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QFT Dr Tasos Avgoustidis (Notes based on Dr A. Moss’ lectures)

Lecture 8: &

!1 QFT (Recap)

µ µ µ • Consider transformation ⇤ ⌫ = ⌫ + ✏w ⌫ µ ⌧ ⌫ µ⌫ • Using definition of LT ⇤ ⌘ ⇤ ⌧ = ⌘ • For terms linear in ✏ then wµ⌫ + w⌫µ =0 • For infinitesimal LT the matrix needs to be anti- symmetric. This has 6 degrees of freedom, corresponding to the 6 transformations of the Lorentz group • Introduce of 6 anti-symmetric 4x4 matrices ( ⇢)µ⌫ = ⌘⇢µ⌘⌫ ⌘µ⌘⇢⌫ M • ⇢ , label which matrix, µ, ⌫ the row/column of each matrix !2 Lecture equations

Adam Moss (Dated: February 28, 2012)

1 d3xd3kd3q Hˆ = × (1) 8 (2π)6E(k)E(q) ! LectureE equations(k)E(q) −(ˆa†(k)eik·x − aˆ(k)e−ik·x)(ˆa†(q)eiq·x − aˆ(q)e−iq·x) (2) † ik·x −ik·x † iq·x −iq·x (−ik aˆ (k")e + ik aˆ(k)e )(−iq aˆ (q)e + iq aˆ(q)e #) (3) Adam2 Moss† ik·x −ik·x † iq·x −iq·x (Dated: February"m (ˆa 28,(k 2012))e +ˆa(k)e )(ˆa (q)e +ˆa(q)e ) # (4) " #

1 d3k Hˆ = × (5) 1 d3xd3kd3q 3 2 Hˆ × 8 (2π) E(k) = 6 ! (1) 8 (2π) E(k)E(q) k k ! (−E(k)2 + k2 + m2)(ˆa†(k)ˆa†(−k)e2iE( )t +ˆa(k)ˆa(−k)e−2iE( )t) (6) E(k)E(q) −(ˆa†(k)eik·x − aˆ(k)e−ik·x)(ˆa†(q)eiq·x − aˆ(q)e−iq·x) (2) 2 2 2 † † † ik·x −$(ikE·x(k) + k† + miq·x)(ˆa (k)ˆa(k)+ˆ−iq·xa(k)ˆa (k)) % (7) (−ik aˆ (k")e + ik aˆ(k)e )(−iq aˆ (q)e + iq aˆ(q)e #) (3) 2 † ik·x −ik·x † iq·x −iq·x "m (ˆa (k)e +ˆa(k)e ")(ˆa (q)e +ˆa(q)e ) # # (4) " # 1 d3k Hˆ = (ˆa†(k)ˆa(k)+ˆa(k)ˆa†(k)) (8) π 3 1 d3k 4 (2 ) Hˆ = × ! (5) 8 (2π)3E(k)2 ! k k (−E(k)2 + k2 + m2)(ˆa†(k)ˆa†(−k)e2iE( )t +ˆa(k)ˆa(−k)e−2iE( )t) (6) φˆ(x)φˆ(y), if x0 >y0 T φˆ(x)φˆ(y)= (9) $ 2 2 2 † † ˆ ˆ % 0 0 2 (E(k) + k + m )(ˆa (k)ˆa(k)+ˆa(k)ˆa (k)) &φ(y)φ(x), if y >x (7) " #

