Solution to sunset diagram problem

Transcription

1 Solution to sunset diagram problem The sunset diagram provides the leading contribution to the p 2 )/ and hence Z, where I am simplifying notation by using Z Z φ.. First, use Feynman parameters to write the product of 3 propagators as to separate the ɛ of dimensional regularization from the Feynman propagator iɛ, I use i0 for the latter) 3 j= i qj 2 m 2 + i0 = 2i3 ) D) 3 where D = xq 2 + yq2 2 + zq2 3 m2 + i0. 2) Then substitute q 3 = p q q 2. Expanding q 2 3 and collecting all the terms containing the q momentum into a full square, we find xq 2 + yq2 2 + zp q q 2 ) 2 = x+z) q + We identify the first term here as αk 2, where z ) 2 x + z q xz 2 p) + x + z q 2 p) 2 + yq2 2. 3) α = x + z), k = q + z x + z q 2 p). 4) For the other two terms on the right hand side of Eq. 3), we expand q 2 p) 2 and collect all terms containing the q 2 momentum into another full square, obtaining xz x + z q 2 p) 2 + yq 2 2 = Consequently, we define β = xy + xz + yz x + z xz + yx + z) x + z, γ = q 2 ) 2 xz xz + yx + z) p + xy + xz + yz, k 2 = q 2 xzy xz + yx + z) p2. 5) xz p, ) xy + xz + yz after which the right hand side of Eq. 5) takes the form βk γp2. Altogether, we have xq 2 + yq zq 2 3 = αk 2 + βk γp 2 7) We will shift to these new diagonal momenta. The Jacobian for replacing the original independent loop momenta q and q 2 with k and k 2 is, given Eqs. 4) and ), k, k 2 ) q, q 2 ) z ) = det x+z 0 =, and therefore dk dk 2 = dq dq 2, dimension by dimension. In other words, for fixed Feynman parameters, d 4 q d 4 q 2 d 4 k d 4 k 2 = 2π) 4 2π) 4 2π) 4 2π). 8) 4

2 And, our final denominator form is Note, we must not set p 2 = m 2 at this stage. D = αk 2 + βk γp 2 m 2 + i0. 9) 2. Assembling all the factors of the two loop amplitude and making use of the above results, we have iσp 2 ) = iλ)2 d 4 q d 4 q 2 i i 3! 2π) 4 2π) 4 q 2 m 2 + i0 q3 2 m 2 + i0 = iλ2 d 4 k 2π) 4 i q 2 3 m 2 + i0 d 4 k 2 2π) 4 2 D 3 0) where D is as in Eq. 9). In particular, the dependence on the external momentum p comes solely through the γp 2 term in D. Hence, d 4 k d 4 k 2 γ). ) 2π) 4 2π) 4 D 4 Note that for large loop momenta k and k 2, D grows like k 2. Consequently, the integrand of the 8 dimensional momentum integral Eq. 0) behaves like /k, and the integral diverges quadratically. On the other hand, the integrand of Eq. ) behaves like /k 8, so the divergence of this integral is only logarithmic. 3. Rotating both loop momenta k and k 2 into Euclidean momentum space, we have d 4 k id 4 k E, d4 k 2 id 4 k E 2, and D αk E ) 2 βk E 2 ) 2 + γp 2 m 2, 2) hence = + λ2 d 4 k E d 4 k2 E 2π) 4 2π) 4 γ) [αk E ) 2 + βk E 2 ) 2 + m 2 γp 2 ] 4. 3) 4. Next, we need dimensional regularization to actually perform the momentum integrals. Changing d 4 k d 4 k 2 d µ 24 n) n k d n k 2 4) 2π) 4 2π) 4 2π) n 2π) n where we employ Euclidean signature for all dimensions and have introduced the µ dependence in a way equivalent to pulling out the µ from the coupling constant), we have d µ 8 2n n k d n k 2 2π) n 2π) n [αk E ) 2 + βk2 E ) 2 + m 2 γp 2 ] 4 using eq. )

