Dirac Spinors. Kevin Cahill for 523 November 28, A 4-component spin-one-half Majorana of Dirac field ψ with action density

Transcription

1 Dirac Sinors Kevin Cahi for 53 November 8, A 4-comonent sin-one-haf Majorana of Dirac fied ψ with action density (where ψ ψ iγ ) obeys the Dirac equation L ψ (γ a a + m) ψ ψ ( + m) ψ () (γ a a + m) ψ ( + m) ψ. () Two Majorana fieds of the same mass ψ and ψ have the Fourier exansions ψ (i) (x) [u (, s)a(, s, i)e ix + v (, s)a (, s, i)e ix] d 3 (3) (π) 3/ whie a Dirac fied made of them has the Fourier exansion ψ (x) ( ψ () (x) + iψ () ) (x) [u (, s)b(, s)e ix + v (, s)c (, s)e ix] d 3 (4) (π) 3/

2 in which the Dirac index runs from to 4, x x t, + m, and the annihiation a i and creation a j oerators satisfy the anticommutation reations {a(, s, i), a(, s, j)} a(, s, i) a(, s, j) + a(, s, j) a(, s, i) {a(, s, i), a (, s, j)} δ i,j δ s,s δ (3) ( ). The state, s, i a (, s, i) reresents a artice of tye i, 4-momentum, and sin s The artice annihiation b(, s) and antiartice creation c (, s) oerators are (5) in the z-direction. b(, s) (a(, s, ) + ia(, s, )) and c (, s) ( a (, s, ) + ia (, s, ) ). (6) They satisfy the anticommutation reations {b(, s), b(, s )} {c(, s) c(, s )} {b(, s), c (, s )} {b(, s), b (, s )} δ s,s δ (3) ( ) {c(, s), c (, s )}. (7) The state, s, b b (, s) reresents a artice of 4-momentum and sin s in the z-direction, whie the state, s, c c (, s) reresents an antiartice of 4-momentum and sin s in the z-direction. The Majorana fied wi satisfy the Dirac equation () if the sinors obey the rues (iγ a a + m) u(, s) ( iγ a a + m) v(, s) (8) which the sinors wi obey if their momentum-deendence is u(, s) (m iγ a a ) u(, s) and v(, s) (m + iγ a a ) v(, s). (9) For we then have (iγ a a + m) u(, s) (iγ a a + m) ( iγ b b m ) u(, s) ( γ a a γ b b + m ) u(, s) [( {γa, γ b } + [γa, γ b ] ) a b + m ] u(, s) ( η ab a b + m ) u(, s) ( ( ) + + m ) u(, s) ()

3 as we as ( iγ a a + m) v(, s) ( iγ a a + m) ( iγ b b m ) v(, s) ( γ a a γ b b + m ) v(, s) [( {γa, γ b } + [γa, γ b ] ) a b + m ] v(, s) ( η ab a b + m ) v(, s) ( ( ) + + m ) v(, s). () The zero-momentum sinors obey the rues (8) at ( iγ k + m ) u(, s) ( iγ m + m ) u(, s) ( iγ k + m ) v(, s) ( iγ m + m ) v(, s). () That is, or since γ i iγ u(, s) u(, s) and iγ v(, s) v(, s) (3) ( ) and incidentay γ i the zero-momentum sinors satisfy ( ) iγ u(, s) u(, s) u(, s) ( ) iγ v(, s) v(, s) v(, s). The natura choices are u(, +), u(, ), v(, +) ( ) σ σ, v(, ) (4) (5). (6) 3

4 They are suitaby orthonorma u (, s)u(, s ) δ s,s v (, s)v(, s ) δ s,s u (, s)v(, s ) v (, s)u(, s ). (7) Because the zero-momentum sinors are eigenvectors of γ, we can rewrite the momentum deendence (9) of the sinors as u(, s) ( iγ a a + m) u(, s) ( iγ a a iγ + m ) u(, s) ( γ a a γ + m ) u(, s) v(, s) (+iγ a a + m) v(, s) ( iγ a a ( iγ ) + m ) v(, s) ( γ a a γ + m ) v(, s). We may use the orthonormaity (7) of the zero-momentum sinors, their deendence (8) uon their momentum, and the reations γ γ and γ γ to show that the sinors u(, s) are orthonorma (8) u (, s)u(, s ) u (, s) (m i a γ a ) ( m i b γ b) u(, s )/[ ( + m)] u (, s) ( i γ + i γ + m ) ( i γ + i γ + m ) u(, s )/[ ( + m)] u (, s) [ (m + i γ ) + ( γ ) ] u(, s )/[ ( + m)] u (, s) [ m + + mi γ + ( ) ] u(, s )/[ ( + m)] u (, s) [ m + + m + ( ) ] u(, s )/[ ( + m)] u (, s)u(, s ) δ s,s. More generay, one has (9) u (, s)u(, s ) δ s,s v (, s)v(, s ) δ s,s u (, s)v(, s ) v (, s)u(, s ). () 4

