7 Veltman-Passarino Reduction

Transcription

1 7 Veltman-Passarino Reduction This is a method of expressing an n-point loop integral with r powers of the loop momentum l in the numerator, in terms of scalar s-point functions with s = n r, n. scalar meaning an integral with no powers of loop-momenta r = 0 in the numerator - i.e. a product of propagator denominators. 7.1 Notation p l + q 1 l l + q j 1 l + q j p j p 1 p n where I n [fl] i4π d/ fl π d l m 0l + q 1 m 1 l + q n 1 m n 1, q i p 1 + p p i, p i being the incoming external momenta. d = 4 ɛ is the number of dimensions in which we perform the loop integral in order to regularise either ultraviolet UV or infrared/collinear IR divergences. We assume that all external momentum are in four dimensions Note that masses can be included in these manipulations but add to the algebraic complexity. The case fl = 1 is what we call the scalar integrals - the integrals that would be obtained in a theory of scalar particles only. We also define the integrals I j n 1: I j n 1[fl] i4π d/ π d fl 1 l m 0l + q 1 m 1 l + q j 1 m j 1l + q j+1 m j+1 l + q n m n, i.e. the n 1-point integral obtained by pinching out the j th propagator. 37

2 p l + q 1 l l + q j 1 l + q j+1 p j p j+1 p 1 p n Similarly we can define integrals such as I j 1,j n [fl], etc. 7. One power of loop momentum in numerator fl = l µ I n [l µ ] = i4π d/ l µ n 1 π d l m 0l q 1 m 1 l q n m n = C n;i p µ i, i=1 using the fact that the vector quantity on the LHS must be constructed from the vectors p 1 p n 1 by conservation of momentum p 1 + p + p n 1 = p n so that p n is not independent Contracting both sides with p µ j we get I n [l p j ] = i4π d/ π d n 1 l p j l m 0l + q 1 m 1 l + q n m n = C n;i ij, i=1 where ij = p i p j is the Gram matrix. Since p j = q j q j 1 with q 0 = 0 we can write the numerator of the integral as l p j = 1 l + qj m j l + q j 1 m j 1 + m j m j 1 q j + q j 1 This is the Veltman-Passarino VP reduction formula. The terms l + q j m j and l + q j 1 m j 1 in the numerator can be used to cancel or pinch the j th and j 1 th propagators respectively and so we end up with a set of n 1 linear equations for the coefficients C n;i. 38

3 n 1 i=1 C n;i ij = 1 I j n 1[1] I j 1 n 1 [1] + m j m j 1 q j + q j 1I n [1], where the RHS is expressed in terms of scalar n-point and n 1-point integrals. This set of linear equations is readily solved C n;i = 1 j 1 ij I j n 1[1] I j 1 n 1 [1] + m j m j 1 q j + q j 1 I n[1] However, care is needed if we are dealing with n > 4. The reason for this is that in this case only the first four of the external momenta p i can be linearly independent. In fact by attempting to write the integral out as the sum of n 1 terms, we have over-parametrised the integral and the coefficients are not unique - the Gram matrix is not invertible as its determinant vanishes. In such cases, it is sufficient to choose the first four external momenta only provided the Gram determinant for those four does not vanish. det 0. In special cases, known as exceptional momenta this determinant may indeed vanish and we must then choose a different four. We then have with C n;i = 1 j=1 1 ij I n [l µ ] = C n;i p µ i, i=1 I j n 1[1] I j 1 n 1 [1] + m j m j 1 q j + q j 1 I n[1] Example: In this example we will consider the three-point function and for simplicity set the internal masses but not the external square momenta to zero, so that we have i4π d/ with q 1 = p 1 and q = p 1 + p. Contracting both sides with p µ 1 gives π d l µ l l + q 1 l + q = C 3;1 p µ 1 + C 3; p µ, i4π d/ π d l p 1 l l + q 1 l + q = C 3;1 p 1 + C 3; p 1 p, 39

