x 3 x 1 ix 2 x 1 + ix 2 x 3

Transcription

1 Peeter Joot PHY2403H Quantum Field Theory. Lecture 19: Pauli matrices, Weyl spinors, SL2,c, Weyl action, Weyl equation, Dirac matrix, Dirac action, Dirac Lagrangian. Taught by Prof. Erich Poppitz DISCLAIMER: Very rough notes from class, with some additional side notes. These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall Fermions: R 3 rotations. Given a real vector x and the Pauli matrices [ ] σ =, σ 2 = 1 0 [ ] [ ] 0 i, σ =. 1.1 i We may form a Pauli matrix representation of a vector [ σ x = x 3 x 1 ix 2 x 1 + ix 2 x 3 ], 1.2 where σ = σ 1, σ 2, σ 3. This matrix, like the Pauli matrixes, is a 2 2 Hermitian traceless matrix. We find that the determinant is detσ x = x3 2 x1 2 x = x 2. We may form Uσ xu, 1.4 where U is a unitary 2 2 unit determinant matrix, satisfying U U = 1 det U =

2 Further detuσ xu = det U detσ x det U = detσ x. Moral: Uσ xu = σ x, where x has the same length of x. We may use this to represent an arbitrary rotation Uσ xu = R i jx j σ i We say that U SU2 and R SU3, and SU2 is called the universal cover of SO3. Pauli figured out that, in non-relativistic QM, that this type of transformation also applies to spin wave functions spinors Ψx Ψ x = UΨx 1.8 where and R T R = 1. Here Ψ is a two element vector x x = Rx, 1.9 Ψx = [ ] Ψ x, Ψ x so the transformation should be thought of as a matrix operation [ ] [ ] Ψ x Ψ x [ ] Ψ x Ψ x Ψ = U x Ψ x Having seen such representations and their SU2 transformations in NRQM, we want to know what the relativistic generalization is. 1.2 Lorentz group Let This has determinant We therefore identify x 0, x as a four vector x 0, x = x 0 σ 0 + σ x [ x = 0 ] + x 3 x 1 ix 2 x 1 + ix 2 x 0. x 3 detx 0, x = x 0 2 x 1 2 x 2 2 x 3 2 = x µ x µ. x 0, x = x µ σ µ We say that SL2, C is a double cover of SO1, 3. Note that the matrix U can be built explicitly. For example, it may be built up using Euler angles as sketched in fig or algebraically U = e iψσ3/2 e iθσ1/2 e iφσ3/

3 Figure 1.1: Euler angle rotations. 1.3 Weyl spinors. We will see that there is generalization of Pauli spinors, called Weyl spinors, but we will have to introduce 4 component objects. We d like to argue that there is a correspondence also 2 1 between SL2, C SO1, 3. Here: S : special L : linear 2 : 2 2 C : complex. and we say that M SL2, C if det M = 1, where M is a complex 2 2, but not necessarily unitary. The SU2 group is a subset of SL2, C. In this representation SU2 matrices are SL2, C matrices, but not necessarily the opposite. We introduce a special notation for the identity matrix [ ] σ and can now form four vectors in a matrix representation x σ x µ σ µ [ x 0 σ 0 + σ x x = 0 + x 3 x 1 ix 2 ] x 1 + ix 2 x 0 x

4 Such 2 2 matrices are Hermitian. Notice that the space of 2 2 Hermitian matrices is 4 dimensional. We found that detx µ σ µ = x 0 2 x The transformation x µ σ µ M x µ σ µ M, 1.19 maps 2 2 Hermitian matrices to 2 2 Hermitian matrices using a unit determinant transformation M. Note that M is not unitary, as it is an arbitrary Hermitian matrix. In particular MM = 1! Also note that the determinant of the transformed object is det M x µ σ µ M = 1 det x µ σ µ 1, 1.20 since det M = 1, so that we see that the Lorentz invariant length is preserved by such a transformation. This can be expressed as x σ Mx σm = x σ, 1.21 where x 2 = x 2. Motivated by this SL2, C SO1, 3 correspondence, postulate that we study two component objects [ ] U1 x Ux =, 1.22 U 2 x where x = x 0, x 1, x 2, x 3 is a four-vector, and assume that such objects transform as follows in SO1, 3 Ux U x = M Ux x µ x µ = Λ µ ν x ν, 1.23 where M is the one giving rise to Λ. To understand what is meant by giving rise to, consider Mx µ σ µ M = x ν σν = σ ν Λ ν µx µ, 1.24 and this holds for all x µ, we must have Mσ µ M = σ ν Λ ν µ Theorem 1.1: Transformation of U xσ µ Ux U xσ µ Ux transforms as a four vector. 4

