Particle Physics 2011
Luis Anchordoqui
Friday, October 21, 2011 Dirac Equation today
we will construct wave equation for spin-1/2 relativistic particles
Following Dirac we proceed by analogy with non-relativistic QM
write equation which -- unlike Klein-Gordon equation-- is linear in ∂t In order to be covariant it must also be linear in ! ∇ Hamiltonian has general form Hψ(x)=("α.p" + βm) ψ(x)
4 coefficients β ,α1,α2 and α3 are determined by requirement that a free particle -- of mass m --
must satisfy relativistic energy momentum relation
2 H ψ =(αipi + βm)(αjpj + βm)ψ 2 2 2 2 =(αi pi +(αiαj + αjαi) pipj +(αiβ + βαi) pi m + β m ) ψ 1 0 0 1
Friday, October 21, 2011 !"#$ ! "# $ ! "# $ !"#$ Anticommutation Relations From we see that all the coefficients α i and β anticommute with each other and hence they cannot simply be numbers k We are lead to consider matrices α (k =1, 2, 3)and β which are required to satisfy the condition
αkαl + αlαk αk,αl =2δkl, αk,β =0, and β2 =1 ≡{ } { } is the unit matrix It turns out that the lowest dimensionality matrices which guarantee relativistic energy momentum relation also holds true 4 4 × A four-component quantity ψα(x) which satisfies the Dirac equation k called i∂t ψρ(x)= i [αρσ] ∂xk ψσ(x)+mβρσ ψσ(x) − a Spinor
Its transformation properties are different from that of a 4-vector
Friday, October 21, 2011 dirac-Pauli & Weyl representations
k Specific representation of matrices α and β
Dirac Pauli representation
0 !σ 1 0 !α = and β = !σ 0 0 1 ! " ! − "
Weyl or chiral representation 2 x 2 block form
!σ 0 0 1 !α = − and β = 0 !σ 1 0 ! " ! "
Friday, October 21, 2011 Covariant form of Dirac Equation On multiplying Dirac's equation by β from the left we obtain iβ ∂ ψ = iβ $α. $ ψ + mψ t − ∇ which can be rewritten as
0 k iγ ∂ ψ + iγ ∂ k ψ mψ =0 t x −
or equivalently
(iγµ∂ m) ψ =0 µ −
we omit spinor subscripts whenever there is no danger of confusion
Friday, October 21, 2011 Dirac matrices
µ we have introduced four Diracγ -matrices γ ( β , β#α ) ≡ which satisfy the anticommutation relations
γµ,γν =2gµν ⊠ { }
We can now unequivocally see that Dirac's equation
is actually 4 differential equations
4 i [γ ]µ ∂ mδ ψ =0 ρσ µ − ρσ σ σ=1 " µ # ! !
which couple the four components of a single Dirac spinor ψ
Friday, October 21, 2011 Lorentz invariance general Lorentz transformation contains rotations and because ψ(x) is supposed to describe a field with spin under the transformation µ µ µ ν x x! =Λ x We allow for a rearrangement of ψ ( x ) components → ν Because both Dirac equation and Lorentz transformation of coordinates are themselves linear we ask transformation between ψ & ψ! be linear
1 ψ!(x!)=ψ!(Λx)=S(Λ) ψ(x)=S(Λ)ψ(Λ− x!) ⦁
