Srednicki Chapter 36

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1 Srednicki Chapter 36 QFT Problems & Solutions A. George March 1, 13 Srednicki Using the results of problem.9, show that, for a rotation by an angle θ about the z axis, we have: DΛ = exp iθs 1 and that, for a boost by rapidity η in the z direction, we have DΛ = exp iηs 3 In problem.9, we showed this is true for the vector representation. It must therefore be true for all other representations, including this spinor representation. Srednicki 36.. Verify that equation is consistent with equation Equation is: Using equation 36.39: γ 5 = i I I σ1 σ 1 γ 5 = iγ γ 1 γ γ 3 σ σ σ3 σ 3 Doing the multiplication: σ1 γ 5 = i σ 1 σ σ 3 σ σ 3 This gives: Now recall that iσ 1 σ σ 3 = I. Then: which is equation σ1 σ γ 5 = i σ 3 σ 1 σ σ 3 γ 5 = I I 1

2 Srednicki a Prove the Fierz Identities χ 1σ µ χ χ 3σ µ χ 4 = χ 1χ 3χ χ 4 χ 1σ µ χ χ 3σ µ χ 4 = χ 1σ µ χ 4 χ 3σ µ χ Recall that the right-handed fields are always written as Hermitian Conjugates of left-handed fields. Thus, we insert the indices: χ 1σ µ χ χ 3σ µ χ 4 = χ 1ȧ σµȧa χ a χ 3ċ σċc µ χ c 4 This is index notation, so we can move these around as we like: χ 1σ µ χ χ 3σ µ χ 4 = σċc µ σ µȧa χ 1ȧ χ aχ 3ċ χc 4 Now we use equation 35.4: χ 1σ µ χ χ 3σ µ χ 4 = ε ac εȧċ χ 1ȧ χ aχ 3ċ χc 4 Now we use the Levi-Cevita symbol to raise the indices recall that the Levi-Cevita symbol is the spinor analog of the metric. We also move them around at will. Then: χ 1σ µ χ χ 3σ µ χ 4 = χ ċ 1 χ 3ċ χ a χ a 4 Now we drop the indices since this are just multiplied: χ 1σ µ χ χ 3σ µ χ 4 = χ 1χ 3 χ χ 4 Now we want to get back where we started with 4. We know that χ χ 4 = χ 4 χ. [If you re confused about why the fields seem to commute when fermions should obviously anticommute, see equation 35.5.] Then: χ 1σ µ χ χ 3σ µ χ 4 = χ 1χ 3 χ 4 χ This shows that the right hand side of equation is invariant under this relabeling. It follows that the left hand side of equation is also invariant under this relabeling which is exactly what equation claims. b Define the Dirac fields Ψ i = χi ξ i, Ψ C ξi i = χ i Use equations and to prove the Dirac Form of the Fierz identities, Ψ 1 γ µ P L Ψ Ψ 3 γ µ P L Ψ 4 = Ψ 1 P R Ψ C 3 Ψ C 4 P L Ψ

