2(a1 · a2 )Tr {ˆ a3 aˆ4 } − Tr {ˆ a2 aˆ1 aˆ3 aˆ4 }

=

8(a1 · a2 )(a3 · a4 ) − Tr {ˆ a2 (2a1 · a3 − aˆ3 aˆ1 )ˆ a4 }

=

8(a1 · a2 )(a3 · a4 ) − 2(a1 · a3 )Tr {ˆ a2 aˆ4 } + Tr {ˆ a2 aˆ3 aˆ1 aˆ4 }

=

8[(a1 · a2 )(a3 · a4 ) − (a1 · a3 )(a2 · a4 )]+ +Tr {ˆ a2 aˆ3 (2a1 · a4 − aˆ4 aˆ1 )} =

=

8 [(a1 · a2 )(a3 · a4 ) + (a1 · a4 )(a2 · a3 ) − (a1 · a3 )(a2 · a4 )] −Tr {ˆ a2 aˆ3 aˆ4 aˆ1 } =

=

8 [(a1 · a2 )(a3 · a4 ) + (a1 · a4 )(a2 · a3 ) − (a1 · a3 )(a2 · a4 )] −Tr {ˆ a1 aˆ2 aˆ3 aˆ4 }

i.e. Tr {ˆ a1 aˆ2 aˆ3 aˆ3 } = 4 [(a1 · a2 )(a3 · a4 ) + (a1 · a4 )(a2 · a3 ) − (a1 · a3 )(a2 · a4 )]

QED

November 8, 2017

(28)

11 / 125

Properties of γ matrices IV Theorem 4 Tr {ˆ a1 aˆ2 . . . aˆ2n } = 4

0 X

σP (ai1 · ai2 ) . . . (ai2n−1 · ai2n )

(29)

P

where P is a permutation of the set (1, 2, . . . , 2n) and σP is the sign of the permutation. The prime over the sum indicates that only the distinct combinations must be included. Proof: Using the anticommutation properties the initial trace is reduced to a sum of traces with less matrices, (as in the previous case) Tr {ˆ a1 aˆ2 . . . aˆ2n } =

2a1 · a2 Tr {ˆ a3 . . . aˆ2n } − 2a1 · a3 Tr {ˆ a2 aˆ4 . . . aˆ2n } + . . .

−

Tr {ˆ a2 aˆ3 . . . aˆ2n a1 }

(30)

Tr {ˆ a1 aˆ2 . . . aˆ2n } = a1 · a2 Tr {ˆ a3 . . . aˆ2n } − a1 · a3 Tr {ˆ a2 aˆ4 . . . aˆ2n } + . . . .

(31)

i.e.

By successive application of the previous method one obtains the desired result.

QED

November 8, 2017

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Properties of γ matrices V Theorem 5 Tr {γ5 } = 0

(32)

Proof: We define the four-vectors n(0) = (1, 0, 0, 0), n(1) = (0, −1, 0, 0), n(2) = (0, 0, −1, 0), n(3) = (0, 0, 0, −1), and we write γ5 as γ5 = i nˆ(0) nˆ(1) nˆ(2) nˆ(3)

(33)

and iTr {ˆ n(0) nˆ(1) nˆ(2) nˆ(3) } =

Tr {γ5 } = =

h

4 (n

(0)

(1)

· n )(n

(2)

(3)

· n ) − (n

(0)

+(n =

(2)

· n )(n

(0)

(1)

(34)

(3)

· n )+ i · n )(n · n(3) ) (3)

(2)

0

as the 4 vectors n(µ) are orthogonal.

QED

November 8, 2017

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Properties of γ matrices VI Theorem 6 Tr {γ5 aˆ1 aˆ2 } = 0

(35)

Tr {γ5 aˆ1 aˆ2 } = a1µ a2ν Tr {γ5 γµ γν }

(36)

Proof:

Calculation of γ5 γµ γν : if µ = ν we use γµ2 ∼ 1 and γ5 γµ γν ∼ γ5 and it gives no contribution to the results because Tr {γ5 } = 0. For µ 6= ν we choose a λ 6= µ, ν and write Tr {γ5 γµ γν } = Tr {γ5 γµ γλ γλ γν }

(37)

without summation over λ. Next, because γλ anticommutes with all the others matrices under the Tr . performing successive commutations we obtain Tr {γ5 γµ γλ γλ γν } = −Tr {γ5 γµ γλ γλ γν } = 0

(38)

i.e. the terms with µ 6= ν do not contribute. We obtained the desired result Tr {γ5 aˆ1 aˆ2 } = 0 QED

(39) November 8, 2017

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Properties of γ matrices VII Theorem 7 Tr {ˆ a1 aˆ2 . . . aˆ2n } = Tr {ˆ a2n aˆ2n−1 . . . aˆ1 }

(40)

Proof: ∃ a matrix C such that C T C = I and C aˆC T = aˆT ; then n o Tr {ˆ a1 aˆ2 . . . aˆ2n } = Tr aˆ1 C T C aˆ2 C T C . . . aˆ2n C T C n o = Tr C aˆ1 C T C aˆ2 C T C . . . aˆ2n C T n o n o T = Tr aˆ1T aˆ2T . . . aˆ2n = Tr (ˆ a2n aˆ2n−1 . . . aˆ1 )T Tr {ˆ a2n aˆ2n−1 . . . aˆ1 }

=

(41)

Theorem 8 γµ aˆγ µ = −2ˆ a

(42)

γµ aˆγ µ = aν γµ γν γ µ = aν (2gµν − γν γµ )γ µ = 2ˆ a − 4ˆ a = −2ˆ a

(43)

Proof:

QED

November 8, 2017

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Properties of γ matrices VIII Theorem 9 γµ aˆ1 aˆ2 γ µ = 4(a1 · a2 )

(44)

Proof: γµ aˆ1 aˆ2 γ µ =

a1ν γµ γν aˆ2 γ µ = a1ν (2gµν − γν γµ )ˆ a2 γ µ

=

2ˆ a2 aˆ1 + 2ˆ a1 aˆ2 = 4a1 · a2

(45)

Theorem 10 γµ aˆ1 aˆ2 aˆ3 γ µ = −2ˆ a3 aˆ2 aˆ1

(46)

Proof: γµ aˆ1 aˆ2 aˆ3 γ µ =

a1ν γµ γν aˆ2 aˆ3 γ µ = a1ν (2gµν − γν γµ )ˆ a2 aˆ3 γ µ

=

2ˆ a2 aˆ3 aˆ1 − 4ˆ a1 (a2 · a3 ) = 2ˆ a2 aˆ3 aˆ1 − 2(ˆ a2 aˆ3 + aˆ3 aˆ2 )ˆ a1 =

=

−2ˆ a3 aˆ2 aˆ1

QED

November 8, 2017

(47) (48)

16 / 125

Properties of γ matrices IX

Theorem 11 γµ aˆ1 aˆ2 aˆ3 aˆ4 γ µ = 2ˆ a2 aˆ3 aˆ4 aˆ1 + 2ˆ a1 aˆ4 aˆ3 aˆ2

(49)

Proof: γµ aˆ1 aˆ2 aˆ3 aˆ4 γ µ =

a1ν γµ γν aˆ2 aˆ3 aˆ4 γ µ = a1ν (2gµν − γν γµ )ˆ a2 aˆ3 aˆ4 γ µ

=

2ˆ a2 aˆ3 aˆ4 aˆ1 + 2ˆ a1 aˆ4 aˆ3 aˆ2

QED

November 8, 2017

(50)

17 / 125

Calculation of sums over electrons spins I The typical form of the sum over initial and final electron spins of the transition probability X u¯i (p2 )Quj (p1 ) 2

(51)

i,j=1,2

where Q is a 4 × 4 matrix, p1 and p2 are the initial, respectively final electron momenta. We use the identities: X X u¯i (p2 )Quj (p1 ) 2 = u¯i (p2 )Quj (p1 ) (u¯i (p2 )Quj (p1 ))∗ (52) i,j=1,2

i,j=1,2

=

X

 ∗ ui† (p2 )γ 0 Quj (p1 ) ui† (p2 )γ 0 Quj (p1 )

i,j=1,2

=

X

ui† (p2 )γ 0 Quj (p1 )uj† (p1 )(γ 0 Q)† ui (p2 )

i,j=1,2

=

X

¯ i (p2 ) u¯i (p2 )Quj (p1 )u¯j (p1 )Qu

i,j=1,2

QED

November 8, 2017

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Calculation of sums over electrons spins II with Q¯ = γ 0 Q † γ 0 .