S(Λ) is a 4 4 matrix which operates on ψ × To figure out S demand Dirac eq. has same form in any inertial frame
µ (iγ ∂ ! m)ψ!(x!) = 0 µ −
or equivalently
(iγµΛ ν ∂ m)S(Λ) ψ(x)=0 µ ν − Friday, October 21, 2011 Lorentz invariance 1 If we multiply by S − ( Λ ) from left we get 1 µ ν (iS− γ S Λ ∂ m) ψ(x)=0 µ ν − Dirac eq. is form-invariant provided we can find S ( Λ ) such that
1 µ ν ν ✣ S− (Λ) γ S(Λ)Λµ = γ consider infinitesimal Lorentz Transformation i S(Λ) = 1 ω Σµν − 2 µν after a bit of algebra ✣ reduces to the condition
[Σµν ,γβ]= i(gµβγν gνβγµ) − − A solution is seen to be 1 i Σµν σµν = [γµ,γν ] ⟡ ≡ 2 4 See lecture notes for details Friday, October 21, 2011 Hints for the calculation
i i i (1 + ω Σρσ)γµ(1 + ω Σρσ) = (1 ω ρσ)µγν 2 ρσ 2 ρσ − 2 ρσJ ν
ρ ρ σ ρ σ with ☛ ( σ) = i(δ δ δ δ ) J µν µ ν − ν µ
This equation is just the infinitesimal form of
µ µ ν S−1(Λ) γ S(Λ)=Λν γ
Friday, October 21, 2011 Lorentz Algebra
By repeated use of ⊠ it is easily seen that ⟡ satisfies commutation relations of Lorentz algebra [Σµν , Σρσ]=i(gνρΣµσ gµρΣνσ gνσΣµρ + gµσΣνρ) − −
0 1 0 Incidentally S†(Λ) = γ S− (Λ) γ
The form for S(Λ) when Λ is not infinitesimal is
(i/2) ω Σµν S(Λ) = e− µν
Friday, October 21, 2011 Rotations & Boosts ij 1 ijk k For a rotation ω =0and ω = θ and because Σ = ! σ i0 ij k 2 (i/2) θ.σ we get S(Λ) = e− which shows the connection between ωij and parameters characterizing rotation (i, j, k =1, 2, 3)
For a pure Lorentz transformation ωij =0 and ωi0 = ϑi i and because Σ0i = αi we have 2
S(Λ) = e(1/2) ϑ.α 1 1 ϑ2 1 ϑ2 ϑ.α = 1 + ϑ.α + + + ... 2 2! 4 3! 4 2 ! " ! " ϑ ϑ = cosh + ϑˆ.α sinh 2 2
Friday, October 21, 2011 Hints for the calculation
0 σ 10 0 σ αβ = = σ 0 0 1 σ −0 ! "! − " ! " 01 0 σ 0 σ βαβ = − = − = α 10 σ 0 σ 0 − ! − "! " ! − " i i i [γ0,γi]= β2α βαβ = α 4 4 − 2 ! " i i Σ0i = α Σi0 = α ω = ω 2 i −2 i 0i − i0
ϑ.α = ϑ1α1 + ϑ2α2 + ϑ3α3
(σ )2 =(σ )2 =(σ )2 (ϑ α + ϑ α + ϑ α )2 = ϑ2 1 2 3 ⇒ 1 1 2 2 3 3
Friday, October 21, 2011 Rotation through imaginary angle For special case we may find connection between ϑ and velocity !v characterizing pure Lorentz transformation by looking at ✣ consider Lorentz transformation in which new -prime- frame moves with velocity v along x 3 axis of original -unprimed- frame
t! = cosh(ϑ ) t sinh(ϑ ) x 3 − 3 3 x! = sinh(ϑ ) t + cosh(ϑ ) x 3 − 3 3 3 with x and y unchanged 1 "v here ☛ cosh(ϑ3)= and ϑˆ = kˆ √1 v2 ≡ v −
Because cos(iϑ3) = cosh(ϑ3) and sin(iϑ3) = sinh(ϑ3) We see that Lorentz transformation may be regarded as a rotation through imaginary angle iϑ in it x plane 3 − 3
Friday, October 21, 2011 Adjoint Dirac Equation To construct currents we duplicate Klein Gordon calculation taking account that Dirac eq. is matrix Eq. and thus we must consider hermitian rather than complex conjugate eq.
