Quantum Field Theory Chapter 1, Homework & Solution

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1 Quantum Field Theory Chater, Homework & Solution. Show that the combination d 3 E, with E + m which occurs frequently in hase sace calculation integration is invariant under Lorentz transformation. Solution: Under the Lorentz transformation in z-direction, we have the relations z γ z βe E γ E β z Then From energy momentum relation, we get Hence Alternatively, we can make use of the identity, d z γ d z βde EdE z d z d z γ d z β z E d z γ d z E E β z d z E d 3 E d z E d3 E d δ + m θ + m E to write d 3 E d4 δ m which is clearly Lorentz invariant. x. Consider a system where articles interacting with eac other through otential energy V x so that the Lagrangian is of the form, L m d x dt + m d x dt x V x a Show that this Lagrangian is invariant under the satial translation given by x x x + a, x x x + a, where a is an arbitrary vector. b Use Noether s theorem to construct the conserved quantity corresonding to this symmetry. Solution : a It is obvious that from d x dt d x dt, d x d x dt dt x x x x

2 that L is invariant under translation. b For infinitesmal translation δ x a, From Noether s theorem, the conserved charge is Or J i a i δ x a L x j δx L j + x j δx j m x j a j + m x j a j J i m x i + m x i This is the usual total momentum of this article system. 3. Comute the following hysical quantities in the right units. a The total cross section for e + e µ + µ at high energies is of the form, σ e + e µ + µ 4πα 3s, s 4E, E :energy of e in cm frame, α fine structure constant Comute the cross section for the energies E Gev, 7T ev b The formula for the µ decay is given by Γ µ eν ν G F M 3 µ 9π 3, G F is the Fermi constant, M µ the roton mass Comute the muon lifetime in seconds. Solution : a E Gev σ 4πα 3s Use c.973 Mev cm we get Gev Gev 37 σgev Gev.973 Mev cm cm E 7T ev σ 7T ev σgev Gev cm 7T ev 7 b Γ µ eν ν G F m5 µ π 3 Gev 5Mev Gev Use 6.58 Mev sec Γ.9 9 Gev 6.58 Mev sec sec /Γ. 6 sec 4. Construct the Lorentz transformation for motion of coordinate axis in arbitrary direction by using the fact that coordinates erendicualr to the direction of motion remain unchanged. Solution : Let v be the velocity of the coordinate system x. We can decomose x as x x + x, with x x ˆvˆv, x x x x x ˆvˆv Under the Lorentz transformation with v, we have x γ x vx, x γx vx, x x

3 Or These can be written as x x ˆv γ x ˆv vx ], x γx v x ˆv], x x x + x x ˆvˆv + x γ x ˆv vx ]ˆv + x x ˆvˆv x vx ˆv + γ x ˆvˆv 5. Electric magnetic fields, E, B, combine into an antisymmetric second rank tensor unde the Lorentz transformtion, F µν µ A ν ν A µ with F i A i i A E i, F ij i A j j A i ɛ ijk B k These Minkowski tensors have the following roerty under the Lorentz transformation, F µν F µν Λ µ αλ ν βf αβ, Λ µ α : matrix element of Lorentz transformation Suose an inertial frame O moves with resect to O with velocity v in the ositive x-direction. a Find the relations between the electric magnetic fields, E, B,in the O b Show that the combination E B, does not change from O to O frames. c Show that the combination E B, does not change either. those in the O frame. Solution: a For convenience write the Lorentz transformation as Λ µ ν Then we get the transformation relation γ βγ βγ γ with γ β E F Λ Λ F + Λ Λ F γ β γ E E E F Λ Λ F + Λ Λ F γe βγ B 3 γe + βγb 3 E 3 F 3 Λ 3 3Λ F 3 + Λ 3 3Λ F 3 γe 3 βγb γe 3 βγb B F 3 Λ Λ 3 3F 3 B B F 3 Λ 3 3Λ F 3 + Λ 3 3Λ F 3 γb βγe 3 b B 3 F Λ Λ F + Λ Λ F γb 3 βγ E γb 3 + βγe B E E B + γe + βγb 3 γb βγe 3 + γe 3 βγb γb 3 + βγe E B + E B + E 3 B 3 B E c E B E + γe + βγb 3 + γe 3 βγb B γb βγe 3 γb 3 + βγe E + E + E 3 B B B 3 E B 3

