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Particle Physics 2011

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Particle Physics 2011 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.
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