Dirac eq. hermitian conjugate is
0 k i∂tψ† γ i∂ k ψ† ( γ ) mψ† =0 − − x − −
To restore covariant form we need to flip plus sign in 2nd term while leaving 1st term unchanged 0 k k 0 Since γ γ = γ γ this can be accomplished by multiplying − 0 from the right by γ 0 Introducing the adjoint --row-- spinor ψ¯ ψ†γ we obtain ≡ µ i∂µψγ¯ + mψ¯ =0
Friday, October 21, 2011 Dirac Lagrangian we pause to discuss transformation properties of ψ¯(x) γµ ψ(x) We have
µ 1 µ ψ¯!(x!) γ ψ!(x!)=ψ¯(x) S− (Λ) γ S(Λ) ψ(x) µ ¯ α =Λ α ψ(x) γ ψ(x)
under a Lorentz transformation µ bilinear combination ψ¯(x) γ ψ(x)
transforms like a contravariant four-vector
we can write down a Lagrangian for spin-1/2 relativistic particles
= ψ¯(iγµ∂ m)ψ LDirac µ −
Friday, October 21, 2011 Continuity Equation
By adding multiplied from left by ψ ¯ and from right by ψ µ µ µ we obtain ψγ¯ ∂µψ +(∂µψ¯)γ ψ = ∂µ(ψγ¯ ψ)=0 µ µ showing that probability and flux densities j = ψγ¯ ψ satisfy continuity equation 4 0 0 2 is now positive Moreover ρ j = ψγ¯ ψ = ψ†ψ = ψi ≡ | | definite i=1 ! In this respect quantity ψ(x) resembles Schrodinger wave function Dirac eq. may serve as a one particle eq. 0 In that role --however-- coefficient of ix in Fourier decomposition − ip . x ψ(x)= dpψ(p) e− ! plays role of energy and there is no reason why negative energies should be excluded
Friday, October 21, 2011 Plane wave solutions Next we discuss plane wave solutions of Dirac equation We will treat positive and negative frequency terms separately and ipx ipx therefore write ψ(x)=u(p)e− + v(p) e
Since ψ also satisfies Klein Gordon equation µ 2 0 it is necessary that p pµ = m so that p =+ p! 2 + m2 E iEx ≡ We will call e 0 positive frequency solution − ! From Dirac eq. it follows that
µ ipx µ ipx [iγ ( ip ) m] u(p) e− +[iγ (ip ) m] v(p) e =0 − µ − µ − or equivalently µ (γ pµ m) u(p)=0 µ − (γ pµ + m) v(p)=0
because positive and negative frequency solutions are independent
Friday, October 21, 2011 Positron spinors
(3,4) two negative energy solutions u are to be associated with an antiparticle say the positron
Using antiparticle prescription positron of energy E and momentum p!
is described by one of E and p! electron solutions − − (3,4) i[ p] .x (2,1) ip . x u ( p) e− − v (p)e − ≡
0 where p E>0 ≡ positron spinors v are defined just for notational convenience
Friday, October 21, 2011 Solution for free particles at rest µ µ It is useful to introduce the notation γ pµ = γµp =p µ ν ! µ “slash” quantities satisfy a, b = a b γ ,γ =2a b 2a.b {! ! } µ ν { } µ ≡ Dirac eq. for a plane wave solution may thus be written as (p m) u(p)=0 ! − (p + m) v(p)=0 ! It is easily seen that u¯(p)(p m)=0 ! − v¯(p)(p + m)=0 ! (γ0 1) mu(0) = 0 − When p! =0and p0 = m equations take form (γ0 + 1) mv(0) = 0
There are two positive and two negative frequency solutions 1 0 0 0 0 1 0 0 u(1)(0) = u(2)(0) = v(2)(0) = v(1)(0) = 0 0 1 0 0 0 0 1 Friday, October 21, 2011 Solution for arbitrary momentum Since (p + m)(p m)=p2 m2 =0 ! ! − − we may write the solution for arbitrary p in the form u(r)(p)=C (m+ p) u(r)(0) (r) ! (r) v (p)=C! (m p) v (0) −! r =1 , 2 and C and C ! are normalization constants For fermions we choose covariant normalization in which we have 2E particles/unit volume just as we did for bosons