(53)

Next we write the product of matrices explicitly as sums over the indices α, β, γ, δ ∈ {1, 2, 3, 4}, X u¯i (p2 )Quj (p1 ) 2 i,j=1,2

=

X

X

[u¯i (p2 )]α Qαβ [uj (p1 )]β [u¯j (p1 )]γ Q¯γδ [ui (p2 )]δ

i,j=1,2 α,β,γ,δ=1,2,3,4

=

X

X

[ui (p2 )]δ [u¯i (p2 )]α Qαβ [uj (p1 )]β [u¯j (p1 )]γ Q¯γδ

i,j=1,2 α,β,γ,δ=1,2,3,4

(54) The completeness property of the Dirac spinors: X

ui (p)u¯i (p) =

i=1,2

QED

pˆ + mc 2mc

(55)

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Calculation of sums over electrons spins III leads to X

u¯i (p2 )Quj (p1 ) 2

i,j=1,2

=

X α,β,γ,δ,=1,2,3,4

=



   pˆ2 + mc pˆ1 + mc Q¯γδ Qαβ 2mc 2mc δα βγ      pˆ2 + mc pˆ1 + mc ¯ Tr Q Q 2mc 2mc

If in the initial or final state are positrons, an analogous calculation leads to      X pˆ2 + mc pˆ1 − mc ¯ u¯i (p2 )Qvj (p1 ) 2 = Tr Q Q 2mc 2mc i,j=1,2      X pˆ2 − mc pˆ1 + mc ¯ v¯i (p2 )Quj (p1 ) 2 = Tr Q Q 2mc 2mc i,j=1,2      X pˆ2 − mc pˆ1 − mc ¯ Q Q . v¯i (p2 )Qvj (p1 ) 2 = Tr 2mc 2mc i,j=1,2

QED

November 8, 2017

(56)

(57) (58) (59)

20 / 125

Example: Klein’s paradox I 1D motion of a particle in the presence of the presence of the potential step  0, z 0

(60)

Stationary solutions i

Ψ(z, t) = ψ(z)e − ~ Et

(61)

for z < 0:   i ψinc = ae ~ p1 z  

1 0 cp1 E +mc 2

   

(62)

0   i ψrefl = be − ~ p1 z  

1 0





   + b 0 e − ~i p1 z  cp1   − E +mc 2 0 QED



0 1 0 cp1 E +mc 2

  

November 8, 2017

(63)

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Example: Klein’s paradox II with c 2 (p12 + (mc)2 ) = E 2

(64)

for z > 0:   i ψtrans = de ~ p2 z  



1 0 cp2 E −V0 +mc 2



   + d 0 e ~i p2 z   

0



0 1 0 2 − E −Vcp+mc 2 0

  

(65)

with c 2 (p22 + (mc)2 ) = (E − V0 )2

(66)

a+b =d

(67)

p2 E + mc 2 d p1 E − V0 + mc 2

(68)

Continuity conditions:

a−b =

QED

November 8, 2017

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Example: Klein’s paradox III b0 = d 0 = 0

(69)

(i.e. no spin flip) T =

4r , (1 + r )2

R = 1 − T,

r=

p2 E + mc 2 p1 E − V0 + mc 2

(70)

Three cases: 0 < V0 < E : “normal” scattering on the potential step 0 < E < V0 ,

V0 − E < mc 2 , → p2 = i|p2 |: ψtrans → 0

0 < E < V0 , right!

V0 − E > mc 2 : R > 1, T < 0, i.e. “incoming” particle from the

The case of smooth potential: V must increase over the Compton wavelenght λ = The Schwinger field: Es = 1.3 × 1018 V/m

QED

November 8, 2017

~ . mc

23 / 125

The non-relativistic propagator I Schrodinger equation: i~

∂ψ = Hψ(x, t) = [H0 + V (x, t)]ψ(x, t) ∂t

(71)

There must be a linear relation between ψ(x, t) and ψ(x, t 0 ) Z ψ(x 0 , t 0 ) = i dx 0 G (x 0 , t 0 , x, t)ψ(x, t)

(72)

Propagate only forward in time: G + (x 0 , t 0 , x, t) =



or θ(t 0 − t)Ψ(x, t 0 ) = i

Z

G (x 0 , t 0 , x, t) 0

t0 > t t0 < t

(73)

dx 0 G + (x 0 , t 0 , x, t)Ψ(x, t)

G + (x 0 , t 0 , x, t) is the retarded propagator. The advanced propagator:  −G (x 0 , t 0 , x, t) t 0 < t G − (x 0 , t 0 , x, t) = 0 t0 > t QED

November 8, 2017

(74)

(75) 24 / 125

The non-relativistic propagator II

θ(t − t 0 )Ψ(x, t 0 ) = −i

Z

dx 0 G − (x 0 , t 0 , x, t)Ψ(x, t)

(76)

Properties of the propagator: for t 0 > t1 > t G + (x 0 , t 0 , x, t) = i

Z

dx1 G + (x 0 , t 0 , x1 , t1 )G + (x1 , t1 , x, t)

(77)

for t 0 < t1 < t G − (x 0 , t 0 , x, t) = −i

Z

dx1 G − (x 0 , t 0 , x1 , t1 )G − (x1 , t1 , x, t)

QED

November 8, 2017

(78)

25 / 125

The non-relativistic propagator

for t > t1

Z δ(x − x) =

for t < t1

Z δ(x − x) =

dx1 G + (x 0 , t, x1 , t1 )G − (x1 , t1 , x, t)

(79)

dx1 G − (x 0 , t, x1 , t1 )G − (x1 , t1 , x, t)

(80)

QED

November 8, 2017

26 / 125

Integral equation for Ψ I Assume the free particle ∂ψ = H0 ψ(x, t) ∂t and the potential V (x, t) acts between t1 and t1 + ∆t1 . In that interval   ∂ i~ − H0 ψ(x1 , t1 ) = V (x1 , t1 )ψ(x1 , t1 ) ∂t1 i~

(81)

(82)

Up to t = t1 we have  ψ(x, t) = φ(x, t),

i~

 ∂ − H0 φ(x1 , t1 ) ∂t1

(83)

after t1 ψ(x1 , t) = φ(x1 , t) + ∆ψ(x1 , t) i.e.

 i~

or

(84)

 ∂ − H0 [φ(x1 , t) + ∆ψ(x1 , t)] = V (x1 , t1 ) [φ(x1 , t) + ∆ψ(x1 , t)] ∂t1   ∂ i~ − H0 ∆ψ(x1 , t1 ) = V (x1 , t1 ) [φ(x1 , t) + ∆ψ(x1 , t)] ∂t1 QED

November 8, 2017

(85)

(86) 27 / 125

Integral equation for Ψ II

t1Z +∆t1

dt 0 V (x1 , t 0 )φ(x1 , t 0 )

(87)

−i V (x1 , t1 )φ(x1 , t1 )∆t1 ~

(88)

i~∆ψ(x1 , t1 + ∆t1 ) = t1

(NB: we neglected ∆t∆ψ.) or ∆ψ(x1 , t1 + ∆t1 ) =

After the potential is turned off the total solution evolves as a free particle Z ψ(x 0 , t 0 ) = i dx2 G0+ (x 0 , t 0 , x2 , t2 )ψ(x2 , t2 ) ψ(x 0 , t 0 ) = i

Z

But ∆ψ(x2 , t2 ) = i

dx2 G0+ (x 0 , t 0 , x2 , t2 ) [φ(x2 , t2 ) + ∆ψ(x2 , t2 )]

−i ~

Z

dx1 G0+ (x2 , t2 , x1 , t1 )V (x1 , t1 )φ(x1 , t1 )∆t1

QED

November 8, 2017

(89) (90)

(91)

28 / 125

Integral equation for Ψ III Compare with the general relation ψ(x 0 , t 0 ) = i

Z

dx2 G + (x 0 , t 0 , x2 , t2 )ψ(x2 , t2 )

(92)

and find that for the potential acting only a small time interval we have G + (x 0 , t 0 , x, t)

G0+ (x 0 , t 0 , x, t) Z V (x1 , t1 ) + + dx1 G0+ (x 0 , t 0 , x1 , t1 ) G0 (x1 , t1 , x, t)∆t1 ~

=

(93)

If we take into account all the possible moments of switching on the interaction we get G + (x 0 , t 0 , x, t)

=

G0+ (x 0 , t 0 , x, t) Z V (x1 , t1 ) + G0 (x1 , t1 , x, t) d 4 x1 G0+ (x 0 , t 0 , x1 , t1 ) ~

(94)

Up to now we considerred only one interaction. If multiple scattering are included G + (x 0 , t 0 , x, t)

=

G0+ (x 0 , t 0 , x, t) Z V (x1 , t1 ) + + d 4 x1 G0+ (x 0 , t 0 , x1 , t1 ) G (x1 , t1 , x, t) ~ QED

November 8, 2017

(95)

29 / 125

Integral equation for Ψ IV or ψ(x 0 , t 0 )

=

φ(x 0 , t 0 ) + Z V (x1 , t1 ) ψ(x1 , t1 ) d 4 x1 G0+ (x 0 , t 0 , x1 , t1 ) ~

(96)

Lowest order approximation G + (x 0 , t 0 , x, t)

=

G0+ (x 0 , t 0 , x, t) Z V (x1 , t1 ) + G0 (x1 , t1 , x, t) d 4 x1 G0+ (x 0 , t 0 , x1 , t1 ) ~

(97)

ψ(x 0 , t 0 )

=

φ(x 0 , t 0 ) + Z V (x1 , t1 ) d 4 x1 G0+ (x 0 , t 0 , x1 , t1 ) φ(x1 , t1 ) ~

(98)

or

QED

November 8, 2017

30 / 125

Integral equation for Ψ

The transition probability: if a particle is at t → −∞ in the (free) state φi , which evolves at in ψi (t), at t → ∞ we calculate the probability to find the particle in the (free) state φ2 . P21 = |hφ2 (t → ∞)|ψ(t → ∞)i|2

(99)

Assume that hφ2 |φ1 i = 0 and, in the LOPT one defines the S matri element Z V (x1 , t1 ) φ1 (x1 , t1 ) S21 = −i d 4 x1 φ∗2 (x1 , t1 ) ~

QED

November 8, 2017

(100)

31 / 125

The Dirac propagator I The free Dirac propagator (Feynman propagator) is the Green function of the free Dirac equation, solution of the non-homogeneous equation (Pˆ − mc)SF (x − y ) = ~Iˆδ(x − y )

(101)

where Iˆ is the identity 4 × 4 matrix and the components of the four-vector momentum operator are Pµ = i~∂µ . We solve the previous equation in the momentum space, i.e. we calculate the Fourier transform of the propagator SF . Starting from the expression of SF (x − y ) in terms of its Fourier transform SF (p) Z i 1 SF (x − y ) = d 4 pe − ~ p·(x−y ) SF (p) (102) (2π~)4 and using the equation verified by SF (x − y ) and also the Fourier representation of the δ function Z i 1 d 4 pe − ~ p·(x−y ) (103) δ(x − y ) = (2π~)4 we obtain 1 (2π~)4