T [φ φ 1φ φ ]=:d3k φ φ φ φ :+ (10) Hˆ 1= 2 3 4 (ˆa1†(k2)ˆa3(k4)+ˆa(k)ˆa†(k)) (8) 0001 4 (2π)3 − − − ! ∆F (x1 x2):φ3φ4 :+∆F (x1 x3):φ2φ4 :+∆F (x1 x4):φ2φ3 :+ (11) ⎛ 0000⎞ ∆F (x2 − x3):φ1φ4 :+∆F (x2 − x4):φ1φ3 :+∆F (x3M−03x4):µ φ1φ2 :+ (12) QFT ( ) ν = (16) − − Lorentz− Group− −⎜ ⎟− ∆F (x1 x2)∆F (x3 x4)+∆F (x1 x3)∆F (x2 x4)+∆F (x1 ⎜x00004)∆F (x2 ⎟x3)(13) φˆ(x)φˆ(y), if x0 >y0 ⎜ ⎟ T φˆ(x)φˆ(y)= (9) ⎜ ⎟ φˆ(y)φˆ(x), if y0 >x0 ⎜ 1000⎟ & ⎜ ⎟ 2 ⎝ ⎠ ⇢0100µ ⇢µ µ ⇢ T [φ1φ2φ3φ4]=:φ1φ2φ3φ4 :+ (10) 00 0 0 • Lower one index ( ⎛ 1000) ⌫ =⎞ ⌘ ⌫ ⌘ ⌫ ∆ (x − x ):φ φ :+∆ (x − x ):φ φ :+∆ (Mx 01− xµ ):φ φ 0001:+ (11) F 1 2 3 4 F 1 3 2 4 F( 1 ) 4Mν = 2 3 ⎛ 00−10⎞ (14) ⎜ ⎟ M12 µ ∆F (x2 − x3):φ1φ4 :+∆F (x2 − x4):φ1φ3 :+∆F (x3 − x4):φ⎜1⎛φ00002 :+⎟⎞ ( ) (12)ν = (17) ⎜ 0000⎟ ⎜ ⎟ − • −Matrices− are now− (M 03no-longer)µ =⎜− antisymmetric⎟− ⎜ 01 on 0 0 µ,⎟ ⌫ (16) 2 ∆F (x1 x2)∆F (x3 x4)+∆F (x1 x3)∆F (x2 x4)+∆νF (x⎜1⎜0000x4)∆F (x⎟2⎟ x3)(13)⎜ ⎟ ⎜⎜ 0000⎟⎟ ⎜ ⎟ ⎝⎜ ⎠⎟ ⎜ 00 0 0⎟ • Infinitesimal boosts: ⎜ ⎟ ⎜ ⎟ ⎜ 1000⎟ ⎜ ⎟ 2 ⎝ ⎠ ⎝ ⎠ 0100 0010 0001 000 0 ⎛ 1000⎞ ⎛ 000000 0 0⎞ ⎛ 0000⎞ M01 µ M02 µ M03 µ ( ) ν = 0001 ( ) ν = ( ) ν(14)= ⎛ 000−1 ⎞ (15) (16) ⎜ 0000⎟ ⎜⎛ 001000−10⎟⎞ (M13)µ =⎜ 0000⎟ (18) ⎜⎛ ⎞⎟ M12 µ ⎜ ⎟ ν ⎜ ⎟ ⎜ 0000⎟ ( ) ν = ⎜ ⎟ ⎜⎜ 000⎟ 0⎟ (17) (M03)µ = ⎜ ⎟ ⎜ ⎟⎟ (16) ⎜⎜ ⎟ ⎟ ν ⎜⎜ 0000⎟ ⎜ 010000 0 0⎟⎟ ⎜⎜1000⎟ ⎟ ⎜⎜ 0000⎟ ⎜ ⎟⎟ ⎜⎜ ⎟ ⎟ ⎝⎜ ⎟⎠ ⎝⎜ ⎠⎟ ⎜ 010 0⎟ ⎜ ⎟ ⎜ 00 0 0⎟ ⎝⎜ ⎠ ⎟ ⎜ 1000⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ • Infinitesimal⎝ ⎠rotations: 0010 00 0 0 000 0 ⎛ 0000⎞ 000 0 M02 µ 00 0 0 ⎛ 00−10⎞ ( ) ν = M12 µ (15) ⎛ ⎞ ⎜ ⎟ ⎛ 000−1 ⎞( ) ν = 000 0 (17) ⎛⎜ 1000− ⎟⎞ 13 µ M23 µ ⎜ ⎟ 12 µ 00 10 M ( ) = 01 0 0 (19) (M ) = ⎜ ⎟ ( ) ν = (17)ν ⎜ ⎜ ⎟ ⎟ (18) ν ⎜⎜ ⎟⎟ ⎜ 000 0⎟ ⎜ ⎜ 000−⎟1 ⎟ ⎜⎜010000 0 0⎟⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ ⎜ ⎟ ⎜ ⎜00 0 0⎟ ⎟ ⎜⎝ ⎠⎟ ⎜ ⎟ ⎜ ⎜ ⎟ ⎟ ⎜ 00 0 0⎟ ⎜ 010 0⎟ ⎜ 001 0⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎜ ⎠ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ !3 000 0 000 0 000 0 ⎛ 000−1 ⎞ ⎛ 000−1 ⎞ ⎛ 000 0⎞ M13 µ (M13)µ = (M23)µ = ( ) (18)ν = (19) (18) ν ⎜ ⎟ ν ⎜ ⎟ ⎜ 000 0⎟ ⎜ 000 0⎟ ⎜ 000−1 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 010 0⎟ ⎜ 001 0⎟ ⎜ 010 0⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