3 Ψ 1 γ µ P L Ψ Ψ 3 γ µ P L Ψ 4 = Ψ 1 γ µ P L Ψ 4 Ψ 3 γ µ P L Ψ All we actually have to do is prove that these forms of the Fierz Identities are equivalent to those in part a. Let s start with equation 36.58: χ 1σ µ χ χ 3σ µ χ 4 = χ 1χ 3χ χ 4 On both sides, let s write this as a bra times a ket. [ ] [ ] [ ξ 1, χ 1 σ µ ξ χ 3, χ 3 = ξ σ µ χ 1, χ 1 4 χ 3 ] [ χ 4, ξ χ 4 where in the last term we remember, as in part a, that χ χ 4 = χ 4 χ. We can use equation 36.6 to identify some of these terms: [ ] [ ] [ ] [ ] Ψ 1 σ µ Ψ χ 3 = Ψ σ µ χ 1 4 χ Ψ C χ 4 3 ] On the left-hand side, let s separate the σs into their own matrix: [ ] [ ] [ σ µ χ σµ χ4 Ψ 1 σ µ Ψ 3 = Ψ σ µ 1 χ 3 ] [ Ψ C 4 χ ] These are gamma matrices: [ ] [ Ψ 1 γ µ χ χ4 Ψ 3 γ µ ] [ = Ψ 1 χ 3 ] [ Ψ C 4 χ ] Now let s rewrite the remaining kets as projections: [ ] [ ] [ Ψ 1 γ µ χ χ4 ξ3 P L ξ Ψ 3 γ µ P L ξ = Ψ 1 P R 4 χ 3 ] [ Ψ C 4 P L χ ξ ] Using 36.6 again: [ Ψ1 γ µ P L Ψ ] [ Ψ3 γ µ P L Ψ 4 ] = [ Ψ1 P R Ψ C 3 ] [ Ψ C 4 P L Ψ ] which is To prove 36.6, let s go back to equation and write it as a bra times a ket as before, but this time we will write the last term differently: [ ] [ ] [ ξ 1, χ 1 σ µ ξ χ 3, χ ] [ 3 = ξ σ µ χ 1, χ ] 1 4 χ χ, ξ χ 4 3 The left-hand side has not changed; the right hand side is the same up to an arbitrary relabeling 4. We can therefore read off the result from equation : [ ] [ ] [ ] [ ] Ψ1 γ µ P L Ψ Ψ3 γ µ P L Ψ 4 = Ψ1 P R Ψ C 3 Ψ C P L Ψ 4 3

4 This shows that the right-hand side of equation is invariant under 4. It follows that the left-hand side must also be invariant under that transformation, ie that: [ Ψ1 γ µ P L Ψ ] [ Ψ3 γ µ P L Ψ 4 ] = [ Ψ1 γ µ P L Ψ 4 ] [ Ψ3 γ µ P L Ψ ] which is equation Note: This may seem like a brilliant but completely unintuitive solution. If so, just start with the result, apply the projection and gamma matrices, and arrive at Then reorder your results to get this solution or announce that all your steps are reversible, so you re already done. c By writing both sides out in terms of Weyl fields, show that Ψ 1 γ µ P R Ψ = Ψ C γ µ P L Ψ C 1 Ψ 1 P L Ψ = Ψ C P L Ψ C 1 Ψ 1 P R Ψ = Ψ C P R Ψ C 1 Writing both sides as instructed, we have: ξ 1, χ σ µ? 1 σ µ = χ ξ σ µ σ µ ξ ξ1 Doing the matrix multiplication: ξ 1 σ µ ξ? = ξ σ µ ξ 1 To see if this can be true, let s insert the indices. As usual, the daggers represent righthanded fields, the non-daggered fields are left handed that s our convention. We give dotted indices to right-handed fields, non-dotted indices to left-handed fields. Further, σ takes raised indices, σ takes lowered indices. Then: ξ a 1σ µ aȧ ξ ȧ? = ξ ȧ σµȧa ξ 1a On the left-hand side, let s use the Levi-Cevita symbol to raise our indices. Remember we have to contract with the second index of the Levi-Cevita symbol, otherwise we get a minus sign. Then, ε ab εȧḃξ 1b σ µ aȧ ξ? = ξ ḃ ȧ σµȧa ξ 1a Now let s use equaton which is a definition: ξ 1b σ µḃb ξ ḃ? = ξ ȧ σµȧa ξ 1a In order to be able to drop the indices on the left-hand side, we need to reverse the order of the two terms. However, note that the function indicated by the index of a Weyl Field is a fermionic function, and these always anticommute. Thus, ξ ḃσµḃb ξ 1b? = ξ ȧ σµȧa ξ 1a 4