4 Using we have l p 1 = 1 p 1 C 3;1 + p 1 p C 3; = 1 l + q1 l p 1 I 1 [1] I 0 [1] p 1 I 3[1], with and I 0 [1] = i4π d/ I 1 [1] = i4π d/ π d 1 l l + q d d l 1 = i4π d/ π d l + q 1 l + q π d 1 l l + p. In the last step we have performed a shift of integration variable l l q 1. Likewise, if we contract with p µ we get i4π d/ π d l p l l + q 1 l + q = C 3;1 p 1 p + C 3; p, Using we have l p = 1 p 1 p C 3;1 + p C 3; = 1 l + q l q 1 + p 1 p 1 + p I [1] I 1 [1] p 1 p + p I 3 [1], The solutions are C 3;1 = C 3; = 1 p 1 p 1p p 1 p + p 1 p I [1] p I0 [1] p 1 p I [1] p 1 p p 1 p p p 1 p I 3 [1], 1 p 1 p 1p p 1 p 1 + p 1 p I [1] + p 1 I [1] p 1 p I 0 [1] p 1 p + p 1 p 1 p I 3 [1] 7.3 Two powers of loop momentum in numerator For the case fl = l µ l ν, the integral is a rank-two tensor which can be formed out of the outer products of external momenta p µ i p ν j and the metric gµν. Again for n > 5 we only use i, j =

5 I n [l µ l ν ] = i4π d/ π d l µ l ν l m 0l + q 1 m 1 l + q n m n = C n:00g µν + i,j C n;ij p µ i p ν j, The first equation we can derive relating the coefficients is obtained by contracting both sides with g µν, remembering that we need to work in general in d dimensions so that we get I n [l ] = i4π d/ π d l l m 0l + q 1 m 1 l + q n m n = C n:00d+ i,j C n;ij ij Writing we get the relation l = l m 0 + m 0, I 0 n 1[1] + m 0 I n[1] = d C n;00 + i,j C n;ij ij Another relation is obtained by contracting both sides with 1 kh pµ k pν h, to obtain 1 kh I n[p k l p h l] = min4, n 1C n;00 + C n;kh kh k,h We note here that if n > 5 and therefore UV finite and the integral contains no infrared divergences, then we have two equivalent expressions for the combination of coefficients 4C n;00 + i,j C n;kl kl. This means that the term C n;00 is redundant not uniquely defined and we can set it to zero. However, in the general case we need to keep this coefficient separate. The remaining coefficients are obtained by contracting with p µ k pν h, to obtain I n [p k l p h l] = C n;00 kh + i,j C n;ij ik jh For n > 5 this provides 11 equations for the 10 coefficients C n;ij symmetric in {i, j} and C n,00 The LHS is most easily handled by using the VP reduction formula once only say for p h l and leaving the other scalar product alone. p h l = 1 so that we get l + qh m h l + q h 1 m h 1 + q h 1 q h + m h m h 1 1 h I n 1[p k l] I h 1 n 1 [p k l] + qh 1 q h + m h m h 1 In [p k l] = C n;00 kh + C n;ij ik jh 41 i,j

6 We the use the results previously obtained for the case where fl = p k l to reduce the integrals on the LHS to scalar integrals. Care must be taken for the case where h = 0 or h 1 = 0 since in that case it is the propagator denominator l m 0 that has been cancelled and we have the integral I 0 n 1[p k l] = i4π d/ p k l π d l + q 1 m 1l + q m l + q n 1 m n 1 To get this into the standard form we must make a shift of integration variable l l q 1 This affects the numerator also and we arrive at i4π d/ π d p k l p k q 1 l m 1l + p m l + q,n 1 m n 1, which is the linear combination of an r = 1 integral and a scalar integral. A simplification occurs when n < 5. The equation 1 1 h kh I n 1[p k l] I h 1 n 1 [p k l] + qh 1 q h + m h m h 1 In [p k l] = n 1C n;00 + C n;kh kh, where k, h are summed from 1 to n 1, reduces to since 1 0 I n 1[1] + 1 kh q h 1 qh + m h h 1 m In [p k l] = n 1C n;00 + C n;kh kh 1 h kh I n 1[p k l] I h 1 n 1 [p k l] = 1 I0 n 1[1] this is best seen by considering the examples n = 3 and n = 4 separately. The LHS can be rewritten as 1 I 0 n 1 n 1[1] + q h 1 qh + m h m h 1 Cn;h, h=1 exploiting the expression for the r = 1 coefficients C n;h. Together with the equation obtained by contracting with g µν we end up with I 0 n 1[1] + m 0I n [1] = d C n;00 + C n;kh kh, d + 1 nc n;00 = 1 I0 n 1[1] + m 0 I n[1] 1 4 n 1 h=1 k,h q h 1 q h + m h m h 1 Cn;h. k,h