5 Proof: U xσ µ Ux U x σ µ U x = M Ux σ µ M Ux = U x Mσ µ M Ux = U xσ ν UxΛ ν µ 1.26 so we find that U xσ µ Ux transforms as a four vector as claimed. Theorem 1.2: Transformation of partials. The four-gradient coordinates transform as a four vector µ = Λ 1 σ µ σ. Proof 1 : Inverting the transformation relation x µ = Λ µ ν x ν, 1.27 gives so x σ = Λ 1 σ µ Λµ νx ν = Λ 1 σ µ x µ, 1.28 µ µ = x µ = xσ x µ x σ = Λ 1 σ µ x σ = Λ 1 σ µ σ Theorem 1.3: Transformation of U σ µ µ U U σ µ µ U transforms as a four vector. 1 In class we proved this by considering the transformation properties of a direction derivative dx µ µ, but that isn t the method that seems most intuitive to me. 5

6 Proof: We can now define Definition 1.1: Weyl action name? U xσ µ x µ Ux U x σ µ x µ U x = Λ µ νu xσ ν x µ Λ 1 µ µ Ux = U xσ ν x µ δµ νux = U xσ ν x ν Ux 1.30 We may construct the following Lorentz invariant action where Ux is a Weyl spinor. S Weyl = d 4 xiu xσ µ µ Ux, The i factor here is so that the action is real. This can be seen by noting that iσ µ = iσ µ and integrating the Hermitian conjugate by parts iσ 0 [ ] 0 i = = iσ 0 i a iσ 1 [ ] 0 i = = iσ 1 i b iσ 2 [ ] 0 1 = = iσ c iσ 3 [ ] i 0 = = iσ 3 0 i 1.31d S Weyl = d 4 x µ U xiσ µ Ux = d 4 x µ U xiσ µ Ux = d 4 x µ U xiσ µ Ux + = d 4 xu xiσ µ µ Ux = S Weyl, d 4 xu xiσ µ µ Ux

7 where it was assumed that any boundary terms vanish. Theorem 1.4: Weyl equation. Variation of the action definition 1.1 gives rise to the equations of motion which is called the Weyl equation. σ µ x µ U = 0 Proof: δs = i = i = i = d 4 x δu σ µ µ U + U σ µ µ δu d 4 x δu σ µ µ U + µ U σ µ δu d 4 x δu σ µ µ U µ U σ µ δu d 4 x δu iσ µ µ U + δu iσ µ µ U. µ U σ µ δu 1.33 Requiring this to vanish for all variations δu proves the result. Written out explicitly in matrix form, the Weyl equation is [ ] [ ] i 2 U1 = 0, 1 + i U or U i 2 U 2 = a 1 + i 2 U U 2 = b Theorem 1.5: Weyl equation relation to the massless KG equation. The Weyl equation is equivalent to a set of massless KG equations. for k = 1, 2. µ µ U k = 0, 7

8 Proof: Multiplying eq. 1.35a by 1 + i 2 gives 1 + i U i 2 U 2 = i 2 U i 2 1 i 2 U 2 = U i 2 1 i 2 U 2 = U 2 = U 2 = µ µ U Similarly, multiplying eq. 1.35b by 1 i 2 we find 0 = 1 i i 2 U U 2 = U i 2 U 2 }{{} = U 1 = U 1 = µ µ U Because S Weyl results in a massless KG equation, this is no good for electrons, and we have to look for a different action. Claim: U T σ 2 U is the only bilinear Lorentz invariant that we can add to the action. An action like: L mass = 1 2 mut σ 2 U m U σ 2 U T, 1.38 may exist in nature we don t know, and are called Majorana neutrino masses. The problem with such a Lagrangian density is that it breaks U1 symmetry. In particular U e iα U symmetry of the kinetic term. This means that the particle associated with such a Lagrangian cannot be charged. Recall that we introduced electromagnetic potentials into NRQM with i h t Ψ = 1 2m ea2 Ψ which is a gauge transformation. We d like to have this capability. What we can do instead and maintain U1 symmetries, is to introduce two U s, like L mass = 1 2 mut 1 σ 2U m U 2 σ 2 U 1 T What we are really doing is assembling a four component spinor out of the two U s. 8