5 Dropping the indices, we find: ξ σ µ ξ 1 = ξ σ µ ξ 1 Expanding the second equation, we have: ξ 1, χ χ 1? = χ ξ ξ 1 This gives: ξ 1 χ = χ ξ 1 The third equation is evaluated in the same manner as the second; the result is χ 1ξ = ξ χ 1 by is true see equation Srednicki Consider a field φ A x in an unspecified representation of the Lorentz Group, indexed by A, that obeys UΛ 1 φ A xuλ = LA B Λφ B Λ 1 x For an infinitesimal transformation: L B A 1 + δω = δ B A + i δω µνs µν B A a Following the procedure of section, show that the energy-momentum tensor is T µν = g µν L L µ φ A ν φ A The derivation in chapter still holds: everything is the same except that we ve replaced the scalar field φ a with the representation-independent field φ A. We therefore read off the energy-momentum tensor from equation.9; the result is equation b Show that the Noether current correpsoning to a Lorentz transformation is M µνρ = x ν T µρ x ρ T µν + B µνρ where Equation.7 gives the Noether Current: L B µνρ = i µ φ A Sνρ A B φ B j µ x = Lx µ φ A x δφ Ax K µ x

6 We need δφ A, which we get from the symmetry. Our Lorentz Symmetry is, in differential form: φ A LA B 1 + δωφ B x µ x ν δ ω µν We use for L B A Doing the multiplication: and make a Taylor Series of φ B. Then: [ φ A δa B + i ] δω µνs µν A B [φ B x µ φ A x ν δω µν ] Hence: φ A φ A µ φx ν δω µν + i δω µνs µν B A φ B δφ = [ ] i Sµν A B φ B µ φ A x ν δω µν Now for K µ. To determine this, we need to look at the change in the Lagrangian: Therefore: We can write this as: Lx LΛ 1 x Lx Lx µ δω µν x ν Lx Lx µ Lδω µν x ν δl = µ Lδω µν x ν δl = µ x ν Lδω µν Why? We can recover the original equation by doing the derivative with the product rule. One derivative will give a metric, which we use to change the differential to δω µ µ. This represents the trace of an antisymmetric matrix, which vanishes. The other term is the original equation. Next, we bring δω inside the derivative as well. We can treat this as a constant term because it is a differential: taking the derivative of a differential will effectively give a second-order differential, which we neglect. Then, Next we swap the indices: which gives, by the definition of K µ : δl = µ x ν Lδω µν δl = µ x ν Lδω µν K µ = δω µν x ν L Finally, we combine and into , the result is changing the indices as needed: j ρ = Lx [ ] i ρ φ A x Sµν A B φ B µ φ A x ν δω µν + Lδω ρν x ν 6

7 We distribute the first term: j ρ = Lx i ρ φ A x Sµν A B φ B µ φ A x ν Lx δω µν ρ φ A x + Lδωρν x ν This first term is B: j ρ = 1 Bρµν δω µν µ φ A x ν Lx δω µν ρ φ A x + Lδωρν x ν Using the result of part a, we have: j ρ = 1 Bρµν δω µν + T ρµ x ν δω µν g ρµ Lx ν δω µν + Lδω ρν x ν After manipulating the indices, these last two terms cancel: j ρ = 1 Bρµν δω µν + T ρµ x ν δω µν Rewriting this: j ρ = 1 [Bρµν δω µν T ρµ x ν δω µν T ρµ x ν δω µν ] In this last term, µ and ν are just dummy indices. We ll exchange them: j ρ = 1 [Bρµν δω µν T ρµ x ν δω µν T ρν x µ δω νµ ] δω is antisymmetric, so we can switch those indices: j ρ = 1 [Bρµν δω µν T ρµ x ν δω µν + T ρν x µ δω µν ] Now we factor out the differential: We recognize the term in the brackets as M: j ρ = 1 [Bρµν T ρµ x ν + T ρν x µ ] δω µν j ρ = 1 [Mρµν ] δω µν It is customary to drop the differential as well as the constant terms. Then, j ρµν = M ρµν as expected. c Use the conservation laws µ T µν = and µ M µνρ = to show that T νρ T ρν + µ B µνρ = 7