000 0 000 0 ⎛ ⎞ 23 µ 000 0 ⎛ ⎞ (M ) = 23 µ(19) 000 0 ν ⎜ − ⎟ (M ) = (19) ⎜ 000 1 ⎟ ν ⎜ ⎟ ⎜ ⎟ ⎜ 000−1 ⎟ ⎜ 001 0⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 001 0⎟ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ QFT Lorentz Group

• Can write any infinitesimal LT in terms of this basis 1 wµ = ⌦ ( ⇢)µ ⌫ 2 ⇢ M ⌫ • Here ⌦ ⇢ are six real numbers specifying the LT 1 • Any finite LT can be written as ⇤µ =exp ⌦ ( ⇢)µ ⌫ 2 ⇢ M ⌫ ✓ ◆ • The six basis matrices obey the Lie algebra

[ ⇢, ⌧⌫]=⌘⌧ ⇢⌫ ⌘⇢⌧ ⌫ + ⌘⇢⌫ ⌧ ⌘⌫ ⇢⌧ M M M M M M • Here the row/column index is suppressed. This equation encapsulates the properties of the Lorentz group. We are interested in other matrices which satisfy this algebra

!4 2

2 0001 0 00001 ( 03)µ = (16) 0001M ⌫ B C B 0000C B C 0 1 B C 03 µ 0000 B 1000C ( ) = B C (16) M ⌫ B C B 0000C @ A B C B C B 1000C B C 00 0 0 @ A 0 00 101 ( 12)µ = (17) M ⌫ B C 00 0 0 B 01 0 0C B C B C 0 1 B 00 0 0C 12 µ 00 10 B C ( ) = @ A (17) M ⌫ B C B 01 0 0C B C B C B 00 0 0C 000 0 B C @ A 0 000 1 1 ( 13)µ = (18) QFT M ⌫ B C RepresentationB 000 0C 000 0 B C B C B 010 0C 0 000 1 1 B C ( 13)µ = @ A (18) M ⌫ B C B 000 0C B C • Interested in findingB otherC representations000 0 of the Lorentz B 010 0C group B C 0 000 01 @ ( A23)µ = (19) M ⌫ B C B 000 1 µC ⌫ µ⌫ • The is definedB as C, =2⌘ 1 000 0 B C B 001{ 0C } µ B C • with µ =0 0 , 000 1 , 2 , 3 0 are1 a set@ of 4 matrices,A so ( 23)µ = (19) M ⌫ B C 0 2 B 000i 2 1 C 01 µ ⌫ ⌫ µ B C 0 ⌫ = µ ( ) =1 (B ) = C1 = = (20) B 001 0C 0 1 6 B C 10 • Simplest representation@ A is 4x4@ matricesA