4 . The Dirac Hamiltonian for free article is given by The angular momentum oerator is of the form, Quantum Field Theory Homework set, Solution H α + βm L r a Comute the commutators, L, H ] Is L conserved? b Define α S i 4 α show that ] L + S, H c Show that S satisfy the angular momentum algebra, i.e. S i, S j ] iε ijk S k S Solution : a L, α α + βm] x 3 x 3, α + α 3 3 ] iα 3 iα 3 i We can generalize this to L, α + βm ] α i So L is not conserved. b S, H] i α α 3, α ] + βm It is easy to verify that Similarly, Then Or α α 3, α ] α α 3 α α α α 3, α α 3, α ] α α 3 α α α α 3 α 3, α α 3, α 3 ] α, α α 3, β] S, H] i α S, H] i α So ] L + S, H i.e. the total angular momenta is conserved. c i S, S ] α α 3, α 3 α ] 4 α α 3 α 3 α α 3 α α α 3 α α is 3

5 Then S S i α α 3 α α , S. The Dirac sinors are of the form, u, s χ s, v, s χ s s, where χ, χ a Show that u, s u, s mδ ss, v, s u, s, v, s v, s mδ ss u, s v, s v, s u, s, u, s v, s b Show that Solution : a u α, s u β, s / + m αβ s v α, s v β, s / m αβ u, s u, s χ s s χ s χ s mδ ss χ s σ v, s v, s χ s χ s u, s v, s χ s v, s u, s χ s v, s u, s χ s u, s v, s χ s χ s mδ ss χ s χ s χ s χ s χ s

6 bthe sin sum for u sinors, s u α, s u β, s σ E+m χ s χ s / + m s σ E+m σ σ E+m E+m E+m where we have used Similarly for the v sinor, s v α, s σ v β, s E+m χ s χ s E m s χ s χ σ s E+m / m E+m σ E+m σ E+m 3. Suose a free Dirac article at t, is described by a wavefunction, ψ, x πd ex r 3/4 d ω where d is some constant ω Comute the wavefunction for t. What haens when d is very small? Solution : Ex ψ, x in terms of sinors ψ, x s d 3 b, s u, s e i x + d, s υ, s e i x ] π 3 E We can comute the exansion coeffi cients by the orthogonality roerties b, s d 3 xe i x x, u π, s ψ 3 E d 3 xe i x u π, s ω ex r 3 E πd 3/4 d The Fourier transform of the Gaussian wave acket can be calculated as follows, d 3 xe i x x ex d d 3 x ex d x + i d d ] where we have used d ex dxe x d 3 x ex d d x ] ex πd 3 π Similarly d, s b, s d E + mδ s ex E π 3/4 d 3 xe i x x, υ π, s ψ 3 E d 3 xe i x v π, s ω ex r 3 E πd 3/4 d 3

7 For non-zero t we get ψ t, x s v, s ω χ s d, s d E + m ex E π 3/4 χ s Note that d, s b, s χ s z x + i y z x + i y d 3 b, s u, s e iet e i x + d, s υ, s e iet e i x ] π 3 E This shows that the negative energy amlitude becomes areciable when m. 4. Consider a hermitian matrix defined by X x + x where σ σ, σ, σ 3 are Pauli matrices x, x are sace-time coordinates. a Comute the determinant of X b Suose U is a matrix with det U. Define a new matrix by a similarity transformation, Show that X can be written as c Show that the relation between d Suose U is of the form, Find the relation between x, x Solution: a X UXU X x + x x, x x, x is a Lorentz transformation. e iα U e iα x, x. X x + x 3 x ix x + ix x x 3 b Note that X is Hermitian, X det X x x x x 3 x + x x + x So is X X UXU UX U UXU X Ex X in terms of comlete set of Hermitian matrices, X x + x cfrom the invariance of the determinant det X detuxu det U det X det U det X we see that x x x x 3 x x x x 3 4