Z

i

d 4 pe − ~ p·(x−y ) (pˆ − mc)SF (p) = ~ QED

1 (2π~)4

Z

i

d 4 pe − ~ p·(x−y ) November 8, 2017

(104) 32 / 125

The Dirac propagator II

i.e. SF (p) = ~(pˆ − mc)−1 = ~ and SF (x − y ) = ~

1 (2π~)4

Z

i

pˆ + mc p 2 − (mc)2

d 4 pe − ~ p·(x−y )

(105)

pˆ + mc p 2 − (mc)2

(106)

Note that the four variables of integration are independent, i.e. the four vector p is not on mass shell. 2 2 2 The integrand p has poles at p0 = p + (mc) . We will denote the two poles by p0 = ±ζ, with ζ = p 2 + (mc)2 > 0. We change the integration contour in p0 as in the figure

QED

November 8, 2017

33 / 125

The Dirac propagator III

QED

November 8, 2017

34 / 125

The Dirac propagator IV and rewrite the propagator as SF (x − y ) = ~

1 (2π~)4

Z

i

dpe ~ p·(x−y )

Z dp0 C

pˆ + mc p02 − p 2 − (mc)2

(107)

We distinguish two cases: for x 0 > y 0 the integration contour is closed in the lower half-plane and we use the Cauchy theorem taking into account the contribution of the pole p 0 = ζ. We obtain Z 0 i p·(x−y )−ζ(x 0 −y 0 ) γ ζ − γ · p + mc 1 ] ~[ SF (x − y ) = −i dpe (108) (2π~)3 2ζ or, if we define a four vector p on the mass-shell, p = (ζ, p) we can write Z i 1 pˆ + mc SF (x − y ) = −i dpe − ~ p·(x−y ) (2π~)3 2p0

QED

November 8, 2017

(109)

35 / 125

The Dirac propagator V for x 0 < y 0 the calculation is similar, except for integration contour is closed in the upper half-plane. The result is Z 0 i p·(x−y )+ζ(x 0 −y 0 ) −γ ζ − γ · p + mc 1 ] ~[ dpe SF (x − y ) = −i (110) (2π~)3 2ζ or,if we define a four vector p on the mass-shell, p = (ζ, −p) we can write Z i −pˆ + mc 1 dpe ~ p·(x−y ) SF (x − y ) = −i (2π~)3 2p0

(111)

Written in compact form, the propagator is SF (x − y )

=

Z i 1 pˆ + mc dpe − ~ p·(x−y ) (2π~)3 2p0 Z i p·(x−y ) −p 1 ˆ + mc 0 0 −iθ(y − x ) dpe ~ (2π~)3 2p0 −iθ(x 0 − y 0 )

QED

November 8, 2017

(112)

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The Dirac propagator VI An equivalent form can be written as a superposition of free Dirac solutions XZ ¯ i (y , p) SF (x − y ) = −iθ(x 0 − y 0 ) dpΨi (x, p)Ψ

(113)

i=1,2

+iθ(y 0 − x 0 )

XZ

¯ i (y , p) dpΨi (x, p)Ψ

i=3,4

where Ψ1,2 (x, p)

=

Ψ3,4 (x, p)

=

r 1 mc 2 − ~i p·x e u1,2 (p) 3/2 E (2π~) r i p·x 1 mc 2 ~ e v1,2 (p) 3/2 E (2π~)

(114) (115) (116)

NB: instead of changing the integration contour we can use the same result by slightly changing the position of the poles in the complex plane ζ → ζ − i,

−ζ → −ζ + i, QED

(117) November 8, 2017

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The Dirac propagator VII

or, equivalently, we can write the propagator as Z i 1 1 SF (x − y ) = ~ d 4 pe − ~ p·(x−y ) (2π~)4 pˆ − mc + i

(118)

The Feynman propagator SF propagates the positive energy solutions forward in time, and the negative energy solutions backward in time.

QED

November 8, 2017

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The Klein-Gordon equation Energy-momentum relation: pα p α = (mc)2 , pα → i~∂α , i.e. −~2 ∂α ∂ α φ = (mc)2 φ [∂α ∂ α + µ2 ]φ = 0,

µ=

(119) mc ~

(120)

Plane wave solutions φ(x) ∼ e ±ik·x ,

k · k = µ2 ,

k=

p

µ2 + k 2 , k



(121)

p = ~k: energy-momentum four vector; negative/positive energy solutions. The four current density [∂α ∂ α + µ2 ]φ = 0 α

2

∗

[∂α ∂ + µ ]φ = 0

φ∗ |

(122)

φ

(123)

φ∗ [∂α ∂ α + µ2 ]φ − φ[∂α ∂ α + µ2 ]φ∗ = 0

(124)

∂α [φ∗ ∂ α φ − φ∂ α φ∗ ] = 0

(125)

QED

November 8, 2017

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The Klein-Gordon equation The four current density vector jα =

i~ ∗ α [φ ∂ φ − φ∂ α φ∗ ] 2m

(126)

i i~ h ∗ ˙ ˙∗ φ φ − φ φ 2mc 2

(127)

Correct dimensions + NR limit. ρ˜ =

has arbitrary sign (because φ and φ˙ are arbitrary). Define charge density   ∂φ∗ ie~ ∗ ∂φ ρ= φ − φ 2mc 2 ∂t ∂t The total charge ie~ Q= 2mc 2

Z

  ∂φ∗ ∗ ∂φ dx φ −φ ∂t ∂t

(128)

(129)

is conserved

QED

November 8, 2017

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The Klein-Gordon equation

Normalized solutions

i

f (±) (p; x) =

1 e ∓ ~ p·x p (2π~)3/2 2Ep

(130)

The Klein-Gordon scalar product Z hf (x), g (x)i

= ≡

↔

dr f ∗ (x)i~∂0 g (x)   Z ∂f ∗ (x) ∂g (x) − g (x)i~ . dr f ∗ (x)i~ ∂t ∂t

(131)

The orthogonality relations at constant time hf ± (p1 ; x), f ± (p2 ; x)i = ±δ(p1 − p2 ),

QED

hf + (p1 ; x), f − (p2 ; x)i = 0;

November 8, 2017

(132)

41 / 125

The Klein-Gordon equation For a superposition of positive energy solutions Z φ(+) = dpa+ (p)f (+) (p; x)

(133)

the total charge Z e~ dp|a+ (p)|2 2 2mc similar, for a superposition of positive energy solutions Z φ(−) = dpa+ (p)f (+) (p; x) Q=

(134)

(135)

the total charge e~ Q=− 2mc 2

Z

dp|a− (p)|2

(136)

(different signs).

QED

November 8, 2017

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The Klein-Gordon equation The Klein-Gordon equation in al electromagnetic field: P µ → Πµ = P µ − eAµ or h i ∂µ ∂ µ + µ2 φ(x) = V φ(x) (137) with V (x) = ie(∂µ Aµ + Aµ ∂µ ) − e 2 Aµ Aµ

(138)

The density current becomes ↔

jµ ∼ φ∗ ∂µ φ − e 2 Aµ Aµ φ∗ φ Z hf (x), g (x)i

= Z



(139)

  ↔ dr f ∗ (x) i~∂0 − 2ceA0 (φ) g (x)

dr f ∗ (x)i~

(140)

∂g (x) ∂f ∗ (x) − g (x)i~ − 2ceA0 (φ)f ∗ (x)g (x) . ∂t ∂t 

QED

November 8, 2017

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The Klein-Gordon propagator

h

i ∂ α ∂α + µ2 ∆F (x − y ) = −δ(x − y ) Z i 1 1 ∆F (x − y ) = ~ d 4 pe − ~ p·(x−y ) 2 4 (2π~) p − (mc)2 + i

(141) (142)

In momentum space ∆F (p) =

1 p 2 − (mc)2 + i

(143)

Expansion in free solutions ∆F (x, x 0 )

= −

Z −i~θ(t − t 0 ) dpf+ (p, x)f+∗ (p; x 0 ) Z i~θ(t 0 − t) dpf− (p; x)f−∗ (p; x 0 ) ;

QED

November 8, 2017

(144)

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The Klein-Gordon propagator

∆F (x 0 , x)φ(x) = Z φ(x) =

Z

↔

dr ∆F (x 0 , x)[i~ ∂ 0 ]φ(x),

(145)

Z dp c+ (p)f+ (p; x) +

dp c− (p)f− (p; x) ≡ φ+ (x) + φ− (x)

∆F (x 0 , x)φ(x) = −i~θ(t 0 − t)φ+ (x 0 ) + i~θ(t − t 0 )φ− (x 0 )

(146) (147)

i.e. propagates forward in time the positive energy solutions and backward in time the negative energy solutions.

QED

November 8, 2017

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Derivation and properties of Gordon-Volkov solutions

The Klein-Gordon Hamiltonian for a particle of mass m and electric charge e < 0 in the laser field A(φ) is, H(x) = −c 2 Πµ Πµ + m2 c 4 ,

Πµ = i~∂µ − eAµ (φ).

(148)

and the Volkov solutions of the corresponding equation H(x)ψ(x) = 0,

(149)

are i

i

Rφ

∓ ~ p·x± 2~(n·p) dχ[e 1 φ0 e ψ± (p; x) = √ 2E (2π~)3/2

QED

2 2

A (χ)∓2eA(χ)·p ]

.