ρdV = ψ† ψdV = u†(p) u(p) = 2E !unit vol. ! 4 where we have used 0 0 2 ρ j = ψγ¯ ψ = ψ†ψ = ψ ≡ | i| i=1 ! ipx ipx ψ(x)=u(p)e− + v(p) e
Friday, October 21, 2011 Orthogonality Relations This leads to the orthogonality relations (r) (s) (r) (s) u †(p) u (p)=2Eδrs ,v†(p) v (p) = 2Eδrs 0 µ u¯(p) γ (γ pµ m) u(p)=0 By summing − u¯(p)(γµp m) γ0 u(p)=0 µ − we obtain 2¯u(p) p u(p) 2 mu†(p) u(p)=0 0 − 0 k k 0 where we have used relation γ γ = γ γ −
orthogonality relations then become
(r) (s) m (r) (s) u¯ (p) u (p)= u †(p) u (p)=2mδ E rs and (r) (s) m (r) (s) v¯ (p) v (p)= v †(p) v (p)= 2mδ − E − rs
Friday, October 21, 2011 Normalization Constant Using p p = p 2 we obtain ! ! u¯(r)(p) u(s)(p)= C 2 u¯(r)(0) (m+ p)(m+ p) u(s)(0) | | ! ! =2m C 2 u¯(r)(0) (m+ pu(s)(0) | | ! =2m C 2 u¯(r)(0) (m + γ0p + αkp β) u(s)(0) | | 0 k =2m C 2 (m + E)¯u(r)(0) u(s)(0) | | =2m C 2 (m + E) δ | | rs 1 and determine the normalization constant C = √m + E
A straightforward calculation leads to 1 C! = √m + E
Friday, October 21, 2011 Hints for the calculation
(p + m)(p + m)=m2 +2m p+ p2 =2m2 +2m p ! ! ! ! !
E 0 0 0 0 0 0 p1 0 E 00 00p 0 γ0p + γkp = + 1 0 k 00 E 0 0 p 00 − 1 0 0 0 E p 0 0 0 − 1 000ip2 00p3 0 00ip −0 0 0 0 p + 2 + 3 0 ip 00 p 0 0− 0 − 2 3 ip 000 0 p 00 2 − 3 E 0 p3 p1 ip2 0 Ep+ ip −p = 1 2 3 p p ip E −0 3 1 − 2 − p + ip p 0 E 1 2 − 3 −
Friday, October 21, 2011 Hints for the calculation (cont’d)
1 0 0 0 0 1 0 0 u¯(1)(0)(γ0p + γkp )u(1)(0) = 1000 0 k 00 10 − ! " 0 0 0 1 − E 0 p3 p1 ip2 1 0 Ep+ ip −p 0 1 2 − 3 × p3 p1 ip2 E 0 0 − − p + ip p 0 E 0 1 2 − 3 − 1 0 0 0 E 0 1 0 0 0 = 1000 00 10 p − 3 ! " 0 0 0 1 p1 + ip2 − E 0 = 1000 p − 3 ! " p1 ip2 − −
Friday, October 21, 2011 General Solution of Dirac Equation (r) 1 1 2 0 Introducing two-component spinors χ where χ =( 0)and χ =(1) we may examine explicit form of solution of Dirac equation For we have m+ p E>0 u(r)(p)= ! χ(r) √m + E m + σ E iσ σ.p# = 3 − 2 χ(r) √m + E 1 m + E σ.p# χ(r) = − √m + E σ.p# m E 0 ! − "! " χ(r) = √E + m (E + m) 1 σ.p# χ(r) ! − " and so positive-energy four spinor solutions of Dirac’s equation are 1 0 u (E,p! )=√m + E 1 $ % 1 (m + E) 1 σ.p! − 0 $ % 0 √ 1 u2(E,p! )= m + E $ % 1 0 (m + E)− σ.p! For low momenta 1 $ % upper two components are a great deal larger than the lower ones
Friday, October 21, 2011 negative energy solutions
For the E<0
1 m + E σ.p" 0 u(r+2)(p)= − √m + E σ.p" m E χ(r) ! − "! "
solutions of Dirac eq. are
1 1 (m E)− σ.p! − − 0 u3(E,p! )=√m E $ % − 1 0 $ %
1 0 (m E)− σ.p! − − 1 u4(E,p! )=√m E $ % − 0 1 $ % Friday, October 21, 2011 Projection operators & completeness
To obtain completeness properties of solutions consider positive and negative solutions separately We use the explicit solutions already obtained