8 So the relation between x, x x, x is a Lorentz transformation. d X UXU e iα x + ix x x 3 e iα e iα x + x 3 x ix e iα x + x 3 e iα x ix e iα x + ix x x 3 This corresond to a rotation z axis. Note that U is not necessarily unitary. In fact the Lorentz boost corresond to e α U e α which gives This imlies UXU e α x + x 3 x ix e α e α x + ix x x 3 e α e α x + x 3 x ix x + ix e α x x 3 x cosh α x + sinh α x 3 x 3 sinh α x + cosh α x For the Lorentz boost along x axis, we can first rotate π about y-axis using cos β sin β U y β sin β cos β for β π hen X U y β XU y β x + x 3 x ix x + ix x x 3 x + x ix x 3 ix x 3 x x X 3 U e α x + x X U ix x 3 e α ix x 3 x x e α x + x ix + x 3 x 3 ix e α x x e α e α X 4 U y β X 3 U y β x 3 + e α x x + eα x + x ix e α x x + eα x + x e α x + x ix + x 3 x 3 ix e α x x eα x + x e α x x ix e α x x x 3 + eα x + x x e α x x + eα x + x cosh αx + sinh αx x 3 x 3, x x x e α x x + eα x + x sinh αx + cosh αx 5

9 5. Dirac article in the resence of electromagnetic field satisfies the equation, Or In the non-relativistic limit, we can write γ µ i µ ea µ m] ψ x i ψ ] t α ea + βm + eφ ψ ψ x e imt u l Show that the uer comonent satisfies the equation, i u ] t ea e B + eφ u m m For the case of weak uniform magnetic field B we can take A B r. Show that i u ] t e L + S B u. m m Solution: In non-relativistic limit, Dirac equation becomes i u t + mu i l t + ml m + eφ ea e A m + eφ u l Or From the nd equation, we get i u t eφ u + ea l i l t e A u + m + eφ l l ea u m where we have neglected eφ l.substitute this into first equation we get t i u t eφ u + ] ea u m Using the identity A B A B A + i B we get ] ea ea + i ea ea Since e A we get A ea ea e + A +ie A ie B Then the equation becomes i u t m e A ] e B + eφ u m 6

10 For weak field, we get ea e A + A e e i B r B e L r B i u ] t e L + S B u. m m 6. a, a, a, a are creation annihilation oerators satifying the commutation relations ] a i, a j δ ij, a i, a j ], i, j, Define J + a a, J J +, J 3 a a a a a Comute the commutators J x, J y ], J y, J z ], J z, J x ] where J x J + + J, J y i J + J b Define the state by Let the state n, n be n, n a i, for i, a n a n n! n! Show that this state is an eigenstate of J 3 comute the eigenvalue. c Show that this is also eigen state of J J + J + J 3 comute the eigenvalue. d Show that the state J + n, n is an eigenstate of J 3. What is the eigenvalue? a J +, J 3 ] ] a a, a a a a a a J + Similarly Also J, J 3 ] J J +, J ] a a, a a ] a a a a J 3 Then J, J 3 ] J + + J, J 3 ] J + + J ij ij The other commutator is J, J 3 ] i J + J, J 3 ] i J + J i J ij Define the number oerators Then we can derive J, J ] 4i J + + J, J + J ] 4i J 3 ij 3 N a a, N a a ] ] n ] n N, a a a, a a, N, a n a n ] n N, a n a 7

11 From this we see that where we have used Similarly, Then We can write J as N n, n N a n a n n! n n n, n! N N n, n n n, n J 3 n, n N N n, n n n n, n J J +J + J J + ] + J 3 We can write J +, J in term of number oerators as J + J a a a a a + a a a N + N So J + J N + N J J +J + J J + ] + +J 3 N + N + N N ] + 4 N N 4 N + N N + N + This imlies that d From we see that Furthermore Use J n, n 4 n + n n + n + n, n J n + n J +, J 3 ] J + J 3 J + n, n n n + J + n, n J + n, n a a n, n n+ a a a n n! n! n] n a, a n a we get From we see that J + n, n n+ n a n a n! n n + n n +, n! J n + n, m n n n J + m, n J m J + n, n J + m + J m n +, n 8

12 . The Dirac equation for free article is given by, Quantum Field Theory Homework 3 solution iγ µ µ m ψ x Under the arity transformation the sace-time coordiante transform as x µ x µ x, x, x, x 3 The Dirac equation in the new coordinate system is of the form, iγ µ µ m ψ x Find the relation between ψ x ψ x. Solution: For the arity transformation, the Lorentz transformation is of the form, Λ ν µ Then from we see that Clearly, It is easy to see that ψ ψ ψψ, ψ γ 5 ψ ψγ 5 ψ, S γ µ S Λ ν µγ ν S γ S γ, S γ i S γ i, i,, 3 S γ, ψ x γ ψ x ψ γ µ ψ ψγ µ ψ, ψ γ µ γ 5 ψ ψγ µ γ 5 ψ, ψ σ µν ψ ψσ µν ψ,. The left-hed right-hed comonents of a Dirac article are defined by, ψ L γ 5 ψ, ψ R + γ 5 ψ where γ 5 is defined by γ 5 γ 5 iγ γ γ γ 3 a Show that {γ 5, γ µ }, γ 5 b Show that ψ L, ψ R are eigenstates of γ 5 matrix. What are the eigenvalues? c Are they eigenstates of arity oerator? d Write the u sinor in the form, u, s N E+m where N is some normalization constant χ s is an arbitrary comonent sinor. Show that if we choose χ s to be eigenstate of, ˆ χ s χ s χ s then u, s is an eigenstate of the helicity oerator λ S ˆ where S is the sin oerator given by σ S σ

13 - Solution: a {γ 5, γ } i{γ γ γ γ 3, γ } iγ γ γ γ 3 γ + γ γ γ γ γ 3 i γ γ γ 3 + γ γ γ 3 γ 5 i γ γ γ γ 3 γ γ γ γ 3 γ γ γ 3 γ γ γ 3 γ γ 3 γ γ 3 γ γ 3 γ γ 3 γ 3 b Similarly γ 5 ψ L γ 5 γ 5 ψ γ 5 ψ ψ L γ 5 ψ R ψ R c Under the arity we have Then P ψ L γ ψ L ψ R, P ψ γ ψ P ψ R γ ψ R ψ L d In the stard reresentation Thus γ 5 u L γ 5 u, N χ E+m N χ N E + E E χ N χ 3. Consider a one-dimensional string with length L which satisfies the wave equaiton, φ x φ v t a Find the solutions of this wave equation with the boundary conditions, φ, t φ L, t b Find the Lagrangian density which will give this wave equation as the equation of motion. c From the Lagrangian density find the conjugate momenta imose the quantization conditions. Also find the Hamiltonian. d Find the eigenvalues of the Hamiltonian. - Solution : awrite φ x, t ψ x e iet.then ψ x E v Plane wave solution φ e ikx, gives Or E ±ω, E v k ω kv For the boundary condition ψ ψl,we take nπx ψ n x sin L L

14 so that Note that the energy eigenvalues are bthe Lagrangian density is Euler-Lagrange Eq cconjugate mometua Hamiltonian Commutation relations d Mode exansion x L ψ n x ψ m x dx δ nm E n ±ω n, ω n nπv L L tφ v xφ L x φ L t t φ, φ x φ v t φ x, t n π x, t L t φ tφ H πφ L tφ + v xφ φ x, t, t φ y, t] iδ x y a n sin nπx L L e iωnt + a n sin nπx L eiωnt] ωn t φ x, t n L iω n a n sin nπx L e iωnt a n sin nπx L eiωnt] ωn Then a n i L nπx dx iω n φ x, t + t φ x, t] sin ωn L L eiωnt an, a m] L dxdy ωn ω m sin nπx L L e iωnt sin nπy L eiωmt iω n φ x, t + t φ x, t, iω m φ y, t + t φ y, t] Or an, a m] L dx sin nπx ωn ω m L L e iωnt sin mπx L eiωmt ω n + ω m δ nm Hamiltonian is The first term is L H L ] dx tφ + v xφ dx tφ L ω L n ω m dx sin nπx mπx sin an e iωnt a ωn ωm n,m L L ne iωnt] a m e iωmt ω n an a n e iωnt + a na ne iωnt a n a n a ] na n the second term is L v dx xφ nπ mπ L L L n,m nπ v L n n ωn ωm L dx cos nπx L ω n an a n e iωnt + a na ne iωnt + a n a n + a na n ] mπx cos an e iωnt + a L ne iωnt] a m e iω 3

15 The Hamiltonian is then H L 4. Consider the Lagrangian density given by ] dx tφ + v xφ ω n an a n + a na n L µφ µ φ µ φ + J x φ, J x arbitray function a Show that the equation of motion is of the form, µ µ + µ φ x J x n b Find the conjugate momenta imose the quantization conditions. c Find the creation annihilation oerators. Solution: a Equation of motion b Conjugate momentum Quantization µ L µ φ L φ, µ µ φ + µ φ Jφ, π L φ φ φ x, t, π y, t] iδ 3 x y Define µ µ + µ x y δ 4 x y Then φ x φ x + x y J y d 4 y φ x + φ cl x where φ x satisfies the homogeneous equaiton µ µ + µ φ x φ cl x x y J y d 4 y So φ x can be exed in terms of lane waves d 3 φ x a e ix + a e ix] π 3 ω Then we can solve for a to write d 3 x a π 3 ω d 3 x π 3 ω { } e ix φ x {e ix φ x } x y J y d 4 y Note that the last term is a c-number will not effect the commutation relation. We can write the action as S d 4 xl d 4 x φ + µ ] φ + Jφ d 4 x φ + φ cl + µ ] φ + φ cl + J φ + φ cl d 4 x φ + µ φ φ + µ φ cl J φ cl + µ ] φ cl + Jφ cl d 4 x φ + µ φ φ cl + µ ] φ cl + Jφ cl 4

16 Note that We get S + µ φ cl J d 4 x φ + µ ] φ + Jφ cl 5. Let φ be a free scalar field satisfying the field equation, µ µ + µ φ x a Show that the roagator defined by can be written as F x y T φ x φ y θ x y φ x φ y + θ y x φ y φ x F x y d 4 k π 4 eik x y i k µ + iε b Show that the unequal time commutator for these free fields is given by d 3 k i x y φ x, φ y] π 3 e ik x y e ik x y] ω k c Show that x y for sace-like searation, i.e. x y, if x y < Solution: a i x, y T φ x φ y θ x y φ x φ y + θ y x φ y φ x Using the mode exansion, we see that d 3 kd 3 k φ x φ y π 3 ake ikx ]a + k e ik y ω k ω k d 3 kd 3 k π 3 δ 3 k k e ikx+ik y d 3 k ω k π 3 e ikx y ω k d 3 k i x, y π 3 θ x y e ikx y + θ y x e ikx y] ω k Note that We then get π d 4 k e ik x π 4 k + iε { dk i k ω + iε e ikx yx ω e iωx y for x > y i ω eiωx y for x < y y i i x, y b From art a we see immediately that i x y φ x, φ y] d 3 k ] π 3 θt t e ikx x + θt t e ikx x ω k d 3 k π 3 e ik x y e ik x y] ω k c For sace like searation x y <, we can chose a frame such that x y has only satial comonent x y, x y.then φ x, φ y] d 3 k π 3 ω k x e i x ] k y e i k y where we have change the integration variable k to k in the second term. 5

17 Quantum Field Theory Ling-Fong Li December 5, 4 Homework set 4, Solution. Dirac equation for electron moving in the electromagnetic field can be obtained from the free Dirac equation by the relacement i µ i µ ea µ, x, γ µ i µ ea µ m] ψ t Then the equation for the ositron is x, γ µ i µ + ea µ m] ψ c t Assume that ψ c is related to ψ by C is called the charge conjugation matrix. a Find C in terms of Dirac γ matrices. b For the v sinor of the form, ψ c Cψ v, s N E+m χ s Comute its charge conjugate v c, s Cv, s c To imlement the charge conjugation for the fermion field, we write ψ c CψC Cψ where C is the charge conjugation oerator. Find the relation between Solution: a Dirac equation for a charged article in em field is of the form, On the other h the equation for ositron is γ µ i µ ea µ m] ψ γ µ i µ + ea µ m] ψ c Take the comlex conjugate of Dirac equation we get γ µ i µ + ea µ m ] ψ ψ c γ µ ψ c ψγ µ ψ. If we assume ψ c is related to ψ by then ψ c Cψ C γ µ C γ µ In the stard notation where γ, γ, γ 3 are real γ is imaginary, C can be taken as iσ C iγ iσ which has the roerties, C C C

18 b From the v sinor of the form, we get v, s N v c, s Cv iσ, s iσ N where we have used the relations N iσ χ s N χ s, s ± σ χ s N iσ σ iσ s χ s s u, s χ s σ σ σ σ, iσ χ s s χ s Note that the sin comonent is flied under charge conjugation. In terms of creation annihilation oerators, we have ψ d 3 b, s u, s e i x + d, s v, s e i x] s π 3 E the charge conjugate field is ψ c C ψ T s s d 3 b, s Cu, s e i x + d, s Cv, s e i x] π 3 E d 3 b, s v, s e i x + d, s u, s e i x] π 3 E Write then c From we get Then Cb, s C d, s, ψ c ψ T C, ψ c CψC ψ c Cψ Cd, s C b, s ψ c ψ T C γ ψ c γ µ ψ c ψ T C γ γ µ Cψ ψ T γ C γ µ C ψ ψ T γ γ µ ψ ψ T γ γ µ ψ ] T ψ γ µ γ ψ ψγ µ ψ where we have used the roerty that fermion fields anti-commute. This means the 4-vector current for the anti articles is negative of that for the article. In other words, the 4-vector current is odd under charge conjugation. Similarly for the scalar current we have ψ c ψ c ψ T C γ Cψ ψ T γ C C ψ ψ T γ ψ ψ T γ ψ ] T ψ γ ψ ψψ for the sudoscalar current ψ c γ 5 ψ c ψ T C γ γ 5 Cψ ψ T γ γ 5 C C ψ ψ T γ γ 5 ψ ψ T γ γ 5 ψ ] T ψ γ 5 γ ψ ψγ 5 ψ

19 This shows that the scalar current is even under charge conjugation. ψ c γ µ γ 5 ψ c ψ T C γ γ µ γ 5 Cψ ψ T γ C γ µ γ 5 C ψ ψ T γ γ µ γ 5 ψ ψ T γ γ µ γ 5 ψ ] T ψ γ 5 γ µ γ ψ ψγ µ γ 5 ψ. Consider a free scalar field φ x where the 4-momentum oerator is of the form, P µ d 3 k k µ a k a k a As a useful tool, show that for two oerators A B, the following identity holds e A Be A B + A, B] + A, A, B]] + A, A, A, B]]] + 3! b Use this identity to show that e ip x a k e ip x a k e ik x P µ, φ x] i µ φ x c Let K be an eigenstate of P µ, satisfying P µ K K µ K. Show that K φ x φ y K K φ x y φ K Solution : a consider the function F λ defined as Then df dλ eλa A, B]e λa, On the other h, Taylor exansion of F λ,gives F λ e λa Be λa d F dλ eλa A, A, B]]e λa, Setting λ, we get F λ F + λ df dλ λ + λ d F! dλ λ + e A Be A B + A, B] + A, A, B]] + A, A, A, B]]] + 3! b From the identity in a we get e ip x a k e ip x a k + ix µ P µ, a k] + xν x µ P ν, P µ, a k]] + Now we calculate the commutators, P µ, a k] d 3 k k µ a k a k, a k] k µ a k Then P ν, P µ, a k]] k µ k ν a k e ip x a k e ip x a k + ik x + ] ik x + e ik x a k Take Hermitian conjugate we get e ip x a k e ip x e ik x a k 3

20 From the formula for P µ φ x d 3 a e ix + a e ix] ω π 3 we get P µ, φ x] d 3 d 3 k k µ a k a k, a e ix + a e ix ] ω π 3 d 3 d 3 k k µ a k e ix a e ix ]δ 3 k ω π 3 d 3 ω π 3 µ a e ix a e ix ] i µ φ From this we can show that e ip a φ x e ip a φ x + ia µ P µ, φ x] + φ x a c Write Then φy e ip y ip y φ e K φ x φ y K K φ x e ip y φ e ip y K K e ip y e ip y φ x e ip y φ K e ik y e ik y K φ x y φ K e ik y K φ x y φ K 3. The roagator for a massless scalar field can be written in the form, d 4 x e ik x F x π 4 k + iε Carrying out the integration to show that - Solution: F x F x d 4 x e ik x π 4 k + iε i 4π x iε d 3 k k x π 4 ei The k integration can be erformed by the stard contour method, We then have Using the identity we get dk e ikx k k + iε dk e ikx k iπ θ x e i k x + θ x e i k x ] k + iε k F x 8π r e ±iατ dτ dk e ikr e ikr θ x e ikx + θ x e ikx ] e ±iα±iετ dτ F x 8π θ x r r x + iε + r + x iε i ] θ x θ x 4π r x + iεx + r x iεx i α ± iε + θ x i 4π x iε r + x + iε + r x iε ] 4

21 4. In thequantization of free electromagnetic fields the mode exansion is of the form, d 3 k A x, t ɛ k, λak, λe ikx + a k, λe ikx ] w k k ωπ 3 where The quantization condition is of the form, λ ɛk, λ, λ, with k ɛk, λ A i x, t, A j x, t] iδ tr ij δ tr ij x x Solve for ak, λ a + k, λ comute the commutator, ak, λ, a k, λ ] Solution: d 3 x ak, λ i eik x ɛk, λ Ax] π3 ω a k, λ i d 3 x π3 ω e ik x ɛk, λ Ax] Commutation relations, ak, λ, a k, λ ] d 3 xd 3 x e ikx e ik x ɛk, λ Ax ik ɛk, λ Ax, ɛk, λ Ax + ik ɛk, λ π 3 w k π 3 w k d 3 xd 3 x e ikx e ik x {ε i k, λik ε j k, λ A i x, A j x ] ik ε i k, λε j k, λ A i x, π 3 w k π 3 w k d 3 xd 3 x e ikx e ik x {ε i k, λε j k, λ i δ tr π 3 w k π 3 ij x x ik + ik } w k Note that l d 3 xd 3 x e i k x e i k x ε i k, λε j k, λ δ tr ij x x d 3 xd 3 x e i k q x e i k q x ε i k, λε j k, λ d 3 xd 3 x e i k x e i k x ε i k, λε j k, λ d 3 qδ ij q iq j q d 3 qδ 3 k q δ δ 3 k k δ ij k ik j k ε ik, λε j k, λ δ 3 k k ɛk, λ ɛk, λ k ɛk, λ k δ λλ δ 3 k k d 3 qe i q x x δ k q ε i k, λε j k, λ ] ɛk, λ k where we have used ɛk, λ ɛk, λ δ λλ, ɛk, λ k 5

22 Quantum Field Theory Ling-Fong Li January 4, 5. Consider the reaction Homework set 5, Solution e + + e µ + k + µ k a The sin averaged robability is of the form 4 sin Me + e µ + µ e4 q 4 T r / m e γ µ / + m e γ ν] T r /k + m µ γµ /k + m µ γ ν] Show that for energies m µ, this can be written as Me + e µ + µ 8 e4 4 q 4 k k + k k ] sin b The hase sace for this reaction is given by ρ π 4 δ 4 + k k d 3 k d 3 k π 3 ω π 3 ω Show that in the center of mass frame. Solution: a 4 sin ρ dω 3π T r / m e γ µ / + m e γ ν] T r / γ µ /γ ν] m T r γ µ γ ν ] 4 µ ν g µν + µ ν ] 4m eg µν T r /k + m µ γµ /k m µ γ ν] T r /k γ µ /kγ ν] m µt r γ µ γ ν] 4 k µ k ν g µν k k + k µ k ν ] 4m µg µν Me + e µ + µ e4 q 4 T r / m e γ µ / + m e γ ν] T r /k + m µ γµ /k + m µ γ ν] 8 e4 q 4 k k + k k ] where m e, m µ have been neglected. b In the center of mass frame, we write the momenta as k µ E, k µ E,,, E,, k µ E, k µ E,,, E, with k ẑ k cos θ Then the hase sace is ρ π 4 δ 4 + k k d 3 k d 3 k π 3 ω π 3 ω 4π δ E ω ω d3 k 4ωω 3π δ E ω k dkdω ω dω 3π

23 . The Lagrangian for the free hoton is of the form, Suose we add a mass term to this Lagrangian a Find the equation of motion. b From the equation of motion show that L 4 F µνf µν, where F µν µ A ν ν A µ L 4 F µνf µν + µ A µ A µ µ A µ use the equation of motion to exress A in terms of other field variables c Carry out the quantization rocedure find the eigenvalues of the Hamiltonian. Solution : a Equation of motion b From equation of motion L µ A ν F µν, µ F µν + µ A ν L A ν µ A ν ν µ F µν + µ ν A ν, ν A ν A µ if i c Conjugate momenta π i L A i F i, A does not have conjugate momenta Commutation relation, π i x, t, A j x, t] iδ ij δ 3 x x, A i x, t, A j x, t], π i x, t, π j x, t] Note that we can write A µ if i µ iπ i Also A x, t, A j x, t] µ iπ i x, t, A j x, t] i µ jδ 3 x x A x, t, π j x, t] µ iπ i x, t, π j x, t] From π i F i A i i A, we get A i i A π i µ j i π j π i, Equation of motion Solutions are A x, t A x, t, A i x, t] A x, t, d 3 k ωπ 3 µ j i π j x, t π i x, t] µ F µi + µ A i, or µ µ + µ A i λ,,3 Suose the wave vector is in the z direction, ɛ k, λak, λe ikx + a k, λe ikx ], ω k k k,,, k k + µ

24 We can choose the olarization vectors as with ɛ µ k,,,,, ɛ µ k,,,,, ɛ µ k, 3 µ k,,, k ɛ k, λ ɛ k, λ δ λλ, λ, λ,, 3 Since A also satisfies the same Klein-Gordon equation, we can extend the exansion as A µ d 3 k x, t ɛ µ k, λak, λe ikx + a k, λe ikx ], ω k k + µ ωπ 3 λ,,3 Then Commutation relations, ak, λ, a k, λ ] ak, λ i a k, λ i d 3 x π3 ω eik x εk, λ Ax] d 3 x π3 ω e ik x εk, λ Ax] d 3 xd 3 x e ikx e ik x εk, λ Ax ik εk, λ Ax, εk, λ Ax + ik εk π 3 w k π 3 w k d 3 xd 3 x e ikx e ik x {ε µ k, λik ε ν k, λ A µ x, A ν x ] ik ε i k, λε j k, λ A i π 3 w k π 3 w k d 3 xd 3 x e ikx e ik x {ε i k, λε j k, λ i δ tr π 3 w k π 3 ij x x ik + ik } w k Note that l d 3 xd 3 x e i k x e i k x ε i k, λε j k, λ δ tr ij x x d 3 xd 3 x e i k q x e i k q x ε i k, λε j k, λ d 3 xd 3 x e i k x e i k x ε i k, λε j k, λ d 3 qδ ij q iq j q d 3 qδ 3 k q δ δ 3 k k δ ij k ik j k ε ik, λε j k, λ δ 3 k k ɛk, λ ɛk, λ k ɛk, λ k δ λλ δ 3 k k d 3 qe i q x k q ε i k, λε j k ] ɛk, λ k where we have used ɛk, λ ɛk, λ δ λλ, ɛk, λ k 3. In the λφ 4 theory the interacting Lagrangian is of the form, L int λ 4! φ4 For the -body elastic scattering we need to comute to second order in λ the following vacuum exectation value τ y, y, x, x i λ d 4 z d 4 z T φ! in y φ in y φ in x φ in x 4! φ4 in z λ 4! φ4 in z Use Wick s theorem to write this matrix element in terms of roagators. 4. The Lagrangian for the free fermion field is of the form, L ψiγ µ µ mψ Comute the free roagator d 4 xe ix T ψ α x ψ β 3

25 Solution : The roagator is defined as, S αβ d 4 xe ix T ψ α x ψ β d 4 xe ix θ x ψ α x ψ β θ x ψ β ψ α x Note that there is a minus sign for the second term due to the fact that fermion fields anti-commute. Write out the mode exansion, ψ α x s π 3 ψ β d 3 s d 3 π 3 E b, s uα, s e i x + d, s υ α, s e i x] b, s u β, s + d, s υ β, s ] E Then we get ψ α x ψ β s s d 3 d 3 u π 3/ E π 3 α, s e i x u β, s b, s b, s E d 3 π 3 u α, s u β, s e i x d 3 / E π 3 + m x E αβ e i where we have used Similarly, ψ β ψ α x s u α, s u β, s / + m αβ s d 3 π 3 v α, s v β, s e i x d 3 / E π 3 m x E αβ ei θ x ψ α x ψ β + θ x ψ β ψ α x where we have used Note that We then get d 4 π 4 π e i x m / + m i + iε v α, s v β, s / m αβ s d 3 π 3 E θ x / + m e i x θ x αβ / m ei x αβ { d i E + iε e it E P e iet for t > i E P e ie P t for > t i i d 3 π 3 θ t e iet e i x E γ E γ + m + θ te iet e i x E γ d 3 π 3 θ t e iet e i x / + m + θ te iet e i x / + m E d 3 θ t e ix π 3 / + m + θ te ix / + m ] E ] So the fermion roagator is of the form, S αβ d 4 xe ix T ψ α x ψ β d 4 e i x π 4 m + iε / + m αβ 4