November 8, 2017

(150)

46 / 125

Derivation and properties of Gordon-Volkov solutions

The inner-product definition:   Z ↔ hf (x), g (x)i = dr f ∗ (x) i~∂0 − 2ceA0 (φ) g (x) (151)   Z ∂g (x) ∂f ∗ (x) dr f ∗ (x)i~ − g (x)i~ − 2ceA0 (φ)f ∗ (x)g (x) . ∂t ∂t Orthogonality hψ± (p1 ; x), ψ± (p2 ; x)i = ±δ(p1 − p2 ),

QED

hψ+ (p1 ; x), ψ− (p2 ; x)i = 0;

November 8, 2017

(152)

47 / 125

Derivation and properties of Gordon-Volkov solutions

The general form of the completeness relation for a set of functions {φi } orthogonal with respect to an inner product, hφi , φk i = di δik , is I =

(153)

X 1 φi hφi , ·i di i

(154)

where I is the unit operator. For us: Z Z I = dpψ+ (p)hψ+ (p), ·i − dpψ− (p)hψ− (p), ·i

(155)

and, using the definition of the Klein-Gordon scalar product one sees that the above relation is equivalent to the pair of relations

QED

November 8, 2017

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Derivation and properties of Gordon-Volkov solutions

Z I1 ≡

  ∗ ∗ dp ψ+ (p; x)ψ+ (p; x 0 ) − ψ− (p; x)ψ− (p; x 0 ) = 0

(156)

and ∗ ∗    ∂ψ+ (p; x 0 ) ∂ψ− (p; x 0 ) dp ψ+ (p; x) i~ − ψ− (p; x) i~ (157) ∂t ∂t 
∗ ∗ 2ceA0 (φ0 ) ψ+ (p; x)ψ+ (p; x 0 ) − ψ− (p, x)ψ− (p; x 0 ) = δ(r − r 0 ) ;

Z I2

≡ −

Using I1 one simplifies I2   ∗  ∗  Z ∂ψ+ (p; x 0 ) ∂ψ− (p; x 0 ) I2 = dp ψ+ (p; x) i~ − ψ− (p; x) i~ . ∂t ∂t K (x 0 , x)φ(x) =

Z

(158)

↔

dr K (x 0 , x)[i~ ∂ 0 − 2ceA0 (χ)]φ(x),

QED

November 8, 2017

(159)

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The transition amplitude “Usual” expression Aif = −

i c~

Z

∗ d 4 xψ+ (p2 ; x)Hint (x)ψ+ (p1 ; x) .

(160)

Q: Is it valid for Klein-Gordon case? The Klein-Gordon Feynman propagator K (x, x 0 )

= −

Z ∗ −i~θ(t − t 0 ) dpψ+ (p, x)ψ+ (p; x 0 ) Z ∗ i~θ(t 0 − t) dpψ− (p; x)ψ− (p; x 0 ) ;

The action of the Feynman propagator on an arbitrary function Z Z φ(x) = dp c+ (p)ψ+ (p; x) + dp c− (p)ψ− (p; x) ≡ φ+ (x) + φ− (x)

QED

November 8, 2017

(161)

(162)

50 / 125

The transition amplitude

K (x 0 , x)φ(x) = −i~θ(t 0 − t)φ+ (x 0 ) + i~θ(t − t 0 )φ− (x 0 ) 0

(163) 0

which means that K (x , x) propagates forward in time any function φ+ (x ) originating from positive energy solutions, and backward in time the functions φ− (x 0 ) originating from negative energy solutions. The role played by the Feynman propagator of the Klein-Gordon equation for the charged particle in a plane-wave electromagnetic field is identical to that of the free propagator in the free particle case. Using the properties of the Volkov solutions, one can easily see that K (x 0 , x) is also a Green function of the Klein-Gordon equation. Indeed, using the Hamiltonian (148) and the expression (161) of K (x, x 0 ), one obtains by direct calculation H(x)K (x, x 0 ) = −~2 δ(t − t 0 )δ(r − r 0 )

QED

(164)

November 8, 2017

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The transition amplitude

The particle in the laser field and an interaction H = −c 2 Πµ Πµ + m2 c 4 + Hint (x) ≡ H0 + Hint (x)

(165)

We follow exactly the same steps as in the case of the free particle under the action of Hint , assumed of finite duration, lim Hint (x) = 0

(166)

t→±∞

but use Volkov states instead of free plane-waves; the S matrix element between the initial state and a Volkov state of positive energy and momentum p2 is Sif = hψ+ (p2 ; x), ψi (x)i

.

(167)

t→∞

QED

November 8, 2017

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The transition amplitude In the previous equation ψi (x) is the solution of the Klein-Gordon equation with the Hamiltonian (165) which evolves from the initial state, assumed a Volkov state of positive energy and momentum p1 state at t → −∞; for the case of a finite laser pulse, ψi (x) reduces at t → −∞ to a plane wave free solution. The solution ψi (x) obeys the integral equation Z 1 ψi (x) = ψ+ (p1 ; x) + 2 d 4 yK (x, y )Hint (y )ψi (y ) (168) ~ c with K (x, y ) is the Green function (161) of the Klein-Gordon equation, analyzed before. In the first order perturbation theory with respect to Hint (x), ψi (x) is approximated as Z 1 ψi (x) ≈ ψ+ (p1 ; x) + 2 (169) d 4 yK (x, y )Hint (y )ψ+ (p1 ; y ) ~ c and the S matrix becomes Sif ≈ δ(p1 − p2 ) + Aif

QED

(170)

November 8, 2017

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The transition amplitude

with

Z 1 d 4 x 0 ψ+ (p1 ; x 0 )Hint (x 0 ) lim hψ+ (p1 ; x), K (x, x 0 )i. (171) t→∞ ~2 c Using in the previous equation the expression (161) of the propagator and the properties of the Volkov solutions one obtains Z i ∗ d 4 xψ+ (p2 ; x)Hint (x)ψ+ (p1 ; x) . (172) Aif = − c~ Aif =

QED

November 8, 2017

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The principle of least action Classical mechanics Zt2 I =

q = {q1 , q2 , . . . qN }

dtL[q(t), q(t)], ˙

(173)

t1

minimum for the real motion. If Q(t) is the real trajectory and q(t) = Q(t) + δq(t) Zt2 I (q) = I (Q) +

dt

δI (Q)δq(t) δq(t)

(174)

t1

and

δI (Q) = 0 δq(t) Zt2

 dt

δI =

(175)

 ∂L ∂L d δq(t) + δq(t) ∂q(t) ∂ q(t) ˙ dt

(176)

t1

QED

November 8, 2017

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The principle of least action ∂L d ∂L − =0 (177) ∂q(t) dt ∂ q(t) ˙ NB: Lagrange function defined only up to the total derivative with respect to t of an arbitrary function f (q, t) Hamiltonian formulation H(p, q) = pi q˙ i (p, q) − L[q, q(p, ˙ q)]

(178)

∂H ∂H , p˙i = − , ∂pi ∂qi df ∂f = + {H, f } dt ∂t Reobtaining the Hamilton equations from the principle of least action: q˙ i =

Zt2 L(q, q)dt ˙ = pdq − Hdt,

(180)

Zt2 [pdq − Hdt]

dtL[q(t), q(t)] ˙ =

I =

(179)

t1

QED

(181)

t1

November 8, 2017

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The principle of least action

Zt2 δI =

 dt

   d ∂H ∂H δp(t) + p δq − δq q˙ − ∂p dt ∂q

(182)

t1

Canonical transformations: (p, q) ↔ (p 0 , q 0 )

(183)

0

such that there is a new function H and q˙ i0 =

∂H 0 , ∂pi0

p˙i = −

∂H 0 , ∂qi0

(184)

A transformation is canonical if the Poisson brackets are invariant  X  X  df dg df dg df dg df dg − 0 0 {f , g } = − = 0 0 dpi dqi dqi dpi dpi dqi dqi dpi i i

QED

November 8, 2017

(185)

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The principle of least action The continuum case: we have a Lagrange density L(r, t) ≡ L(x); assume that the Lagrange denisty depends on the fields φr (x) and the gradients ∂µ φr (x) ≡ φr ,µ L(x) = L(φr (x), φr ,µ (x)) We define the action

Z I =

(186)

d 4 xL(φr , φr ,µ )

(187)

Ω

and assume that δφr = 0 on the frontier Γ(Ω)   Z ∂L ∂L δφr + δφr ,µ δI = d 4 x ∂φr ∂φr ,µ

(188)

Ω

Z δI =

   Z ∂L ∂L ∂L 4 d x − ∂µ δφr + d x∂µ δφr ∂φr ∂φr ,µ ∂φr ,µ 4



Ω

(189)

Ω

The Lagrange equations:   ∂L ∂L − ∂µ =0 ∂φr (x) ∂φr ,µ (x)   ∂L ∂ ∂L − =0 ∂φr (x) ∂x µ ∂φr ,µ (x) QED

(190) (191) November 8, 2017

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The principle of least action At fixed t one divide the space on small cells δxi and the discrete coordinates are the fields in each cell qri = φr (xi , t) ≡ φr (i, t) The Lagrange function X X L(t) = δxi Li ≡ δxi L(φr (i, t), φ˙ r (i, t), φr (i 0 , t)) (192) i

i

NB: i 0 in the Lagrange density comes from the derivatives of teh fields expressed by finite differences. The conjugate momenta pri (t) =

∂L ∂L ≡ ≡ πr (i, t)δxi ˙ ∂ q˙ ri ∂ φr (i, t) πr (i, t) =

(193)

∂L ∂ φ˙ r (i, t)

(194)

The Hamilton function H=

X i

pri q˙ ri − L =

X

h i δxi πr (i, t)φ˙ r (i, t) − Li

(195)

i

QED

November 8, 2017

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The principle of least action

If the volumes δxi are small one take the continuum limit πr (x) =

∂L ∂ φ˙ r

(196)

Z

d 3 xL(φr , φr ,µ ) Z Z h i H = d 3 xH(x) = d 3 x πr φ˙ r − L(φr , φr ,µ ) L=

QED

November 8, 2017

(197) (198)

60 / 125

Canonical quantization

Replace φ and π by operators  1 φr (j, t), πs (j 0 , t) = i~δjj 0 δrs δxj     φr (j, t), φs (j 0 , t) = πr (j, t), πs (j 0 , t) = 0 

(199) (200)

In the continuum limit  φr (x, t), πs (x0 , t) = i~δrs δ(x − x0 )     φr (x, t), φs (x0 , t) = πr (x, t), πs (x0 , t) = 0 

(201) (202)

NB: canonical commutation relations at the same time

QED

November 8, 2017

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Conservation laws For an arbitrary operator O, in the Heisenberg picture i~

dO(t) = [H, O(t)] dt

(203)

then O is a constant of motion if [H, O] = 0. Invariance properties of a system: a unitary transformation: |Ψi → |Ψ0 i = U|Ψi,

O → UOU †

(204)

with UU † = U † U = I . Under a unitary transformation the commutation relations are invariant. Infinitesimal transformation: consider first U = e iαT ,

T = T†

(205)

for small α → δα U ≈ I + iδαT

(206)

O = O + δO = O + iδα[T , O]

(207)

0

QED

November 8, 2017

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Conservation laws

For the therory to be invariant under the transformation H must be invariant, i.e. δH = 0. If one puts O = H in the previous calculation, one obtains that the theory is invariant if [T , H] = 0, i.e. T (the transformatrion generator) is a constant of motion.