01 0 i 0 = i = (20) (21) 0 101 0 i 0 1 @ A @ A • Where i are the 0 i i = (21) 0 i 0 1 !5 @ A QFT Spinor Representation

• There is a “unique” (up to a similarity transformation) irreducible representation of the Clifford algebra. These µ matrices define the chiral (or Weyl) rep

• Consider the commutator of two µ 1 1 1 S⇢ = [⇢,]= ⇢ ⌘⇢ 4 2 2

• Can show these form a representation of the Lorentz group such that

[S⇢,S⌧⌫]=⌘⌧ S⇢⌫ ⌘⇢⌧ S⌫ + ⌘⇢⌫ S⌧ ⌘⌫S⇢⌧

!6 QFT

• Introduce a Dirac spinor, a complex valued object with 4 components which transforms as ↵ ↵ 1 (x) S[⇤] (⇤ x) ! • Here ↵ =1 , 2 , 3 , 4 labels the row/column of the S µ ⌫ matrices and 1 ⇢ 1 ⇢ ⇤=exp ⌦ S[⇤] = exp ⌦⇢S 2 ⇢M 2 ✓ ◆ ✓ ◆

• Particular LT specified by ⌦ ⇢ - these are the same for both ⇤ and S[⇤] • Lets look at S [ ⇤ ] in the chiral representation

!7 2 2

0001 0001 2 0 00001 0 1 03 µ 03 µ 0000 ( ) ⌫ = ( ) = (16) (16) M B CM ⌫ B C B 0000C B 0000C B C B C B C B C B 1000C B 1000C B C 0001B C @ A @ A 0 00001 ( 03)µ = (16) M ⌫ B C 00 0 0 B 000000 0C 0 B C 0 1 B 01000C 1 00 10 12 µ B 00 C10 ( 12)µ = ( ) B= C (17) (17) ⌫ M ⌫ @ B A C M B 01 0 0C B 01 0 0C B C B C B C B C B C 00B 00 0 0 0 0C B 00 0 0C B C B C @ A @ A 0 00 101 ( 12)µ = (17) M ⌫ B C B 01000 0 0C 0 000 0 B C B 000 0 0C 1 0 1 B 000C1 000 (1 13)µ B= C (18) ( 13)µ = M ⌫@ B A C (18) M ⌫ B C B 000 0C B 000 0C B C B C B C B C 000B 010 0 0C B 010 0C B C B C 0 000@ 1 1 A @ ( 13A)µ = (18) M ⌫ B C B 000 0C B 000C 0 000 0 B C B 0100 0C 1 B 000C 0 0 000 0( 123)µ @= A (19) 23 µ M ⌫ B C ( ) ⌫ = B 000 1 C (19) M B C B C B 000 1 C B C B C 000B 001 0 0C B C B C B 001 0C 0 000@ 01 A B ( 23C)µ = (19) @ M A⌫ B C B 000 1 C B 01 C 0B= C (20) 01 B 001 0C 0 B 0 101 C = @ A (20) 0 101 @ A @ A 0 01 i i= 0 (20) i =0 101 (21) i i 0 0 0 1 QFT = @ A (21) Dirac Spinor0 i 0 1 @ A @ A i i 0 1 = i k 0 (21) Sij = [i,0j]=i 0 ✏1ijk (22) 1 i k4 0 2 0 0 k 1 • For rotations Sij = [i, j]= ✏ijk @ A (22) 4 2 0 0 k 1 @ A k ij @1 i j A i ijk 0 S = [ , ]= ei'✏/2 0 (22) k 4 2 · 0 k 1 • Writing rotation as ⌦ ij = ✏ ijk ' S[⇤]= 0 (23) 0 i' /2 1 0 @e · A @ A 3 ei⇡ 0

• For a rotation of ' =(0, 0, 2⇡) S[⇤]= 3 = 1 (23) 0 0 ei⇡ 1 @ A • This means that under 2 ⇡ rotations ↵ ( x ) ↵ ( x ) which is not what happens to a vector - different! rep • For rotations in the chiral representation S [ ⇤ ] is unitary, i.e. S[⇤]†S[⇤] = 1