3 d = µ 8 2n n k d n k 2 dt t 3 exp t[αk 2π) n 2π) n E ) 2 + βk2 E ) 2 + m 2 γp 2 ] ) 0 = µ 8 2n dt t 3 e tm2 γp 2 ) d n k d n k 2 e 0 2π) n 2π) n e tαk2 tβk2 2 using eq. 7) = µ 8 2n dt t 3 e tm2 γp 2) 4παt) n/2 4πβt) n/2 = = 0 µ 8 2n 4π) n αβ) dt t 3 n e tm2 γp 2 ) n/2 0 µ 8 2n 4π) n αβ) Γ4 n/2 n)m2 γp 2 ) n 4. 5) Note the Γ4 n) factor: It has a pole at n = 4 but no poles at n < 4. This is dimensional regularization s way to show that the momentum integrals diverge, but only logarithmically. In obtaining the above, I have used the following formulae, the first two of which provide an alternative way of doing the dimensionally regulated momentum integrals. You should get the same result if you use the approach given in the notes.) = dt t 3 e At, ) A 4 0 d n k = 4πct) n/2, 7) 2π) n e ctk2 Γ2ɛ)X ɛ = 2ɛ γ E + log X. 8) 2 The last formula above is simply the standard Gamma function expansion to be used below. At this point, we may take n = 4 2ɛ for an infinitesimally small ɛ. Hence, the last line of Eq. 5) becomes 4π) 4 αβ) 2 Γ2ɛ) 4πµ 2 αβ m 2 γp 2 ) 2ɛ ɛ 0 4π) 4 αβ) 2 2ɛ γ E + log 4πµ2 ) αβ. 9) m 2 γp 2 Plugging this formula back into Eq. 3) and assembling all the factors, we finally arrive at 3072π 4 { ɛ γ αβ) 2 µ2 m 2 + C + log αβ [ p 2 /m 2 )γ] 2 where C = 2 log4π) 2γ E is a numerical constant while αx, y, z), βx, y, z) and γx, y, z) depend on the Feynman parameters according to Eqs. 4) and ). 5. We are left with one more task, namely integrating over the Feynman parameters. This looks like a daunting task, especially if one wants analytic dependence on the external momentum p 2, but fortunately we are only interested in the particular value of p 2 = physical mass 2. Since } 20)

4 we are working at the leading order of perturbation theory which contributes to the /, we may neglect the difference between the physical and the bare masses as a higher-order correction and set p 2 = m 2. Consequently, Eq. 20) simplifies to γ p 3072π 2 =m 4 αβ) 2 2 { } µ2 ɛ m + C + log αβ 2 γ) 2 ) 3072π 4 xy + xz + yz) { 3 µ2 ɛ m + C + log xy + xz + yz) 3 } ) 2 xy + xz + yz ) 2 where the second equality follows from Eqs. 4) and γx, y, z). At this stage, you have arrived at the promised form: = dxdydz δx + y + z ) F x, y, z) { } ɛ + const + log Gx, y, z; p2 /m 2 ) 2) ) for the αx, y, z), βx, y, z), and for the rational functions F, G of the Feynman parameters and in the case of G, also of p 2 /m 2 = ) that are given explicitly in Eq. 2). The reason that we have set p 2 = m 2 which to the present order is the same as m 2 physical ), is that our goal is the field strength renormalization factor Z = where the derivative is evaluated at p 2 = m 2 physical. [ 22) ] 23). Despite the above simplification, Eq. 2) is a painful mess to evaluate simply by hand. However, if we turn to Mathematica it is fairly straightforward to find that dxdydz δx + y + z ) = xy + xz + yz) 3 2, dxdydz δx + y + z ) xy + xz + yz) log xy + xz + yz) 3 = 3 3 xy + xz + yz ) ) Using Eqs. 24), we find p 44π 2 =m 4 2 { ɛ µ2 m + C 3 } )

5 to the leading order in λ, and therefore Z = p 2 =M 2 = λ 2 { 44π 4 ɛ µ2 m + C 3 } Oλ 3 ). 2) As stated, to get these results in Mathematica it is useful to replace the x, y, z) variables with w, ξ) according to x = ξw, y = ξ)w, z = w, then integrate over the w variable first and over ξ second. The Mathematica program and results follow. 7. Finally, we use the ɛ term to identify c in the expansion for Z φ in our general formulae: obtaining ] λ 0 = µ 2 ) [λ ɛ a k λ) + ɛ k= k ] m 2 0 = m [ 2 b k λ) + k= ɛ k [ ] c k λ) Z φ = +, 27) k= ɛ k c = 44π. 28) 4 From this, we can now compute the anomalous dimension The result is as promised in the lecture notes. γ d λ) = 2 λ2 γ d λ) = λ dc dλ. 29) ) 2 λ + Oλ 3 ), 30) π 2

6 phi4sunsetdiagramintegrals.nb z w w x xi w w xi y xi w w xi jac w w f jac x y z x y x z y z ^3 w w 3 xi xi w w xi w w xi w 2 xi xi 3 f Integrate f, xi, 0,, w, 0, 2 g f Log x y x z y z ^3 x y x z y z x y z ^2 w w 3 xi xi Log w w xi w w xi w 2 xi xi 3 w w xi w w xi w 2 xi xi w w 2 xi xi 2 w w xi w w xi w 2 xi xi 3 g NIntegrate g, w, 0,, xi, 0, NIntegrate::slwcon : Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration being 0, oscillatory integrand, or insufficient WorkingPrecision. If your integrand is oscillatory try using the option Method Oscillatory in NIntegrate. More