8 We have: This gives: µ M µνρ = µ x ν T µρ x ρ T µν + B µνρ = µ M µνρ = g ν µ T µρ + x ν µ T µ ρ g ρ µ T µν x ρ µ T µν + µ B µνρ = The derivative of T is zero, so: T νρ T ρν + µ B µνρ = d Define the improved energy-momentum tensor or Belinfante tensor i Show that Θ µν is symmetric. Θ µν = T µν + 1 ρb ρµν B µρν B νρµ We have: Θ µν = T µν + 1 ρ B ρµν B µρν B νρµ We can simply reorder these last two terms: Θ µν = T µν + 1 ρ B ρµν B νρµ B µρν From the definition of B, we see that the last two indices of B go onto S µν, the generators of the Lorentz Group for spinors. Since S is antisymmetric, it follows that B is antisymmetric in its last two indices. Therefore, we can write: We can even rewrite this as: Θ µν = T µν + 1 ρ B ρνµ B νρµ B µρν Θ µν = T µν ρ B ρνµ + 1 ρ B ρνµ B νρµ B µρν Again using the fact that B is antisymmetric in its last two indices, we write: Using the result from part c, we have: Θ µν = T µν + ρ B ρµν + 1 ρ B ρνµ B νρµ B µρν Θ µν = T νµ + 1 ρ B ρνµ B νρµ B µρν which gives: Θ µν = Θ νµ 8

9 ii Show that Θ µν is conserved, µ Θ µν = We take the derivative: µ Θ µν = µ T µν + 1 µ ρ B ρµν B µρν B νρµ This first term vanishes by Conservation of Energy. In the remaining terms, we will distribute: µ Θ µν = 1 µ ρ B ρµν 1 µ ρ B µρν 1 µ ρ B νρµ For the second term, we will use our usual trick of arbitrarily choosing to swap the dummy indices ρ and ν. Also, we ll divide the last term into two. These two changes give: µ Θ µν = 1 µ ρ B ρµν 1 ρ µ B ρµν 1 4 µ ρ B νρµ 1 4 µ ρ B νρµ Recall that partial derivatives commute; thus, the first and second terms cancel. fourth term, we ll again use our trick of swapping the dummy indices. Then: In the µ Θ µν = 1 4 µ ρ B νρµ 1 4 ρ µ B νµρ Now in the fourth term, we can again commute the partial derivatives. In addition, we have already discussed how B is antisymmetric in its last two indices. Thus, the remaining two terms cancel: µ Θ µν = iii Show that d 3 xθ ν = d 3 xt ν = P ν, where P ν is the energy-momentum four-vector. In general relativity, it is the Belinfante tensor that couples to gravity. We have: Θ ν d 3 x = T ν d 3 x + 1 ρ B ρν B ρν B νρ d 3 x We break this last term into temporal and spatial counterparts: Θ ν d 3 x = T ν d 3 x + 1 i B iν B iν B νi d 3 x + 1 B ν B ν B ν d 3 x The second integral on the right hand side vanishes, as it is an integral over a total divergence assuming reasonable boundary conditions at spatial infinity. In the last integral, the first two terms cancel, and the third term vanishes because B is antisymmetric in its last two indices, as discussed above. We re left with: Θ ν d 3 x = T ν d 3 x = P ν where the last equality follows by equation.35. e We define the improved tensor by: Ξ µνρ = x ν Θ µρ x ρ Θ µν 9