!8 2

0001 0 00001 ( 03)µ = (16) M ⌫ B C B 0000C B C B C B 1000C B C @ A

00 0 0 0 00 101 ( 12)µ = (17) M ⌫ B C B 01 0 0C B C B C B 00 0 0C B C @ A

000 0 0 000 1 1 ( 13)µ = (18) M ⌫ B C B 000 0C B C B C B 010 0C B C @ A

000 0 0 000 01 ( 23)µ = (19) M ⌫ B C B 000 1 C B C B C B 001 0C B C @ A

01 0 = (20) 0 101 @ A

0 i i = (21) 0 i 0 1 @ A

1 i k 0 Sij = [i, j]= ✏ijk (22) 4 2 0 0 k 1 @ A

3 ei⇡ 0 S[⇤]= = 1 (23) QFT Dirac Spinor3 0 0 ei⇡ 1 @ A

i 1 1 0 3 • For boosts S0i = [0, i]= (24) 4 2 0 0 i 1 @ A /2 e · 0 • Writing boost as ⌦ i 0 = i S [⇤ ]= (25) 0 /2 1 0 e · @ A • For boosts in the chiral representation S [ ⇤ ] is not unitary, i.e. S[⇤]†S[⇤] =1 6 • In general there are no finite dimensional unitary representations of the Lorentz group

!9 3

QFT Chiral Spinors

/2 e · 0 S[⇤]= (25) 0 0 e /2 1 • The chiral representation of the · Lorentz group is reducible. It decomposes@ into two irreducibleA representations u+ = (26) 0 u 1 @ A • 2 component objects u are called Weyl spinors ±

i' /2 • Under rotations u u e · ± ! ± ' /2 • Under boosts u u e± · ± ! ±

!10 QFT Dirac Action

• Want an action which is Lorentz invariant

? T • Define adjoint in usual way †(x)=( ) (x)

• Try and form a Lorentz scalar from † with the spinor index summed over

• Under LT 1 1 (x) S[⇤] (⇤ x) †(x) †(⇤ x) S[⇤]† ! !

• Therefore † is not a Lorentz scalar since S [ ⇤ ] is not unitary

!11 QFT Dirac Action

• If we choose a representation of the Clifford algebra 0 0 i i 0 µ 0 µ which satisfies ( ) = ( ) † = then =( )† †

0 1 0 • Can show this gives S[⇤]† = S[⇤]

0 • With this in mind define the Dirac conjugate ¯(x)= †(x)

• Can form Lorentz invariant objects from Dirac spinor and its conjugate, e.g. scalars and vectors 1 1 ¯(x) (x)= ¯(⇤ x) (⇤ x) µ µ 1 ⌫ 1 ¯(x) (x)=⇤ ⌫ ¯(⇤ x) (⇤ x)

!12 QFT Dirac Equation

• Can construct a Lorentz invariant action S = d4x ¯(x)(iµ@ m) (x) µ Z • After this theory will describe particles of m and spin-1/2 • Varying with respect to ¯ gives the Dirac equation (iµ@ m) (x)=0 µ • First order in derivatives but Lorentz invariant • Mixes up components of spinor but can show each individually solves Klein-Gordon equation µ (@µ@ + m) =0 !13 3

/2 e · 0 S[⇤]= (25) 0 /2 1 0 e · QFT Weyl@ EquationA

u+ = (26) 0 u 1 • Let’s decompose the Dirac@ LagrangianA into chiral spinors

i 01 0 @0 + @i u+ =(u+† ,u† ) i m (27) L 0 1 2 0 i 1 3 0 1 10 @0 @i 0 u @ A 4 @ A 5 @ A µ µ = iu† @µu + iu+† ¯ @µu+ m(u† u+ + u+† u ) L where µ =(1,i) ¯µ =(1, i) • For a massless the chiral spinors decouple and they satisfy the Weyl equations of motion µ µ i @µu =0 i¯ @µu+ =0