7 Assuming that the r = 1 coefficients have already been calculated, the expression for C n00 can be inserted into the expressions of the other r = coefficients, C n;ij and the set of nn 1 linear equations can be solved. 1 Example: Once again we consider a triangle integral with two powers of loop momentum in the numerator n = 3, r =, and internal masses set to zero. This has a UV divergence by power counting and will also be infrared divergent if any of the external square momenta vanish. I 3 [l µ l ν ] i4π d/ l µ l ν π d l l + q 1 l + q = C 3;00 g µν + C 3;11 p µ 1p ν 1 + C 3;p µ p ν + C 3;1p µ 1p ν + pµ p ν 1 We can obtain four equations for the four independent unknowns as follows: Contract with g µν I 0 [1] = dc 3;00 + p 1 C 3;11 + p C 3; + p 1 p C 3;1 1 Contract with p µ 1p ν 1 I 3 [p 1 l p 1 l] = p 1C 3;00 + p 1 C 3;11 + p 1 p C 3; + p 1p 1 p C 3;1 Using p 1 l = 1 l + q1 l q 1, this may be written 1 1 I [p 1 l] I 0 [p 1 l] p 1I 3 [p 1 l] = p 1C 3;00 +p 1 C 3;11 +p 1 p C 3; +p 1p 1 p C 3;1 Contract with p µ p ν I 3 [p l p l] = p C 3;00 + p 1 p C 3;11 + p C 3; + p p 1 p C 3;1 Using p l = 1 l + q l + q 1 + q 1 q, this may be written 1 I [p l] I 1 [p l] + q1 q I 3[p l] = p C 3;00+p 1 p C 3;11 +p C 3; +p p 1 p C 3;1 3 43

8 Contract with p µ 1p ν I 3 [p 1 l p l] = p 1 p C 3;00 + p 1 p p 1C 3;11 p 1 p p C 3; + p 1p + p 1 p C 3;1 Using p 1 l = 1 l + q1 l q 1, this may be written 1 1 I [p l] I 0 [p l] q 1I 3 [p l] = p 1 p C 3;00 +p 1 p p 1C 3;11 +p 1 p p C 3; +p 1 p C 3;1 4 Alternatively we could use p l = 1 to obtain an equivalent equation l + q l + q 1 + q 1 q 1 I [p 1 l] I 1 [p 1 l] + q1 q I 3 [p 1 l] = p 1 p C 3;00 + p 1 p p 1C 3;11 + p 1 p p C 3; + p 1 p C 3;1 4, It is useful to define the function Bq by i4π d/ π d 1 l l + q Bq from which we can easily derive i4π d/ π d l µ l l + q = 1 Bq q µ With this definition the four equations become where we have used q q 1 = p, Bp = dc 3;00 + p 1C 3;11 + p C 3; + p 1 p C 3;1 1, 1 Bq 4 p 1 + p 1 p + Bp p 1 + p 1 p p 1 I 3 [p 1 l] = p 1 C 3;00 + p 1 C 3;11 + p 1 p C 3; + p 1 p 1 p C 3;1, where we have shifted the integration variable l l q 1, and set p 1 q = p 1 + p 1 p, 1 Bp 4 1 p p 1 + Bq p + p 1 p + p 1 q I 3[p l] = p C 3;00 + p 1 p C 3;11 + p C 3; + p p 1 p C 3;1 3, 44