QED

November 8, 2017

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Conservation laws

In the Lagrangean theory we use the invariance of L under a symmetry transformation. In the general case we shall obtain equation of the form ∂f α =0 ∂x α

(208)

with f α : function of the fields and fields derivatives (i.e. one obtains continuity-”like” equations).

QED

November 8, 2017

64 / 125

Conservation laws One defines Fα = and

1 dF 0 =− c dt

Z

Z

d 3 xf α

d 3 x∂j f j ≡ −

Z

(209)

d 3 x∇ · f = 0

(210)

if we assume that the fields vanish at the infinity. Then F 0 is a conserved quantity. The components of the four-vector f are the components of a four-current. Example: consider a transformation φr → φr + δφr which leaves the Lagrange density invariant. Then   ∂L ∂ ∂L ∂L δφr + δφr ,µ = δφ =0 (211) δL = r ∂φr ∂φr ,µ ∂x α ∂φr ,α The conserved four current is fα =

∂L δφr ∂φr ,α

QED

(212)

November 8, 2017

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Conservation laws

Then f0 =

∂L ∂L δφr = c δφr = cπr δφr ∂φr ,0 ∂ φ˙ r

and F0 = c

Z

(213)

d 3 xπr (x)δφr (x)

QED

(214)

November 8, 2017

66 / 125

Charge conjugation If we have complex fields then φ and φ∗ are treated as independent (or Re(φ) and Im(φ) are independent). Assume L is invariant under the transformation φ†r → e −i φ†r

φr → e i φr ,

(215)

For infinitesimal transformation φr → (1 + i)φr ,

φ†r → (1 − i)φ†r

(216)

i.e. δφr = iφr ,

δφ†r = −iφ†r

(217)

The conserved quantity F 0 = ic

Z

iq ~

Z

h i d 3 x πr φr − φ†r πr†

(218)

h i d 3 x πr φr − φ†r πr†

(219)

Up to a constant Q=−

QED

November 8, 2017

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Charge conjugation

By direct calculation   Z   iq d 3 x0 πr (x 0 )φr (x 0 ) − φ†r (x 0 )πr† (x 0 ) , φs (x) [Q, φs (x)] = − ~ [Q, φs (x)] = −

iq ~

Z

  d 3 x0 πr (x 0 ), φs (x) φr (x 0 ) = −qφs (x)

(220)

(221)

In |Q 0 i is the eigenvector of Q for the eigenvalue Q 0 then φr (x)|Q 0 i is eigenvector for the eigenvalue Q 0 − q. I.e. φr creates particles of charge −q and absorbs particles of charge q. Analogous φ†r creates particles of charge q and absorbs particles of charge −q. The conservation of charge means [Q, H] = 0 and the coresponding uinitary transformation is U = e iαQ (222)

QED

November 8, 2017

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Rotational and translational invariance Consider an infinitesimal transformation xα → xα0 = xα + αβ x β + δα

(223)

with αβ = −βα and αβ , δα are small. Assume that 1 φr (x) → φ0r (x 0 ) = φr (x) + αβ Srsαβ φs (x) 2 Invariance of the Lagrange density

(224)

L(φr (x), φr ,µ (x)) = L(φ0r (x 0 ), φ0r ,µ (x 0 ))

(225)

We derive the continuity equations: define δφr (x) = φ0r (x) − φr (x) δT φr (x) =

φ0r (x 0 )

− φr (x) =

φ0r (x 0 )

QED

0

(226) 0

− φr (x ) + φr (x ) − φr (x)

November 8, 2017

(227)

69 / 125

Rotational and translational invariance δT φr (x) = δφr (x 0 ) +

∂φr ∂φr δxβ = δφr (x) + δxβ ∂xβ ∂xβ

(228)

Same calculation: L(φ0r (x 0 ), φ0r ,µ (x 0 )) − L(φr (x), φr ,µ (x)) = δL + Previous result

∂L δxβ ∂xβ

  ∂ ∂L δL = δφr ∂x α ∂φr ,α    ∂ ∂L ∂φr δL = δ φ − δx r T β ∂x α ∂φr ,α ∂xβ

(229)

(230) (231)

The conserved curent in the continuity equation ∂f α = 0, ∂x α

fα =

with T αβ =

∂L δT φr − T αβ δxβ ∂φr ,α

(232)

∂L ∂φr − Lg αβ ∂φr ,α ∂xβ QED

(233) November 8, 2017

70 / 125

Translations For a translation αβ = 0, i.e. δT φr = 0. f α = −T αβ δβ

(234)

with arbitrary δβ . The continuity equations (4 conserved quantities) ∂T αβ =0 ∂x α with

(235)

∂L ∂φr − Lg αβ ∂φr ,α ∂xβ   Z Z ∂φr cP β = d 3 xT 0β = d 3 x cπr − Lg 0β ∂xβ T αβ =

(236) (237)

Z

cP 0

=

Pj

=

h i Z d 3 x πr (x)φ˙ r (x) − L(x) = d 3 xH(x) = H   Z ∂φr (x) d 3 x πr (x) ∂xj

(238) (239)

(energy-momentum tensor) QED

November 8, 2017

71 / 125

Translations Pj =

Z

  ∂φr (x) d 3 x πr (x) ∂xj

(240)

the commutators of P with the field operators are   h i Z ∂φr (x) ∂φs (x 0 ) P j , φs (x 0 ) = d 3 x πr (x) , φs (x 0 ) = −i~ ∂xj ∂xj0 analogous calculation (with an integration by parts)   h i Z ∂φr (x) ∂πs (x 0 ) P j , πs (x 0 ) = d 3 x πr (x) , πs (x 0 ) = −i~ ∂xj ∂xj0

(241)

(242)

Then, for an arbitrary operator F (x) ≡ F (φr (x), πr (x)) we have [P j , F ] = −i~

∂F ∂xj

(243)

[P 0 , F ] = −i~

∂F ∂x0

(244)

and

QED

November 8, 2017

72 / 125

Rotations

For a rotation δα = 0, δT φr (x) = 21 αβ Srsαβ φs (x). The continuity equation ∂Mαβγ =0 ∂x α with Mαβγ =

(245)

  ∂L βγ Srs φs + x β T αγ − x γ T αβ ∂φr ,α

(246)

with six conserved quantities (Mαβγ = −αγβ) Z cM βγ = d 3 xM0βγ Z h i = d 3 x x β T 0γ − x γ T 0β + cπr (x)Srsβγ φs (x)

QED

November 8, 2017

(247) (248)

73 / 125

The real Klein-Gordon field Consider L=

1 (φ,α φ,α − µ2 φ2 ) 2

(249)

with the Lagrange equations   ∂L ∂L =0 − ∂α ∂φ(x) ∂φ,α (x)

(250)

i.e. [∂α ∂ α + µ2 ]φ = 0,

[ + µ2 ]φ = 0

(251)

(the Klein-Gordon equation). Energy-momentum relation: pα p α = (mc)2 , pα → i~∂α , i.e. −~2 ∂α ∂ α φ = (mc)2 φ [∂α ∂ α + µ2 ]φ = 0, The canonical momentum π=

µ=

(252) mc ~

(253)

∂L 1 = 2 φ˙ c ∂ φ˙ QED

(254)

November 8, 2017

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The real Klein-Gordon field Canonical quantization  φ(x, t), π(x 0 , t) = i~δ(x − x 0 )     φ(x, t), φ(x 0 , t) = π(x, t), π(x 0 , t) = 0 

(255) (256)

h

i ˙ 0 , t) = i~c 2 δ(x − x 0 ) φ(x, t), φ(x i   h ˙ ˙ 0 , t) = 0 φ(x, t), φ(x 0 , t) = φ(x, t), φ(x

(257) (258)

Expand φ(x) in plane wave solutions (discrete sums, periodic conditions in a box of volume V ) φ(x) = φ+ (x) + φ− (x) (259)   1/2 2 X ~c φ+ (x) = a(k)e −ik·x (260) 2V ωk k

φ− (x) =

X  ~c 2 1/2 † a (k)e ik·x 2V ωk

(261)

k

with ωk = ck 0 = c

p µ2 + k 2 , QED

Ek = ~ωk

(262) November 8, 2017

75 / 125

The real Klein-Gordon field The commutation relations are obeyed if h i a(k), a† (k 0 ) = δk,k 0 

(263)

i  h a(k), a(k 0 ) = a† (k), a† (k 0 ) = 0

(264)

The energy-momentum tensor Z h i Z ˙ cP 0 = d 3 x π(x)φ(x) − L(x) = d 3 xH(x) = H   Z ∂φ(x) j 3 P = d x π(x) ∂xj with

∂L 1 = 2 φ˙ ˙ c ∂φ   Z 1 1 2 2 2 2 3 ˙ H= d x φ + (∇φ) + µ φ , 2 c2 π=

QED

(265) (266)

(267) Z P=−

d 3x

1 ˙ φ∇φ c2

November 8, 2017

(268)

76 / 125

The real Klein-Gordon field

One obtains

  1 a† (k)a(k) + 2 k   X 1 P= ~k a† (k)a(k) + 2

H=

X

(269)

~ωk

(270)

k

a† (k)a(k) interporeted as the number operator. The vacuum energy is infinite E0 =

1X ~ωk 2

(271)

k

simply eliminated by a shift of the zero energy level. X α † P α = (H/c, P) = ~k a (k)a(k)

(272)

k

QED

November 8, 2017

77 / 125

The real Klein-Gordon field

Normal ordering of the operators N[a1 a2 a3† ] = a3† a1 a2

(273)

for ai ≡ a(ki ) N[φ(x)φ(y )] = φ+ (x)φ+ (y ) + φ− (x)φ+ (y ) + φ− (y )φ+ (x) + φ− (x)φ− (y ) h0|N[. . .]|0i = 0

(274) (275)

The normal ordering introduces a constant (e.g. eliminates the infinite zero energy).