!14 3 QFT 5

/2 e · 0 S[⇤]= 3 (25) 0 /2 1 0 e · • The matrices @S [ ⇤ ] came Aout block diagonal in the chiral representation /2 e · 0 u+ S[⇤]= = (25) (26) • How do we define chiral spinors0 in a/ 2general1 0 e · 0 u 1 representation of the Clifford@ algebra?A @ A • Introduce the fifth gamma matrix 5 = i0123 01u+ 0 @ + i@ u = 0 i + (26) =(u+† ,u† ) i m (27) 5 Lµ 0 5 121 2 0 i 1 3 0 1 • This satisfies , =0 (u10) =1@0 @i 0 u { } @ AA 41@ A 5 @ A • Define a projection operator P = (1 5) ± 2 i ± 2 01 0 @0 + @i u+ u+ =(u+†P,u† =) P i P P =0 + = m (27) (28) L 0 1 2 0 + i 10 13 0 1 ± 10± @0 @i 0 0 u • Define chiral spinors@ byA 4 @ = P A@ A5 @ A ± ± u+ 0 • In chiral representation + = = (28) (29) 0 0 1 0 u 1 !15 @ A @ A

0 = (29) 0 u 1 @ A QFT Symmetries

µ • The Dirac Lagrangian = ¯ ( i @ µ m ) enjoys a number of symmetries L µ • For space-time translations spinor transforms = ✏ @µ

• Lagrangian depends on @ µ not @µ ¯ • Recall previous definition of energy- tensor

µ⌫ @ ⌫ µ⌫ T L @ a ⌘ ⌘ @(@µa) L • Conserved currents arise when equations of motion are satisfied - can set to zero L • For Dirac Lagrangian obtain T µ⌫ = i ¯µ@⌫

!16 QFT Symmetries

↵ ↵ 1 • Under LT (x) S[⇤] (⇤ x) ! µ µ µ • Work infinitesimally ⇤ ⌫ = ⌫ + w ⌫ 1 ↵(x)= ↵ + ⌦ (S⇢)↵ + ... (x) wµ x⌫ @ (x)+... 2 ⇢ ⌫ µ  ⇥1 ⇤ ↵ = wµ x⌫ @ ↵ + ⌦ (S⇢)↵ ⌫ µ 2 ⇢ µ 1 ⇢ µ ⇢ µ ⇢µ µ ⇢ • Remember w = ⌦ ( ) , ( ) = ⌘ ⌫ ⌘ ⌫ ⌫ 2 ⇢ M ⌫ M ⌫ • This means that wµ⌫ =⌦µ⌫ 1 • Obtain ↵ = wµ⌫ x @ ↵ (S )↵ ⌫ µ 2 µ⌫ 

!17 QFT Symmetries

• Now apply Noether’s theorem (again setting =0 ) to find conserved current L jµ = w⇢⌫ x T µ i ¯µS ⌫ ⇢ ⇢⌫ ⇥ ⇤ • Left choice of w µ ⌫ explicit. Strip it off to give 6 different currents ( µ)⇢ = x⇢T µ xT µ⇢ i ¯µS⇢ J which satisfy @ ( µ)⇢ =0 µ J • After quantization the final term will be responsible for providing single particle states with internal angular momentum

!18 QFT Symmetries

• Dirac Lagrangian is invariant under rotating phase of i↵ spinor e or = i↵ ! µ ¯ µ • This gives rise to a conserved vector current jV =

• When m=0 Lagrangian has an extra internal 5 5 ei↵ ¯ ¯ei↵ ! ! µ ¯ µ 5 • This gives rise to a conserved axial current jA =

• This conserved quantity does not survive the quantization process - an example of