10 i Show that it obeys µ Ξ µνρ =. We evaluate the derivative: Using the product rule: µ Ξ µνρ = µ x ν Θ µρ x ρ Θ µν µ Ξ µνρ = µ x ν Θ µρ + x ν µ Θ µρ µ x ρ Θ µν x ρ µ Θ µν Since the Belinfante tensor is conserved our result from part d, the second and fourth terms vanish. In the remaining terms, we evaluate the derivative: Using the metric: µ Ξ µνρ = g ν µ Θ µρ g ρ µ Θ µν µ Ξ µνρ = Θ νρ Θ ρν Since the Belinfante Tensor is symmetric another result from part d, we have: µ Ξ µνρ = ii Show that d 3 xξ νρ = d 3 xm νρ = M νρ, where M νρ are the Lorentz generators. We have: [x Ξ νρ = ν Θ ρ x ρ Θ ν] Using the definition of Θ, we have: Ξ νρ d 3 x = [x ν T ρ + 1 [ µ B µρ B µρ B ρµ] x ρ T ν + 1 [ µ B µν B µν B νµ]] d 3 x Distributing: Ξ νρ d 3 x = x ν T ρ + 1 xν µ B µρ B µρ B ρµ x ρ T ν 1 xρ µ B µν B µν B νµ] d 3 x We break the sums over µ into temporal and spatial components: Ξ νρ d 3 x = x ν T ρ + 1 xν B ρ B ρ B ρ + 1 xν i B iρ B iρ B ρi x ρ T ν 1 xρ B ν B ν B ν 1 xρ i B iν B iν B νi] d 3 x The temporal terms vanish as before. Then: Ξ νρ d 3 x = x ν T ρ + 1 xν i B iρ B iρ B ρi x ρ T ν 1 xρ i B iν B iν B νi] d 3 x 1

11 Integrating by parts, we have: Ξ νρ d 3 x = x ν T ρ 1 ix ν B iρ B iρ B ρi x ρ T ν + 1 ix ρ B iν B iν B νi] d 3 x Doing the derivatives: Ξ νρ d 3 x = x ν T ρ 1 g ν i B iρ B iρ B ρi x ρ T ν + 1 g ρ i B iν B iν B νi] d 3 x Using the metric: Ξ νρ d 3 x = x ν T ρ 1 B νρ B νρ B ρν x ρ T ν + 1 B ρν B ρν B νρ] d 3 x The first and last terms cancel, as do the third and fourth. Then: Ξ νρ d 3 x = [x ν T ρ x ρ T ν + B νρ] d 3 x which is: Ξ νρ d 3 x = M νρ d 3 x = M νρ where the last equality follows by equation.4. f Compute Θ µν for a left-handed Weyl field with L given by equation 36., and for a Dirac field with L given by equation The Lagrangian given by equation 36. is: L = iψ ρ ρ ρ ψ 1 mψψ 1 mψ ψ where the m has been made real and positive as per the discussion in the text. We ve also changed the dummy index to ρ to avoid ambiguity later on. Then we can take the derivative: L µ ψ = iψ σ µ Equation gives, therefore: T µν = g µν [ iψ σ ρ ρ ψ 1 mψψ 1 mψ ψ ] iψ σ µ ν ψ Now we note that, since this is a left-handed Weyl Field, we take the generator of the Lorentz Group by equation 36.51: Equation 36.7 gives, therefore: S µν L B A = i 4 [σµ σ ν σ ν σ µ ] B A B µνρ = i 4 ψ σ µ [σ ν σ ρ σ ρ σ ν ] B A ψ B 11