9 where we have written p q = p + p p, and q 1 = p 1, 1 Bq p 4 + p 1 p + Bp p 1 p + p p 1 I 3[p l] = p 1 p C 3;00 + p 1 p p 1C 3;11 + p 1 p p C 3; + p 1 p C 3;1 4 or 1 Bp 4 1 p 1 + Bq p 1 + p 1 p + p 1 q I 3[p 1 l] = p 1 p C 3;00 + p 1 p p 1 C 3;11 + p 1 p p C 3; + p 1 p C 3;1 4 If we construct the combination p 1 + p 1 p 1 p 4 + 4, divide by p 1p p 1 p and use the previously obtained relations I 3 [p 1 l] = p 1 C 3;1 + p 1 p C 3; I 3 [p l] = p 1 p C 3;1 + p C 3;, we arrive at the promised result C 3;00 + p 1C 3;11 + p C 3; + p 1 p C 3;1 = 1 Bp p 1C 3;1 q p 1C 3; which when combined with 1 gives an expression for C 3;00 C 3:00 = 1 Bp d + p 1 C 3;1 + q p 1 C 3; Note that Bq is UV divergent so that we must keep the number of dimnesions, d, away from 4 even in the absence of infrared divergences. C 3;00 is ultraviolet divergent. The remaining coefficients C 3;11, C 3;, C 3;1 can now be obtained by solving the three simultaneous equations 1,, More powers of l This process can be rolled out for any number of powers of loop momentum in the numerator, but the expressions become increasingly longer. For three powers of loop momentum we have I n [l µ l ν l ρ ] = i=1 C n;00i g {µν p ρ} i + i,jk,=1 C n;ijk p {µ i p ν j pρ} k 45

10 and we need to contract with g µν p ρ r or with p µ r p ν sp ρ t to obtain a set of linear equations for the coefficients C n;00i or C n;ijk. For four powers of loop momentum we have I n [l µ l ν l ρ l σ ] = C n;0000 g {µν g ρσ} i,j=1 C n;00ij g {µν p ρ i p σ} j + and we need to contract with g µν g ρσ, g µν p ρ r pσ s, and pµ r pν s pρ t p σ u coefficients C n;0000, C n;00ij and C n;ijkh i,j,k,h=1 C n;ijkh p {µ i p ν j pρ k pσ} h in order to project out the Although this becomes more complicated, a contraction with a single g µν or with a single momentum p µ i, where p i is one of the first four external momenta enables one to write any loop integral I n [l µ1 l µr ] in terms of and and the process can be iterated. I n [l µ1 l µ r 1 ] I j n 1[l µ1 l µ r 1 ], 7.5 Remnant Integrals In QCD a one-loop n-point graph has a maximum of n vertices each carrying a maximum power of one loop momentum and n propagators. Each stage on the reduction reduces the number of loop momenta on the numerator by one and the number of denominators propagators by one. This means that from an ultraviolet convergent loop integral we will generate terms which are ultraviolet divergent. For example a 5-point matrix element will contain the ultraviolet convergent integral I 5 [l µ l ν l ρ l σ l τ ] Applying the reduction once we will get terms of the form I 4 [l µ l ν l ρ l σ ], which by power counting in ultraviolet divergent. Since we started with a UV convergent integral this means that the sum of all the generated UV divergent integrals will end up being finite. But care must be taken to carry out the integrals using a robust regulator such as dimensional regularisation or dimensional reduction in order to be sure that the remaining finite terms are correct. 46

11 The reduction process can be iterated until the only UV divergent integrals are and I [1] = i4π d/ I [l µ ] = i4π d/ I [l µ l ν ] = i4π d/ 1 π d l m 0l + q m 1 π d l µ l m 0l + q m 1 l µ l ν π d l m 0l + q m 1. All the other integrals are scalar triangles, boxes, pentagons, etc. Each of these integral has an imaginary part in some domains of the invariant square momenta q i. However, the converse statement, namely that the integrals can be reproduced from knowledge of these imaginary parts has one unfortunate exception arising from the relation for massless internal particles q I l µ l ν µ q 1 ν 3 gµν I [1] = 1 g µν q q µ q ν, 1 18 which is purely real. This means that precise knowledge of the imaginary part of an integral does not specify the coefficient of this finite combination of two-point integrals. 47