9 1.4 Lorentz symmetry. We want to examine the Lorentz invariance of U T σ 2 U, but need an intermediate result first. Lemma 1.1: Transpose of Pauli vector representation For any x R 3 σ x T = σ 2 σ xσ 2, or more compactly σ T = σ 2 σσ 2. Geometrically, this transposition operation reflects x about the y-axis. Proving lemma 1.1 is well suited to software FIXME: link: diracweylmatrixrepresentationandidentities.nb, but can also be done algebraically with ease. First note that σ T 1 = σ 1 σ T 2 = σ which means that σ3 T = σ 3 σ x T = σ 1 x 1 σ 2 x 2 + σ 3 x 3 = σ 2 σ 2 σ 1 x 1 σ 2 x 2 + σ 3 x 3 = σ 2 σ 1 x 1 σ 2 x 2 σ 3 x 3 σ 2 = σ 2 σ x T σ Now we are ready to proceed. Theorem 1.6: U T σ 2 U invariance U T σ 2 U is Lorentz invariant. Proof: U T σ 2 U U T σ2 U = U T M T σ 2 M U, 1.43 where U = M U and U T = U T M T. Note that if we can show that M T σ 2 M = σ 2, then we are done. 9

10 It is simple to show that any U = e iσ a, 1.44 for a R 3, has eigenvalues ±i a. The determinant of such a matrix is det U = ei a 0 0 e i a = 1, 1.45 so we see that such a matrix has the U U = 1 and det U = 1 properties that we desire for elements of SU2 2. We haven t shown that all matrices U SU2 can be written in this form, but let s assume that s the case. Claim: Generalizing from the exponential form of SU2 elements seen above, we assume that any SL2, C matrix M can be written as M = e iσ a+ib, 1.46 for a, b R 3. The transpose of an exponential of a sigma matrix goes like so e σ u T = = k=0 k=0 = σ 2 k=0 1 σ u k T k! 1 k! σ 2σ uσ 2 k = σ 2 e σ u σ 2, 1 σ uk k! M T σ 2 M = e iσ a+ib T σ2 e iσ a+ib = σ 2 e iσ a+ib σ 2 σ 2 e iσ a+ib = σ 2, σ which is the result required to finish the proof of theorem Dirac matrices. 2 In class the suitability of e iσ a as an element of SU2 was demonstrated with an argument that diagonalizable matrices satisfy det e A = e tr A 3 A slightly different derivation was done in class, but this one makes more sense to me. 10

11 Definition 1.2: Dirac matrices. The Dirac matrices γ µ, µ {0, 1, 2, 3} are matrices that satisfy that is {γ µ, γ ν } = 2g µν, γ µ γ ν + γ ν γ µ = 2g µν, We will use the explicit 4 4 matrix representation γ 0 = [ ] 0 1, 1 0 and [ ] γ i 0 σ i = σ i. 0 The metric relations can also be written explicitly in the handy form γ 0 2 = 1 γ i 2 = 1. Written out explicitly, these matrices are γ 0 = , γ1 = , i γ 2 = 0 0 i 0 0 i 0 0, γ3 = i We will see HW4 that Lorentz transformations take the form where where x γ = Λ 1 1/2 x γ Λ 1/2, Λ 1/2 = e i 2 ω µνs µν, S µν = i 4 [ γ µ, γ ν]

12 In particular S 0k = i 4 [γ 0, γ k] = i 2 γ0 γ k = i [ ] [ ] σ k σ k 0 = i [ ] σ k σ k 1.54 will generate boosts, whereas S jk = i 4 [γ j, γ k] = i 2 γj γ k = i [ ] [ ] 0 σ j 0 σ k 2 σ j 0 σ k 0 [ σ k σ j ] = i 2 0 σ k σ j = 1 [ ] σ l 0 2 ɛjkl 0 σ l, are rotations and in this case, are Hermitian. The explicit expansion of the half Lorentz transformation operator is Λ 1/2 = e i 2 ω µνs µν = e iω 0kS 0k 2 i ω jks jk = exp 1 2 = [ ω0k σ k 0 0 ω 0k σ k ] i [ ωjk ɛ jkl σ l ] ω jk ɛ jkl σ l ] [ e 1 2 ω 0kσ 0 + i 4 ω jke jkl σ l 0 0 e 1 2 ω 0kσ 0 + i 4 ω jke jkl σ l 1.56 where the 1/2 factor of ω 0i vanished because we had a sum over 0i and i0 which have been grouped. Lemma 1.2: Some Dirac matrix identities. γ 0 = γ 0 γ k = γ k 12