1 2 (Λ ) u(r)(p)¯u(r)(p) + αβ ≡ 2m α β r=1 ! 1 = (p + m)u(r)(0)u ¯(r)(0)(p + m) 2m (m + E) " " " r # ! αβ 1 1+γ0 = (m+ p) (m+ p) 2m (m + E) " 2 " $ %αβ 1 1 = m(p + m)+ (p + m)[(m p)γ0 +2E] 2m (m + E) " 2 " −" & 'αβ 1 = (p + m) 2m " αβ
Friday, October 21, 2011 More on projection Operators & completeness 2 1 (r) (r) Similarly if we define Λ by (Λ )αβ = vα (p)¯v (p) − − −2m β r=1 1 ! we get (Λ )αβ = (m p)αβ − 2m −" completness relation is that 2 1 (r) (r) (r) (r) Λ+ +Λ = [uα (p)¯u (p) vα (p)¯v (p)] = 1 − 2m β − β r=1 ! matrices Λ + and Λ have properties of projection operators − 2 because Λ =Λ and Λ+Λ =Λ Λ+ =0 ± ± − − operators Λ project positive and negative frequency solutions but because there± are four solutions there must still be another projector operator which separates r =1, 2 solutions This projector operator h must be such that (r) (s) (r) (r) h h = δrs h and [h , Λ ] = 0 ± Friday, October 21, 2011 Helicity operator Since two solutions have something to do with two possible polarization directions of spin-1/2 particle we may expect the operator to be some sort of generalization of non-relativistic operator that projects out the state polarized in a given direction for two component spinors
On inspection we see that the the helicity operator
1 σk 0 h p.ˆ Σ = pˆ ≡ 2 k 0 σk ! "
satisfies where pˆ p/! p! ≡ | | Helicity operator commutes with H so it shares its eigenstates with H and its eigenvalues are conserved
Friday, October 21, 2011 Eigenvalues of helicity operator
To find the eigenvalues of the helicity operator we calculate
1 (σ.pˆ)2 0 1 pˆ2 0 h2 = = 4 0(σ.pˆ)2 4 0ˆp2 ! " ! "
Thus, the eigenvalues of the helicity operator are
+ 1 positive helicity 2 −→⇒ h = 1 negative helicity − 2 −→⇐ 1 The spin component in the direction of motion p.ˆ σ 2 is thus a good quantum number and can be used to label the solutions
Friday, October 21, 2011 Particle’s spin up and spin down Assuming a particle has a momentum p! and choosing x axis 3 − along the direction of p! we can determine which of the four spinor u1,u2,v1 and v2 have spin up and spin down With these assumptions σ p" = σ3p3, p" = p3 · | | 1 σ pˆ 0 1 σ 0 and the helicity operator simplifies to h = 3 3 = 3 2 0 σ3pˆ3 2 0 σ3 ! " ! " We then find 1 1 √E + m 1 0 hu1 = − ' ( 2 1 1 (E + m) 1 σ.p" 1 − 0 − ' ( 1 √E + m 0 1 = ' ( = u1 2 1 2 (E + m) 1 σ.p" − 0 ' ( 1 0 √E + m 1 1 hu2 = − ' ( 2 1 0 (E + m) 1 σ.p" 1 − 1 − ' ( 0 √E + m 1 1 = ' − ( = u2 2 0 2 (E + m) 1 σ.p" − − 1 ' − ( Friday, October 21, 2011 Antiparticle’s spin up and spin down For antiparticles with negative energy and momentum p,! σ.p! = σ3( p3) − − and the helicity operator simplifies to 1 σ pˆ 0 1 σ 0 h = − 3 3 = − 3 2 0 σ3pˆ3 2 0 σ3 ! − " ! − " 1 0 (E + m) 1 σ.p" √E + m − 1 − 1 hv1 = ' ( We then find 2 1 0 − 1 1 ' ( 0 (E + m) 1 σ.p" √E + m − 1 1 = ' ( = v1 2 0 2 1 ' ( 1 1 (E + m) 1 σ.p" √E + m − 1 − 0 hv2 = ' ( 2 1 1 − 1 0 ' ( 1 (E + m) 1 σ.p" √E + m − −0 1 = ' ( = v2 2 1 −2 − 0 ' ( Friday, October 21, 2011 Homework
Use Euler-Lagrange equation to derive:
1 µ 1 2 2 Klein-Gordon equation from = ∂ φ∂ φ m φ LKG 2 µ − 2
µ and Dirac equation from = ψ¯(iγ ∂ m)ψ LDirac µ −
Friday, October 21, 2011 Bilinear Covariants To construct most general form of Lorentz covariant currents need to tabulate bilinear quantities of form ψ¯(4 4)ψ × which have definite properties under Lorentz transformations
5 0 1 2 3 To simplify the notation ☛ we introduce γ iγ γ γ γ ≡ It follows that 5 5 5 2 5 µ µ 5 γ † = γ , (γ ) = I,γ γ + γ γ =0
In Dirac-Pauli representation 0 γ5 = I I 0 ! " We are interested in behavior of bilinear quantities under: proper Lorentz transformations --that is rotations and boosts-- and under space invertion --parity operation--
Friday, October 21, 2011 Explicit form of Bilinear covariants µ list arranged in increasing order ofγ matrices that are sandwiched between ψ¯ and ψ pseudoscalar is the product of four matrices Table 1.3: Bilinear covariants. The list is arranged in increasing order of Ifthe five number matrices of γµ werematrices used that ☛ areat least sandwiched two would between beψ¯ theand sameψ. The inpseudoscalar which case is product the product will of fourbe reduced matrices. to If fivethree matrice s were used, at andleast be two already would beincluded the same, in inaxial which vector case the product will be reduced to three and be already included in the axial vector. No. of Compts. Space Inversion, P Scalar ψψ¯ 1 + under P Vector ψγ¯ µψ 4 Space compts. under P − Tensor ψσ¯ µν ψ 6 Axial vector ψγ¯ 5γµψ 4 Space compts. + under P Pseudoscalar ψγ¯ 5ψ 1 under P −
To construct the most general form of currents consistent with Lorentz Friday,covariance, October 21, 2011 we need to tabulate bilinear quantities of the form (Ψ¯ )(4 4)(ψ), × which have definite properties under Lorentz transformations, where the 4 4 × matrix is a product of γ-matrices. To simplify the notation, we introduce
γ5 iγ0γ1γ2γ3 . (1.5.103) ≡ It follows that
γ5 = γ5, (γ5)2 = 1, γ5γµ + γµγ5 =0. (1.5.104)
In the Dirac-Pauli representation
0 1 γ5 = . (1.5.105) ! 1 0 "
We are interested in the behavior of bilinear quantities under proper Lorentz transformations (that is rotations and boosts), and under space invertion (the parity operation). An exhaustive list of the possibilities is given in Table 1.3. Because of the anticommutation relations, (1.5.42), the tensor is antisymmetric i σµν = (γµγν γνγµ) . (1.5.106) 2 − 31 Examples µ ν µν Because of anticommutation relations, γ ,γ =2γ { } tensor is antisymmetric i σµν = (γµγν γν γµ) 2 −
µ 1 µ From ψ¯!(x!) γ ψ!(x!)=ψ¯(x) S− (Λ) γ S(Λ) ψ(x) µ ¯ α =Λ α ψ(x) γ ψ(x) it follows immediately that ψψ¯ is a Lorentz scalar The probability density ρ = ψ†ψ is not a scalar, but is the timelike ¯ µ component of the four vector ψγ ψ 5 5 5 Because γ S = S γ the presence of γ gives rise to P − P the pseudo-nature of the axial vector and pseudoscalar i.e., a pseudoscalar is a scalar under proper Lorentz transformations but, unlike a scalar, changes sign under parity
Friday, October 21, 2011 Friday, October 21, 2011