QED

November 8, 2017

78 / 125

The real Klein-Gordon field The field operators in Schrodinger picture. The field operators defined previously are time-dependent (i.e. in the Heisenberg picture). In the Schr¨ odinger picture φ(x) = φ+ (x) + φ− (x) φ+ (x) =

X k

φ− (x) =

~c 2 2V ωk

1/2

(276)

a(k)e ik·x

(277)

X  ~c 2 1/2 † a (k)e −ik·x 2V ωk

(278)

k

Commutation relations 

 φ(x), π(x 0 ) = i~δ(x − x 0 ) i

(279)

i

φ(x) ≡ φ(x, t) = e ~ Ht φ(x)e − ~ Ht φ(x) ≡ φ(x, t) = e

i P·x ~

QED

φ(0, 0)e

(280)

− ~i P·x

(281)

November 8, 2017

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The complex Klein-Gordon field

h i L = N φ†,α φ,α − µ2 φ† φ

(282)

φ and φ† are independent. The Lagrange equations [∂α ∂ α + µ2 ]φ = 0,

[∂α ∂ α + µ2 ]φ† = 0

(283)

∂L 1 ∂L 1 = 2 φ˙ † π † = = 2 φ˙ † ˙ ˙ c c ∂φ ∂φ The equal time commutation relations h i φ(x, t), φ˙ † (x 0 , t) = i~c 2 δ(x − x 0 ) π=

(284)

(285)

i h i   h ˙ 0 , t) = 0 ˙ ˙ 0 , t) = φ(x, t), φ(x φ(x, t), φ(x 0 , t) = φ(x, t), φ(x h i h i φ(x, t), φ† (x 0 , t) = φ˙ † (x, t), φ˙ † (x 0 , t) = 0

QED

November 8, 2017

(286) (287)

80 / 125

The complex Klein-Gordon field The Fourier expansion φ(x) = φ+ (x) + φ− (x) X  ~c 2 1/2 + φ (x) = a(k)e −ik·x 2V ωk

(288) (289)

k

φ− (x) =

X  ~c 2 1/2 † b (k)e ik·x 2V ωk

(290)

k

φ† (x) = φ†+ (x) + φ†− (x) X  ~c 2 1/2 † φ†− (x) = a (k)e ik·x 2V ωk

(291) (292)

k

φ†+ (x) =

X  ~c 2 1/2 b(k)e −ik·x 2V ωk

(293)

k

QED

November 8, 2017

81 / 125

The complex Klein-Gordon field

Commutation relations h

i h i a(k), a† (k 0 ) = b(k), b † (k 0 ) = δk,k 0

(294)

    a(k), a(k 0 ) = b(k), b(k 0 ) = 0 i   h a(k), b(k 0 ) = a† (k), b(k 0 ) = 0

(295) (296)

a, a† , b, b † : creation-absorption operators for two types of particles. Number of particles operators Na (k) = a† (k)a(k), Nb (k) = b † (k)b(k)

(297)

Momentum four vector Pα =

X

h i ~k α a† (k)a(k) + b † (k)b(k)

(298)

k

QED

November 8, 2017

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The complex Klein-Gordon field

Charge current density sα = −

i iq h α † N (∂ φ )φ − (∂ α φ)φ† ~

(299)

with ∂ α sα = 0

(300)

The conserved charge Q=−

iq ~c 2

Z

h i X d 3 xN φ† φ − φφ† = q [Na (k) − Nb (k)]

(301)

k

a and b : particle with identical properties but different sign charge.

QED

November 8, 2017

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Covariant commutation relations Consider the real field. Commutation relations at different moments  +    φ (x), φ+ (y ) = φ− (x), φ− (y ) = 0

(302)

In general  +  φ (x), φ− (y ) 6= 0

(303)

Direct calculation 

 φ+ (x), φ− (y )

=

h i ~c 2 X 1 a(k), a† (k 0 ) e −ik·x+ik·y √ 2V ωk ωk 0 0 k,k

2

=

~c X 1 −ik·(x−y ) e 2V ωk

(304)

k

The continuum limit  +  φ (x), φ− (y ) =

~c 2 2(2π)3

Z dk

QED

1 −ik·(x−y ) e ≡ i~c∆+ (x, y ) ωk

November 8, 2017

(305)

84 / 125

Covariant commutation relations

 +  φ (x), φ− (y ) ≡ i~c∆+ (x − y ) with ∆+ (x) = 

−ic 2(2π)3

Z dk

(306)

1 −ik·x e ωk

(307)

 φ− (x), φ+ (y ) ≡ −i~c∆+ (y − x) ≡ i~c∆− (x − y )

(308)

[φ(x), φ(y )] = i~c∆(x − y )

(309)

with ∆(x) = ∆+ (x) + ∆− (x) =

QED

−c (2π)3

Z dk

1 sin(k · x) ωk

November 8, 2017

(310)

85 / 125

Covariant commutation relations

∆(x) is a solution of the homogeneous KG equation. (x + µ2 )∆(x − y ) = 0

(311)

∆(x) vanishes if x · x < 0. Equivalent form ∆(x) =

−i (2π)3

Z

d 4 kδ(k 2 − µ2 )(k0 )e −ik·x

(312)

where (k0 ) is the sign of k0 . NB: ∆(x) is invariant under Lorentz transformation; then [Φ(x, t), Φ(y , t)] = ∆(x − y , 0) = 0

(313)

and the lorentz invariance require that ∆(x, y ) = 0 for (x − y )2 < 0, i.e. causality.

QED

November 8, 2017

86 / 125

Covariant commutation relations Equivalent forms: as contour integrals ∆± (x) =

−1 (2π)4

Z

d 4k

e −ik·x k 2 − µ2

(314)

C±

with

QED

November 8, 2017

87 / 125

Covariant commutation relations

∆(x) = ∆+ (x) + ∆− (x) =

−1 (2π)4

Z

d 4k

e −ik·x k 2 − µ2

(315)

C

QED

November 8, 2017

88 / 125

The meson propagator

Previous definition  +  φ (x), φ− (y ) = i~c∆+ (x − y )

(316)

  i~c∆+ (x − x 0 ) = h0| φ+ (x), φ− (x 0 ) |0i = h0|φ+ (x)φ− (x 0 )|0i   i~c∆+ (x − x 0 ) = h0| φ(x), φ(x 0 ) |0i

(317)

Then (318)

Define the time-ordered product as T {φ(x)φ(x 0 )} =



φ(x)φ(x 0 ) φ(x 0 )φ(x)

t > t0 t0 > t

(319)

equivalent form T {φ(x)φ(x 0 )} = θ(t − t 0 )φ(x)φ(x 0 ) + θ(t 0 − t)φ(x 0 )φ(x)

QED

November 8, 2017

(320)

89 / 125

The meson propagator

One defines i~c∆F (x − x 0 ) = h0|T {φ(x)φ(x 0 )}|0i

(321)

i~c∆F (x − x 0 ) = θ(t − t 0 )∆+ (x − x 0 ) − θ(t 0 − t)∆− (x − x 0 )

(322)

or Example: for t > t 0 it represents a meson created at x 0 and anihilated at x (later). For t 0 < t is represents a meson created at x 0 and anihilated at x. ∆F is named “the Feynman propagator of the KG field”, of simply the KG propagator. Equivalent forms: Z 1 e −ik·x 4 ∆F (x) = d k (323) (2π)4 k 2 − µ2 + i or as contour integrals.