13 γ 0 iγ µ γ 0 = iγ µ. The first two are clear from inspection of eq For the last, for µ = 0 and for µ = k = 0 which completes the proof. γ 0 iγ 0 γ 0 = γ 0 γ 0 iγ 0 = iγ 0 γ 0 γ 0 = iγ 0, γ 0 iγ k γ 0 = γ 0 i γ k γ 0 = +iγ 0 γ k γ 0 = iγ 0 γ 0 γ k = iγ k, Dirac Lagrangian. We postulate that there is a four-component object ψ 1 Ψ = ψ 2 ψ 3 Ψ = ψ1, ψ 2, ψ3, ψ4, 1.59 ψ 4 where ψ µ s are all complex fields, and assume that the fields transform as Ψx Ψ x = Λ 1/2 Ψx, 1.60 where our vectors transform in the usual x x Lorentz transformation is the usual = Λx fashion, where the incremental form of the Λ µ ν = δ µ ν + ω µ ν + Oω Definition 1.3: Overbar operator name?. Ψ = Ψ γ 0. 13

14 Definition 1.4: Dirac Lagrangian. L Dirac = Ψx iγ µ µ m Ψx. Armed with lemma 1.2 we can now show the following. Theorem 1.7: The Dirac action is a real Lorentz scalar. The action S = d 4 xψ iγ µ µ m Ψ, is a real scalar and is Lorentz invariant. Real: To show that the action is real, we compute it s Hermitian conjugate, apply lemma 1.2 and integrate by parts S = d 4 xψ iγ µ µ m γ 0 Ψ = d 4 xψ iγ µ µ m γ 0 Ψ = d 4 xψ γ 0 iγ µ γ 0 µ m γ 0 Ψ 1.62 = d 4 xψ iγ µ µ m Ψ = d 4 x µ Ψiγ µ Ψ + d 4 xψiγ µ µ Ψ d 4 xψmψ = d 4 xψ iγ µ µ m Ψ = S, where µ without an overarrow means the traditional right acting operator, and assuming that the boundary terms vanish. To show the Lorentz invariance, we will consider just the transformation of the Dirac Lagrangian density. We need a couple additional pieces of information to do so, the first of which is the transformation property 4 Ψ ΨΛ /2, and from HW4 Λ 1 1/2 γµ Λ 1/2 = Λ µ αγ α Not proven here, but there s an argument for that in [1] eq

15 The Lagrangian transforms as Ψx iγ µ µ m Ψx Ψ x γ 0 iγ µ x µ m Ψ x = ΨxΛ 1 iγ µ Λ 1 α 1/2 µ α m Λ 1/2 Ψx = Ψx iλ 1 1/2 γµ Λ 1/2 Λ 1 α µ α m Ψx = Ψ iγ µ µ m Ψ 1.65 We find that ΨΨ = Ψ γ 0 Ψ is a Lorentz scalar, whereas Ψγ µ Ψ is a 4 vector. 1.7 Problems: Exercise 1.1 Answer for Exercise 1.1 Show that ΨΨ is a Lorentz scalar. The Lorentz property follows from eq ΨΨ = ΨΨ. ΨΛ 1 1/2 Λ 1/2 Ψ 1.66 The scalar nature of this product can be seen easily by expansion. ΨΨ = Ψ γ 0 Ψ Ψ 1 = [ Ψ1 Ψ2 Ψ3 Ψ4 ] Ψ Ψ Ψ 4 Ψ 3 = [ Ψ1 Ψ2 Ψ3 Ψ4 ] Ψ 4 Ψ 1 Ψ 2 = Ψ1 Ψ 3 + Ψ2Ψ 4 + Ψ3Ψ 1 + Ψ4 Ψ 2 = 2 Re Ψ1 Ψ 3 + Ψ2Ψ Clearly any individual Ψ γ µ Ψ product will also be a scalar. Exercise 1.2 Show that Ψγ µ Ψ transforms as a four vector. 15

16 Answer for Exercise 1.2 Ψγ µ Ψ = Ψ ΨΛ 1 1/2 γ µ Λ 1/2 Ψ Λ 1 1/2 γµ Λ 1/2 Ψ = Ψ Λ µ νγ ν Ψ = Λ µ νψγ ν Ψ

17 Bibliography [1] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview,