L = 1 2 µφ µ φ m2 2 φ2 λ 0

Transcription

1 Physics 6 Homework solutions Renormalization Consider scalar φ 4 theory, with one real scalar field and Lagrangian L = µφ µ φ m φ λ 4 φ4. () We have seen many times that the lowest-order matrix element for the scattering process φφ φφ is M = λ. Here we consider the next order. Label the two incoming momenta p,k and the outgoing momenta p,k. Consider the diagram where p enters and p exits the upper vertex and k enters and k exits the lower vertex. (There are two similar diagrams with different external momentum labels, we will figure them out based on what happens with this diagram.). The evaluation Show that the contribution to the matrix element from this diagram is i ( iλ ) d D l ( i) (π) D (l +m iǫ)((l+p p ) +m iǫ) and label l and l+p p on a picture of the diagram. Combine the denominators using the Feynman parameter trick. Then Wick rotate the l variable. Writeanexpression forthediagraminterms ofyour Feynman integral, d D l E, m, and Mandelstamm t. Now set m = from now through the rest of the calculation, to make the problem a little simpler. Perform the l-integration in dimensional regularization. Then perform the x integration by using ln[x( x)] =. You should find (using D = 4 ǫ, your /ǫ will differ if you use the convention of the book) ( ) λ µ +ln 3π ǫ t +. (3) ()

2 Add the other two contributions to find M = λ λ 3π ( ) 3 µ6 +6+ln ǫ stu iπ (4) (Don t worry about showing that the imaginary part is correct, just assume that it arises from the difference between ln( s) and ln(s).).. Solution Actually this was already done in assignment 5. In coordinate space the diagram is ( iλ) d 4 vd 4 u ( i) 6 (x v) (y v) (v u) (u z) (u w). (5) The matrix element is i times this, with the external propagators removed and Fourier transformed into momentum space: refer to the solution key for that problem to see how to reach Eq. (). The Feynman parameter trick is (writing q = p p ) (l +m )((l+q) +m ) = = = = [( x)(l +m )+x((l +q) +m )] [( x)l +( x)m +xl +xl q +xq +xm ] [l +xl q +xq +m ] [(l +xq) +(x x )q +m ]. (6) Switch the order of the x,l integrations and shift the l integration to l + xq. Also, write q = t which is positive: M = iλ Wick rotate by replacing l il E ; M = i λ From now on I will write l E = l for simplicity of notation. In general dimension D = 4 ǫ, the integral is λ = λ d D l (π) D (l +m x( x)t iǫ). (7) d D l E (π) D (l E +m x( x)t iǫ). (8) µ ǫ (4π) D Γ(D/) µ ǫ (l ) (D )/ d(l ) (l x( x)t) (4π) D Γ(D/) ( x( x)t) ǫγ(d/)γ(ǫ). (9) Γ()

3 Now we expand in ǫ. Γ(ǫ) = ǫ γ E, which means other terms must be kept to order ǫ. We do this by using that which must be applied to (4πµ /x( x)( t)) ǫ ; M = λ = λ 3π X ǫ = e ǫlnx = +ǫln(x)+o(ǫ ) () ( (4π) 4πµ ) ǫ( ) x( x)t ǫ γ E +O(ǫ) ( ǫ +ln(4π) γ E +ln µ t ln(x( x))+o(ǫ) ). () The only x dependence is the log so we can do the x integral immediately; we also identify /ǫ+ln(4π) γ E = / ǫ. The result of the x integral is M = λ 3π ( ) µ +ln ǫ t +. () The two similar diagramsdiffer inthe replacement of p p with p k andp+k, replacing t with u and s respectively. Summing, M = λ 3π ( ) 3 µ6 +ln ǫ stu +6. (3) The argument of the log is negative, since stu > ; the iπ arises when we figure out the right branch for the log function.. The cross-section Compute the total scattering cross-section which you obtain using the lowest order result M = λ. Make the strict m = approximation to simplify the calculation. Your answer should be a function of λ and s only. [Hint: The whole phase space integral reduces to the somewhat simpler expression σ = dt (4) 6πs s see, eg, Burgess and Moore, page 96, Eq. (6.) and (6.5). Remember that the identical final states mean that you should integrate over half the t-range, or include an extra factor of to avoid double counting the available final states.] Now re-compute the scattering cross-section to order λ 3 by including the corrections you found to M. Now your result should be a function of s, µ /s, and /ǫ as well. 3

4 .. Solution The leading-order matrix element is λ and the leading-order cross section is σ = dtλ 6πs = λ s 3πs. (5) Write t = ys so the integral is (/3πs) dy. Since s + t + u = in the massless limit, u = ( y)s. At the next order M λ ( λ 6π ( )) 3 µ6 +6+ln ǫ s ln[y( y)]. (6) 3 Note that the imaginary part cancels between M and M unless we work to order λ 4. It is not justified to work to this order since at that point two-loop contributions will also arise. The integral over t becomes so the total cross-section is.3 Renormalization dy(a Bln[y( y)]) = (A+B) (7) σ = 3πs ( λ λ3 6π [ ]) 3 µ6 +8+ln. (8) ǫ s 3 Supposing we define λ r (µ ) so that the cross-section for s = µ equals the lowest-order crosssection evaluated using λ r ; that is, equate the order-λ 3 cross-section you found in the last subsection with the leading-order expression but using λ r. Write this relation, which expresses λ r (µ ) in terms of λ. Now, invert this relation to find an expression for λ in terms of λ r, accurate to second order. (Hint: if x = y +y and both x and y are small, then the y can be approximated as x, so y = x x to the desired order of accuracy.) Substitute this relationship into your previous result for M (at general s, t, u), working to second order in λ r (µ ). You should find that all reference to /ǫ disappears. However your answer should depend on µ 6 /stu. You have now successfully renormalized a theory to one loop order..3. Solution So we have λ r(µ ) = λ λ3 6π [ ] 3 µ6 +8+ln. (9) ǫ µ 6 4

5 Therefore at leading order Inverting the relation, λ r (µ ) = λ λ 3π λ = λ r + λ r 3π ( ) 3 µ +8+3ln. () ǫ µ ( ) 3 µ +8+3ln. () ǫ µ Substituting in our expression for the scattering matrix element, we find to second order that M = λ r λ r 3π.4 Scale dependence of the coupling [ ln µ6 stu iπ ]. () As a final exercise, suppose that you use λ r (µ ) but your friend chooses a different value and uses λ r (µ ). Use the expressions from the last section to relate each coupling to λ, and then use the two relations to eliminate λ and directly relate λ r (µ ) to λ r (µ ). You should find an expression of form If µ > µ, which coupling is larger? λ r (µ ) = λ r (µ )+λ (µ r ) (something) ln µ. (3) µ.4. Solution We have and Equating, λ r (µ ) = λ λ 3π λ r (µ ) = λ λ 3π ( ) 3 µ +8+3ln. (4) ǫ µ ( ) 3 µ +8+3ln. (5) ǫ µ λ r (µ ) = λ r (µ ) 3λ(µ r ) ln µ. (6) 3π µ The differential form of this expression is the renormalization group equation to one-loop order. We easily see that if µ > µ then λ r (µ ) > λ r (µ ). 5

6 an elementary QED calculation Consider the theory QED of electrons and muons. There is a gauge field A µ representing electromagnetism, as well as two Dirac fermion fields ψ and ψ, with masses m and m. The Lagrangian is L = 4 F µνf µν + ψ (i/d m )ψ + ψ (i/d m )ψ. (7) Here /D = γ µ ( µ iea µ ).. Scattering Using the book to find the Feynman rules if you prefer, draw the single diagram responsible for the scattering process ψ ψ ψ ψ. Label the ψ and ψ incoming momenta p,k and the ψ and ψ outgoing momenta p and k respectively. Write down the associated scattering matrix element. Find its Hermitian conjugate. Write the product of the two, summed over final and averaged over initial spins. Now take the limit m,m. Then carry out the traces and find an expression for the squared matrix element. You should find M = e 4s +u. (8) t.. Solution The diagram is The matrix element is ψ,k ψ,p ψ,k ψ,p M = i(ie) ū p γ µ u p ( i)g µν q ū k γ ν u k (9) where I suppressed indicating the polarization state dependences of the u and ū s, and I defined q = p p = k k. I also chose Feynman gauge. Using (ū p γ µ u p ) = ( u p γµ βu p ) = (ūp γ µ u p ) (3) we easily find M = e ū p γ α g αβ u p q ūkγ β u k. (3) 6

7 The squared matrix element is then M = e4 t Trγµ u p ū p γ α u p ū p Trγ µ u k ū k γ α u k ū k. (3) Averaging over initial and summing final spin states, M = e4 4t Trγµ ( /p+m )γ α ( /p +m )Trγ µ ( /k +m )γ α ( /k +m ). (33) Dropping the masses and using we find Trγ µ /pγ α /p = 4(p µ p ν +p ν p µ p p g µν ) (34) M = 4e4 t (pµ p ν +p ν p µ p p g µν )(k µ k ν +k ν k µ k k g µν ) = 4e4 t (p k p k +p k p k 4p p k k +4p p k k ) = e 4s +u t. (35) In the last step we used s p k = p k and u +p k = +p k.. Annihilation and crossing Now consider the process in which ψ and its antiparticle ψ annihilate to give ψ and its antiparticle ψ. Call the ψ incoming momentum p and the ψ momentum k, while the outgoing ψ and ψ momenta are p and k. Draw the single diagram contributing to this process. Write out the matrix element and the spin summed-and-averaged squared matrix element but do not evaluate the traces and contract. Look at your two expressions for spin averaged squared matrix elements for the ψ ψ ψ ψ process and the ψ ψ ψ ψ process. Argue that they are identical, except that the four momenta (p,k, p, k ) in the first expression have been switched to p, k,k, p in the second expression. Now write the expressions for s,t,u (setting all masses to zero). Apply the same permutation what does each Mandelstamm variable turn into? Use this information to show, without performing any Dirac traces or computing any farther, that the spin summed-averaged matrix element squared for the annihilation process is M = e 4t +u. (36) s The property you have just used (that changing particles between the initial and final states corresponds to changing the choice and sign of their momenta and leads to a re-assignment of Mandelstamm variables) is called crossing symmetry. 7

8 .. Solution The diagram is ψ,k ψ,p ψ,k ψ,p is Following the same procedure, the spin averaged and summed, squared matrix element M = e4 4 Trγµ ( /p+m )γ α (+/k +m )Trγ µ (+/k +m )γ α ( /p +m ). (37) NowlookatEq. (33). If youtakethatequationandmakethesubstitutions (p,k, p, k ) (p, k,k, p ) then we get this expression including the correct handling of the particle masses. This exchange modifies the Mandelstamm variables as follows: s = (p+k) = (p k ) = u, (38) t = (p p ) = (p+k) = s, (39) u = (p k ) = (p p ) = t. (4) These substitutions change the previous result for the spin summed-averaged matrix element to M = e 4u +t s. (4) 8