QED

November 8, 2017

90 / 125

The meson propagator

QED

November 8, 2017

91 / 125

The meson propagator

The case of complex field i~c∆F (x − x 0 ) = h0|T {φ(x)φ+ (x 0 )}|0i

QED

(324)

November 8, 2017

92 / 125

The Number Representation for Fermions

Previous case (KG): we defined ar ,

ar† ,

[ar , as† ] = δrs

(325)

and the number operator Nr = ar† ar

(326)

From [Nr , as ] = −δrs as

[Nr , as† ] = δrs as†

(327)

if follows that Nr is the operator of number of particles and the definition of the vacuum as the state for which as |0i = 0. A Fock state is built by action of ar† |nr i ∼ (ar† )nr |0i

QED

(328)

November 8, 2017

93 / 125

The Number Representation for Fermions

The case when ar , ar† anticommute: [ar , as† ]+ = δrs ,

(ar )2 = (ar† )2 = 0

(329)

[Nr , as† ] = δrs as†

(330)

The relations [Nr , as ] = −δrs as

are still valid, i.e. we can define the number of particle also in this case. Other relations: Nr2 = Nr ,

Nr (Nr − 1) = 0

(331)

i.e. the occupation number can be 0 or 1. |1r i = ar† |0i

QED

(332)

November 8, 2017

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The Dirac equation

i~

∂ψ = [cα · P + βmc 2 ]ψ(x) ∂t

(333)

or (Pˆ − mc)ψ(x) = 0

(334)

Pˆ = γ µ Pµ = γ µ i~∂µ

(335)

[γ µ , γ ν ]+ = 2g µν

(336)

with and ¯ The adjoint is ψ(x) = ψ (x)γ , solution of the equation †

0

←

¯ ψ(x)[i~ ∂µ γ µ + mc] = 0

QED

(337)

November 8, 2017

95 / 125

The Dirac equation The Lagrange density: ¯ L(x) = c ψ(x) [i~γ µ ∂µ − mc] ψ(x)

(338)

and the fields ψα and ψ¯α are independent (8 components). The conjugate fields πα =

∂L † = i~ψα ∂ ψ˙ α

(339)

∂L =0 ∂ ψ¯˙

(340)

π ¯α =

α

Energy and momentum: according to Z h i Z cP 0 = d 3 x πr (x)φ˙ r (x) − L(x) = d 3 xH(x) = H   Z ∂φr (x) Pj = d 3 x πr (x) ∂xj

(341) (342)

Z H=

h i ¯ d 3 x ψ(x) −i~cγ j ∂j + mc 2 ψ(x) Z P = −i~ d 3 xψ † (x)∇ψ(x) QED

(343) (344) November 8, 2017

96 / 125

The Dirac equation

Angular momentum: Previous general results: For a rotation the continuity equation is ∂Mαβγ =0 ∂x α with Mαβγ =

(345)

  ∂L βγ Srs φs + x β T αγ − x γ T αβ ∂φr ,α

(346)

with conserved quantities (Mαβγ = −Mαγβ ) Z cM βγ = d 3 xM0βγ Z h i = d 3 x x β T 0γ − x γ T 0β + cπr (x)Srsβγ φs (x)

QED

November 8, 2017

(347) (348)

97 / 125

The Dirac equation

In our case: under a rotation 0 ψα → ψα (x 0 ) = ψα −

with

i µν µν σαβ ψβ (x) 4

(349)

i µ ν [γ , γ ] 2

σ µν =

(350)

The angular momentum:   Z Z ~ M = d 3 xψ † (x) [x × (−i~∇)] ψ(x) + d 3 xψ † (x) σ ψ(x) 2

(351)

(angular momentum and spin). We defined σ ≡ (σ 23 , σ 31 , σ 12 )

QED

(352)

November 8, 2017

98 / 125

The Dirac equation The charge conservation: general theory If we have complex fields then φ and φ∗ are treated as independent (or Re(φ) and Im(φ) are independent). Assume L is invariant under the transformation φr → e i φr , φ†r → e −i φ†r (353) For infinitesimal transformation φr → (1 + i)φr ,

φ†r → (1 − i)φ†r

(354)

i.e. δφr = iφr ,

δφ†r = −iφ†r

(355)

The conserved quantity F 0 = ic

Z

Z h i   d 3 x πr φr − φ†r πr† → ic d 3 x πr ψr − ψ¯r π ¯r

(356)

Z h i   iq d 3 x πr φr − φ†r πr† → − d 3 x πr ψr − φ¯r π ¯r ~

(357)

Up to a constant Q=−

iq ~

Z

QED

November 8, 2017

99 / 125

The Dirac equation

Our case:

Z Q=q

d 3 xψ † (x)ψ(x)

(358)

the density four vector: α ¯ s α ≡ (cρ, j ) = cq ψ(x)γ ψ(x)

QED

(359)

November 8, 2017

100 / 125

The Dirac equation Plane wave solutions of the Dirac equation: 1 √ e −ip·x/~ ur (p), V

1 √ e ip·x/~ vr (p), V

r = 1, 2

(360)

with (pˆ − (mc))ur (p) = 0,

(pˆ + (mc))vr (p) = 0

If σp = σ ·

(361)

p |p|

(362)

then σp ur (p) = (−1)r +1 ur (p),

σp vr (p) = (−1)r ur (p),

r = 1, 2

(363)

Normalization condition ur† (p)us (p) = vr† (p)vs (p) = δrs

QED

Ep , mc 2

ur† (p)vs (−p) = 0

November 8, 2017

(364)

101 / 125

Second quantization ψ(x)

=

ψ + (x) + ψ − (x) = i X  mc 2 1/2 h cr (p)ur (p)e −ip·x/~ + dr† (p)vr (p)e ip·x/~ VEp rp

(365)

¯ ψ(x)

=

ψ¯+ (x) + ψ¯− (x) = i X  mc 2 1/2 h dr (p)¯ vr (p)e −ip·x/~ + cr† (p)u¯r (p)e ip·x/~ VEp rp

(366)

Impose the anticommutation [cr (p), cs† (p 0 )]+ = [dr (p), ds† (p 0 )]+ = δrs δpp0

(367)

and the rest are zero. One defines the number of particles operators: Nr (p) = cr† (p)cr (p),

N¯r (p) = dr† (p)dr (p)

(368)

Interpret c, c † , N, d, d † , N¯ as (absorptions/creations/number of particles) operators for independent particles. QED

November 8, 2017

102 / 125

Second quantization The vacuum state is defined as cr (p)|0i = dr (p)|0i = 0

(369)

ψ + (x)|0i = ψ¯+ (x)|0i = 0

(370)

or We want to introduce the normal ordering to ensure zero expectation values of observables on the vacuum state. We must adapt for the anticommutation relations: N[ψα ψβ ]

=

+ − N[(ψα + ψα )(ψβ+ + ψβ− )] = + + ψα ψβ

−

+ ψβ− ψα

+

− + ψα ψβ

+

(371)

− − ψα ψβ

(372)

NB: Compare with the KG expression N[φ(x)φ(y )] = φ+ (x)φ+ (y ) + φ− (x)φ+ (y ) + φ− (y )φ+ (x) + φ− (x)φ− (y )

(373)

With this rule the observables differ up to a constant at most.

QED

November 8, 2017

103 / 125

Second quantization

H

=

X

Ep [Nr (p) + N¯r (p)]

(374)

p[Nr (p) + N¯r (p)]

(375)

rp

P

=

X rp

Q

=

−e

X [Nr (p) − N¯r (p)],

(q → −e < 0)

(376)

rp

One define the longitudinal spin operator (the spin in the direction of motion) as Z ~ Sp = d 3 xN[ψ + (x)σp ψ(x)] (377) 2 One obtains

QED

November 8, 2017

104 / 125

Second quantization

~ Sp cr† (p)|0i = (−1)r +1 cr† (p)|0i 2 † r +1 ~ † Sp dr (p)|0i = (−1) dr (p)|0i 2 Then the states cr† (p)|0i and dr† (p)|0i are stes with positive helicity (r = 1) and negative helicity (r = 2). Sp is the helicity operator. c, d, . . . becomes annihilation/creation operators for particle/antiparticle. (Anti)Commutations of the field operators [ψα (x), ψβ (y )]+ = [ψ¯α (x), ψ¯β (y )]+ = 0  mc  ± ∆± (x − y ) [ψα (x), ψ¯β∓ (y )]+ = i iγ µ ∂µ + ~ αβ Z −1 e −ik·x ∆± (x) = d 4k 2 (2π)4 k − µ2

(378) (379)

(380) (381) (382)

C±

QED

November 8, 2017

105 / 125

Second quantization As a matrix [ψ ± (x), ψ¯∓ (y )]+ = iS ± (x − y ) and

(383)

 mc  ± ∆ (x − y ) S ± (x − y ) = iγ µ ∂µ + ~ ¯ )]+ = iS(x − y ) [ψ(x), ψ(y

(384) (385)

 mc  S(x) = S + (x) + S − (x) = iγ µ ∂µ + ∆(x) ~ Equivalent forms: representation as contour integrals Z −~ pˆ + mc S ± (x) = d 4 pe −ip·x/~ 2 (2π~)4 p − (mc)2

(386)

(387)

C±

S ± (x) =

−~ (2π~)4

Z

d 4 pe −ip·x/~

1 pˆ − mc

(388)

C±

QED

November 8, 2017

106 / 125

The spin statistics theorem

If we impose commutations relations for creation/absorption operators we obtain the energy as X H= Ep [Nr (p) − N¯r (p)] (389) rp

with arbitrary values for N, i.e. no lower bound. For KG. if we impose the anticommutation relations one can not obtain [φ(x), φ(y )]+ = 0,

(x − y )2 < 0

(390)

To avoid difficulties: paticles with integer spin are quantized according to Bose-Einstein, and particles with half integer spin according to Fermi-Dirac.