!19 3

3

/2 e · 0 S[⇤]= (25) 0 /2 1 0 e · /2 e · 0 @ S[⇤]=A (25) 0 /2 1 0 e · u+ @ A = (26) 0 u 1 u+ @ A = (26) 0 u 1 i 01 0 @0 + @i @ A u+ =(u+† ,u† ) i m (27) L 0 1 2 0 i 1 3 0 1 10 @0 @i 0 u i 01 0 @0 + @i u+ @ A=(4 @u+† ,u† ) i A 5 @ A m (27) L 0 1 2 0 i 1 3 0 1 10 @0 @i 0 u QFT @u+ A 4 @ A 5 @ A Plane Wave+ = Solutions (28) 0 0 1 u+ @ A + = (28) 0 0 1 • Want to solve µ (i @µ m0) =0 @ A = (29) 0 1ip x • Make the ansatz = u(p)ue · 0 @ A = (29) • In chiral representation Dirac equation becomes0 u 1 µ @ A µ mpµ ( pµ m)u(p)= u(p) = 0 (30) 0 µ 1 pµ¯ m µ µ mpµ ( pµ m)u(p)= u(p) = 0 (30) µ i @µ i A 0 µ 1 where =(1, ) ¯ =(1, ) pµ¯ m 2 i j @ 2 i A 2 • Use identity (p )(p ¯)=p pipj = p pip = m · · 0 0 pp ⇠ • Can easily check the solution is u(p)= · (31) 0 pp ⇠¯ 1 · • Here ⇠ is a two-component spinor @ A !20 3

/2 e · 0 S[⇤]= (25) 3 0 /2 1 0 e · @ A 3 /2 e · 0 u+ S[⇤]= (25) 0 /2 1 = 0 e · (26) 0 u 1 /2 @ A e · 0 S[⇤]=@ A (25) 0 0 e /2 1 · u+ @ i =A (26) 01 0 @0 + @i 0 u 1 u+ =(u+† ,u† ) i m (27) L 0 1 2 0 i 1@ A3 0 1 10 @0 @i u+ 0 u = (26) @ A 4 @ 0 1 A 5 @ i A 01u 0 @0 + @i u+ =(u+† ,u† ) i m (27) L 0 @ 1A2 0 i 1 3 0 1 10 @0 @i 0 u u+ = @ A 4 @ A 5 @ A (28) + i 01 0 0 01 @0 + @i u+ =(u+† ,u† ) i m (27) L 0 1 2 0 i 1 3 0 1 10 @ @0 A @i 0 u+ u + = (28) @ A 4 @ 0 A0 1 5 @ A 0 @ A = u (29) 0 1 + +u= (28) 0 0 1 0 @ A = (29) @ A 0 u 1 @ A mpµ µ µ0 ( pµ m)u(p)= u(p) = 0 (30) 0 =µ 1 mpµ (29) pµµ¯ 0 m1 µ ( pµ mu)u(p)= u(p) = 0 (30) 0 µ 1 @ A pµ¯ m @ A QFT Plane Wave@ SolutionsA µ µ pp ⇠mpµ ( pµu(pm)=)u(p)= · up(p )⇠ = 0(31) (30) 0 µu(p)=1 · (31) 0 pp ⇠p¯µ¯1 m · 0 pp ⇠¯ 1 @ @ A A · ip x • Also negative frequency solutions@ A= v(p)e · pp ⇠ u(p)=pp ⌘ · pp ⌘ (31) v(p)= 0· v(p)=1 · (32) (32) with pp ⇠¯ 0 1 0 pp ⌘¯ ·1 pp ⌘¯ @· A · @ A @ A

pp ⌘ 1 • Will be convenientv(p)= to introduce· 1 a basis (32) 0 ⇠1= (33) r s prsp ⌘¯ 0r0 1s rs ⇠ †⇠ = · ⌘ †⌘ = @ A @ A

1 0 • For example ⇠1 = ⇠2 = (33) (34) 0 0 1 0 1 1 @ A @ A

0 ⇠2 = !21 (34) 0 1 1 @ A