5 We have seen in the notes on grou theory that the Dirac reresentation of the Lorentz grou is the direct sum of the D (/,) and D (,/) reresentations. The standard boost that takes k (m,,, ) into (, ) Lk is L() R()B()R () () in which B() is the boost in the z-direction that takes the 4-vector (m,,, ) to (,,, ) and the rotation R() rotates the z-axis into the direction ˆ. In the grou-theory notes, we saw that D (/,) (L) + m σ m( + m) and D (,/) (L) + m + σ m( + m). () Thus the 4 4 matrix that reresents the standard boost L is D(L) D (/,) (L) D (,/) (L) ( ) ( ) D D(L) (/,) (L) D (,/) + m σ (L) m( + m). (3) + m + σ As a first homework robem, show that D(L) γa a γ + m m( + m). (4) It is usefu to further scae the sinors u(, s) and v(, s) so that m m u(, s) D(L) u(, s) and v(, s) D(L) v(, s). (5) As a second homework robem, show that the sin sums of the zero-momentum sinors are u (, s)u (, s) ( ) I + iγ ( ) δ + iγ v (, s)v (, s) ( ) I iγ (6) ( ) δ iγ. 5

6 is As a third homework robem, show that the sin sums of the sinors are u (, s)u [ (, s) ( iγ a a + m) iγ ] v (, s)v [ (, s) ( iγ a a m) iγ ]. The effect of a unitary transformation U(Λ) that imements a Lorentz transformation Λ on a state, s, b (Λ) U(Λ), s, b j D (j) s s (W (Λ, )) Λ, s, b (8) in which the (j + ) (j + ) unitary matrix D (j) s s (W (Λ, )) reresents the Wigner rotation s j (7) W (Λ, ) L (Λ)ΛL() (9) where L() is the standard boost () that takes the 4-vector (m,,, ) to. Let s comute the Feynman roagator for a Dirac fied. It is defined as the mean vaue in the vacuum (of the theory without interactions (the free theory)) of the time-ordered roduct defined as T [ψ (x)ψ m(y)] θ(x y )ψ (x)ψ m(y) θ(y x )ψ m(y)ψ (x) θ(x y )ψ + (x)ψ m (y) θ(y x )ψ m + (y)ψ (x) θ(x y ){ψ + (x), ψ m (y)} θ(y x ){ψ m + (y)ψ (x)} θ(x y ){ψ + (x), ψ m (y)} θ(y x ){ψ m + (y), ψ (x)}. We can use the Fourier exansion of the Dirac fied (4) and of its adjoint ψ (x) ( ) ψ () (x) iψ () (x) (3) [u (, s)b (, s)e ix + v (, s)c(, s)e +ix] d 3 (3) (π) 3/ 6

7 as we as the sin sums (7) to comute these anticommutators: {ψ + (x), ψ m (y)} s,t s,t d 3 (π) 3/ d 3 (π) 3/ d 3 q (π) 3/ u (, s)e ix u m( q, t)e iqy {b(, s), b (q, t)} d 3 q (π) 3/ u (, s)e ix u m( q, t)e iqy δ s,t δ (3) ( q) d 3 (π) 3 u (, s)u m(, s)e i(x y) d 3 [ ( iγ a (π) 3 a + m) iγ ] m ei(x y) [ ( γ a a + m) iγ ] m and since {ψ m + (y), ψ (x)} {ψ (x), ψ + m (y)} {ψ (x), ψ + m (y)} s,t s,t d 3 (π) 3/ d 3 (π) 3/ d 3 (π) 3 ei(x y) [ ( γ a a + m) iγ ] m +(x y) d 3 q (π) 3/ v (, s)e ix v m( q, t)e iqy {c (, s), c(q, t)} d 3 q (π) 3/ v (, s)e ix v m( q, t)e iqy δ s,t δ (3) ( q) d 3 (π) 3 v (, s)v m(, s)e i(x y) d 3 [ ( iγ a (π) 3 a m) iγ ] m ei(x y) [ (γ a a m) iγ ] m d 3 (π) 3 e i(x y) [ (γ a a m) iγ ] m +(y x). (3) (33) 7

8 Putting the two anticommutators together, we find T [ψ (x)ψ m(y)] θ(x y )ψ (x)ψ m(y) θ(y x )ψ m(y)ψ (x) θ(x y ){ψ + (x), ψ m (y)} θ(y x ){ψ m + (y), ψ (x)} θ(x y ) [ ( γ a a + m) iγ ] m +(x y) θ(y x ) [ (γ a a m) iγ ] m +(y x) θ(x y ) [ ( γ a a + m) iγ ] m +(x y) + θ(y x ) [ ( γ a a + m) iγ ] m +(y x). (34) We can u the satia derivatives and the mass term across the Heaviside functions. derivatives, we note that the extra terms actuay vanish To u the time ( θ(x y )) + (x y) + ( θ(y x )) + (y x) δ(x y ) + (x y) δ(x y ) + (y x) (35) because at equa times the two invariant functions cance d 3 + (y x) e i ( x y) (π) 3 + (x y). (36) Thus where T [ψ (x)ψ m(y)] [ ( γ a a + m) iγ ] [ m θ(x y ) + (x y) + θ(y x ) + (y x) ] [ ( γ a a + m) iγ ] [ i m F (x y)] Canceing the is, we have F (x y) (37) d 4 q (π) 4 ex(iq(x y)) q + m iɛ. (38) T [ψ (x)ψ m(y)] [ ( γ a a + m) γ ] m F (x y) [ ( γ a a + m) γ ] d 4 q ex(iq(x y)) m (π) 4 q + m iɛ d 4 q [( iγ a q a + m) γ ] m e iq(x y). (π) 4 q + m iɛ 8 (39)

9 As a Thanksgiving vacation homework robem, comute the equa-time anticommutators of the Dirac fied (4) with itsef and with its adjoint (3). {ψ (t, x), ψ m (t, y)} and {ψ (t, x), ψ m(t, y)} (4) 9