12 Combining all these results in equation 36.7, we have: [ Θ µν = g µν iψ σ ρ ρ ψ 1 mψψ 1 ] mψ ψ iψ σ µ ν ψ + 1 [ i ρ 4 ψ A σ ρ [σ µ σ ν σ ρ σ ν ] B A ψ B i 4 ψ A σ µ [σ ρ σ ν σ ρ σ ν ] B A ψ B i 4 ψ A σ ν [σ ρ σ µ σ ρ σ ν ] B A ψ B There s not much we can do to simplify this. We could commute the derivative through the σ, but that would require us to use the product rule, making this more complicated rather than less so; further, nothing would cancel. All we can really do is to pull out some common factors: Θ µν = ig µν ψ σ ρ ρ ψ m gµν ψψ m gµν ψ ψ iψ σ µ ν ψ + i [ 8 ρ ψ A σ ρ σ µ σ ν σ ν σ µ B A ψ B ψ A σ µ σ ρ σ ν σ ν σ ρ B A ψ B ψ A σ ν σ ρ σ µ σ µ σ ρ B A ψ B ] ] As for the Dirac Field, we again have: The energy-momentum tensor is: Θ µν = T µν 1 ρ B ρµν B µρν B νρµ T µν = g µν L The Lagrangian is given by equation 36.8: and so we can compute L µ Ψ ν Ψ L = iψγ µ µ Ψ mψψ L µ Ψ = iψγµ Thus the energy-momentum tensor is given by: Finally, we need B µνρ, which is given by: T µν = ig µν Ψγ µ µ Ψ mg µν ΨΨ iψγ µ ν Ψ B µνρ = Ψ A γ µ S µν B A Ψ B where S µν is given by: S µν = +S µν L c a S µν R ȧċ 1

13 Combining our boxed results into equation gives us the explicit form for the Belinfante tensor. While it is normally worth writing these results explicitly, doing so in this case promises to yield a horrible equation which will yield very little insight. As such, it is perhaps better to leave our result in this form. Srednicki Symmetries of fermion fields. Consider a theory with N massless Weyl fields ψ j L = iψ j σµ µ ψ j where the repeated index j is summed. under the UN transformation, ψ j U jk ψ k This Lagrangian is clearly invariant where U is a unitary matrix. State the invariance group for the following cases: a N Weyl fields with a common mass M. Let s see if it s still true for UN: This becomes: L = iψ σ µ µ ψ m ψψ m ψ ψ L iu ij ψ j σ µ µ U ik ψ k m U ijψ j U ik ψ k m U ijψ j U ik ψ k L iψ U σ µ µ Uψ m ψ ju T jiu ik ψ k m ψ j U jiu ik ψ k Now we need to kill the Us. In the first term, as Srednicki says, the terms vanish if U U = 1. By the way, don t be concerned about commutation with the Pauli matrix: the Pauli matrix is multiplied by the derivative; the result is a ket of derivatives, which is an operator. This will commute with the constant unitary matrix. We ll soon develop a new notation to make this point cleaner. In the second and third terms, we need U T U = I in order to kill the Us; this implies that U T = U 1, which by our first condition implies that U T = U. This implies that U = U, which means that U must be real. A real unitary matrix is called an orthogonal matrix, and so the group is invariant under ON. b N massless Majorana fields L = i ΨT j Cγ µ µ Ψ j We can break the Majorana field into components of its Weyl fields. A Majorana field consists of only one Weyl field the left-handed one on top, and the conjugate of the right-handed 13

14 one on the bottom. Expanding the expression in the Lagrangian, we then get only N unique terms, the same as when we dealt directly with Weyl fields though there is now a factor of two, since the Weyl field appears twice per Majorana Field. The Lagrangian is then substantially unchanged from that of the first term of part a, and so is the symmetry group. UN. c N Majorana fields with a common mass m L = i ΨT j Cγ µ µ Ψ j 1 mψt j CΨ j Same as part b, except now the entire Lagrangian of part a is applicable, not merely the first term. The result is the same. ON. d N massless Dirac fields L = iψ j γ µ µ Ψ j Same as part b, except now there are two N mass terms in the Lagrangian rather than N, since each Dirac field consists of two Weyl spinors. All of these terms can mix together. Hence, the group is enlarged to UN. e N Dirac fields with a common mass m L = iψ j γ µ µ Ψ j mψ j Ψ j Same as part c, except now there are twice as many terms, since each Dirac field consists of two Weyl spinors. ON. 14