QED

November 8, 2017

107 / 125

The Fermion propagator Define the time-ordered product as ¯ 0 )} = T {ψ(x)ψ(x



¯ 0) ψ(x)ψ(x ¯ −ψ(x 0 )ψ(x)

t > t0 t0 > t

(391)

equivalent form ¯ 0 )} = θ(t − t 0 )ψ(x)ψ(x ¯ 0 ) − θ(t 0 − t)ψ(x 0 )ψ(x) ¯ T {ψ(x)φ(x

(392)

NB: difference of signs with respect to KG (as in the case of normal ordering) ¯ 0 )|0i = iS + (x − x 0 ) h0|ψ(x)ψ(x 0

−

(393)

0

¯ )ψ(x)|0i ¯ h0|ψ(x = iS (x − x )

(394)

¯ 0 )}|0i = iSF (x − x 0 ) h0|T {ψ(x)ψ(x

(395)

SF (x) = θ(t)S + (x) − θ(−t)S − (x)

(396)

Define

with

QED

November 8, 2017

108 / 125

The Fermion propagator

Obs: S ± (x) are solutions of the homogeneous Dirac equation and SF is the green function. Integral representation Z ~ pˆ + mc d 4 pe −ip·x/~ 2 (397) SF (x) = (2π~)4 p − (mc)2 + i Interpretation: t 0 < t: creation of a particle in x 0 , anihilation in x t < t 0 : creation of a antiparticle in x, anihilatin in x 0

QED

November 8, 2017

109 / 125

The Electromagnetic Interaction and Gauge Invariance The minimal substitution: i~∂µ → i~∂µ − qAµ , define Dµ = ∂µ −

q = −e

(398)

ie Aµ ~c

(399)

The Lagrange density becomes ¯ L(x) = x ψ(x) [i~γ µ Dµ − mc] ψ(x) = L0 + L1

(400)

¯ L0 (x) = x ψ(x) [i~γ µ ∂µ − mc] ψ(x) ¯ A(x)ψ(x) ˆ L1 = e ψ(x) (∼ jµ Aµ )

(401)

with (402)

One must add also Lrad (to be discussed) 1 1 Lrad = − (∂µ Aν )(∂ µ Aν ) − sµ Aµ 2 c

QED

(403)

November 8, 2017

110 / 125

The Electromagnetic Interaction and Gauge Invariance

Gauge invariance Aµ → Aµ + ∂µ f

(404)

¯ µ ψ∂µ f L → L + e ψγ

(405)

then The invariance is restored if the fields are also transformed as ψ(x) → ψ(x)e ief (x)/~c ,

−ief (x)/~c ¯ ¯ ψ(x) → ψ(x)e ,

(406)

With this transformation L0 + L1 are invariant.

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The classical electromagnetic field

The Maxwell equations are written using the antisymmentric field tensor F µν (x) = ∂ ν Aµ (x) − ∂ µ Aν (x) as

(407)

1 µ s (x) = 0 c ∂α F µν (x) + ∂µ F να (x) + ∂ν F αµ (x) = 0 ∂ν F µν (x) −

(408) (409)

µ

with s the charge-current density s µ ≡ (cρ, j )

(410)

The equivalent equation for the potentials Aµ − ∂ µ ∂ν Aν −

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1 µ s =0 c

(411)

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The classical electromagnetic field Gauge invariance Aµ (x) → Aµ (x) + ∂ µ f (x)

(412)

The Lagrange density changes as L(x) → L −

1 µ ∂ sµ f (x) c

(413)

i.e. by adding a four-divergence, which does not change the Lagrange equations. The Mawxell equations can be obtained from the Lagrange density 1 1 L(x) = − Fµν F µν − sµ Aµ , 4 c

(414)

using the components Aµ (x) as independent fields. The corresponding momentum πµ =

1 ∂L = − F µ0 c ∂ A˙ µ

(415)

leads to π 0 = 0, impossible to impose the quantization and canonical commutation relations.

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The classical electromagnetic field Another form of the Lagrange density 1 1 L(x) = − (∂ν Aµ (x))(∂ ν Aµ (x)) − sµ (x)Aµ (x) 2 c

(416)

leads to the canonical momenta πµ =

∂L 1 = − 2 A˙ µ c ∂ A˙ µ

(417)

and to the field equations 1 µ s =0 c They are equivalent to the Maxwell equations if Aµ −

(418)

∂µ Aµ = 0

(419)

So we must use the second form of the Lagrange density, perform the quantization and then impose the condition ∂µ Aµ = 0 (the Lorentz condition).

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The classical electromagnetic field

Any gauge transformation performed Aµ (x) → Aµ (x) + ∂ µ f (x)

(420)

must lead to new potentials which still obeys the Lorentz condition, i.e. the function f (x0 must obey f (x) = 0 (421) In the case of free electromagnetoc field the Maxwell equation reduces to Aµ (x) = 0

(422)

µ

i.e. the fields A obeys the KG equation for zero mass.

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The classical electromagnetic field We write the expansion in terms of elementary solutions of KG equation Aµ (x) = Aµ+ (x) + Aµ− (x) with Aµ+ (x) =

(423)

X  ~c 2 1/2 µ r (k)ar (k)e −ik·x 2V ωk

(424)

X  ~c 2 1/2 µ r (k)ar† (k)e ik·x 2V ωk

(425)

k,r

Aµ− (x) =

k,r

with ωk = |k|/c and r = 0, 1, 2, 3 (for each k should be four polarization states); the properties of the (real) polarization vectors are orthogonality and completeness r (k) · s (k) = −ζr δrs , ζ0 = −1, ζ1 = ζ2 = ζ3 = 1 X µ µν ζr r (k)nu r (k) = −g

(426) (427)

r

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The classical electromagnetic field

Particular choice µ0 ≡ (1, 0, 0, 0) ≡ nµ , where for r = 3 3 =

µ r = (0, r )

(428)

k |k|

k · 1,2 = 0,

(429)

1 · 2 = 0

(430)

The third vector can be written as k µ − (k · n)nµ µ 3 = p (k · n)2 − k 2

(431)

(we did not use k 2 = 0). The polarization vectors are named transverse for r = 1, 2, scalar for r = 0 and longitudinal for r = 3.

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Covariant quantization The fields becomes operators and we impose the canonical commutation relations [Aµ (x, t), Aν (x, t)] = 0 h i A˙ µ (x, t), A˙ ν (x, t) = 0 i h Aµ (x, t), A˙ ν (x, t) = −i~c 2 g µν δ(x − x 0 )

(432) (433) (434)

Then the covariant commutations relations becomes [Aµ (x), Aν (x 0 )] = i~cD µν (x − x 0 )

(435)

D µν (x) = −g µν ∆(x) , m=0 Z −c 1 ∆(x) = ∆+ (x) + ∆− (x) = dk sin(k · x) (2π)3 ωk

(436)

with

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(437)

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Covariant quantization The photon propagator i~cDFµν (x − x 0 ) = h0|T {Aµ (x)Aν (x 0 )}|0i

(438)

DFµν (x) = −g µν ∆F (x) m=0 Z g µν e −ik·x µν 4 DF (x) = − d k 2 4 (2π) k + i Commutation relations between the elementary operators h i ar (k), as† (k 0 ) = ζr δrs δkk 0

(439)

i   h ar (k), as (k 0 ) = ar† (k), as† (k 0 )

(442)

(440)

(441)

Problem: for r = 0 there is a different sign, i.e. the role of absorbtion-creation operators must be interchanged. Instead one follows the Gupta-Bleuler formalism.

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Covariant quantization Keep the intepretation of a and a† and the definition of the vacuum state as ar (k)|0i = 0,

r = 1, 2, 3, 4

(443)

i.e. Aµ+ (x)|0i = 0,

(444)

The creation operators on the vacuum state create a scalr, transverse, respectivelly longitudinal photon. The Hamilton operator Z X ˙ H = dxN[π(x) · A(x) − L(x)] = ~ωk ζr ar† (k)ar (k) (445) k,r

is positive definite; also define the number operator Nr (k) = ζr ar† (k)ar (k)

(446)

Remaining problem: the norm of one-photon state (or any odd number of particles state) can be negative. h1k,r |1k,r i = ζr (447)

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Covariant quantization For solving the problem: we still have to impose the Lorentz condition. The covariant commutation relation [Aµ (x), Aν (x 0 )] = i~cD µν (x − x 0 ) (448) requires [∂µ Aµ (x), Aν (x 0 )] = i~c∂µ D µν (x − x 0 )

(449)

which is a contradiction. Gupta-Bleuler method: impose only ∂µ Aµ+ (x)|ψi = 0

(450)

for any physical state ψ and only for absorption operators. The condition and its adjoint hψ|∂µ Aµ− (x) = 0

(451)

hψ|∂µ Aµ− (x) + ∂µ Aµ+ (x)|ψi = 0

(452)

lead to i.e. the condition is verified for expectation values on any state.

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Covariant quantization

Equivalent form in term of elementary operators [a3 (k) − a0 (k)|ψi = 0

(453)

hψ|N3 (k) − N0 (k)|ψi = 0

(454)

with the condition and hψ|H|ψi = hψ|

X

~ωk ar† (k)ar (k)|ψi

(455)

k,r =1,2

i.e. longitudinal and scalar photons are never observed.

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The photon propagator The Feynman propagator Z g µν e −ik·x d 4k 2 4 (2π) k + i Z 1 d 4 ke −ik·x DFµν (k) DFµν (x) = (2π)4 X µ 1 1 DFµν (k) = −g µν 2 = 2 ζr r (k)νr (k) k + i k + i r DFµν (x) = −

(456) (457) (458)

In the particular reference frame defined before ( X µ 1 µν r (k)νr (k)+ DF (k) = 2 k + i r =1,2

(459)

(k µ − (k · n)nµ )(k ν − (k · n)nν ) + + (−1)nµ nν (k · n)2 − k 2

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The photon propagator Define µν T DF (k)

=

k2

µν C DF (k) µν R DF (k)

=

k2

X µ 1 r (k)νr (k) + i r =1,2 =

(460)

nµ nν (k · n)2 − k 2

(461)

k µ k ν − (k · n)(k µ nν + k ν nµ ) 1 + i (k · n)2 − k 2

(462)

And DFµν (k) =T DFµν (k) +C DFµν (k) +R DFµν (k)

(463)

The first term: “normal propagator”, describes the interaction of charges with the transverse field. The second term: describes the instantaeous Coulomb interaction; its form in position representation µν C DF (x)

= g µ0 g ν0

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1 δ(x 0 ) 4π|x|

(464)

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The photon propagator

One can prove that the third term does not contribute to any transition amplitude (of the general form) Z Z Z dx dys1µ (x)DF µν (x − y )s2ν y ∼ dks1µ (−k)DF µν (k)s2ν (k) (465) with s obeying the condition ∂µ s µ = 0, i.e. kµ sµ = 0.

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