# I Tensor integrals in dimensional regularisation

1. Mar 4, 2017

### CAF123

Consider a d dimensional integral of the form, $$\int \frac{d^d \ell}{(2\pi)^d} \frac{\ell^{\sigma} \ell^{\mu}}{D}\,\,\,\text{and}\,\,\, \int \frac{d^d \ell}{(2\pi)^d} \frac{\ell^{\sigma}}{D}$$ where $D$ is a product of several propagators. One can reduce this to a sum of scalar integrals by writing e.g the 1 tensor $\ell^{\mu}$ in terms of two linearly independent vectors p and n and the 2 tensor $\ell^{\sigma} \ell^{\mu}$ as a linear combination of all possible lorentz structure with two indices in the basis {p,n} (so we have e.g $g^{\sigma \mu}, p^{\mu}p^{\sigma}, p^{\sigma}n^{\mu},$etc...

My question is simply, why bother with this 2 tensor decomposition if I can perhaps simply rewrite $\ell^{\sigma} \ell^{\mu}$ as a multiplication of two 1 tensors? This avoids having to deal with a 2 tensor decomposition and just using the 1 tensor decomposition repetitively?

Explicitly, can I not write $\ell^{\sigma} \ell^{\mu} = (p^{\sigma} (\ell \cdot n) + n^{\sigma} (\ell \cdot p)) ( \sigma \rightarrow \mu)$ thereby avoiding the use of the 2 tensor decomposition and only dealing with the 1 tensor decomposition $p (\ell \cdot n) + n(\ell \cdot p) ?$

Thanks!

2. Mar 4, 2017

### Dr.AbeNikIanEdL

Can you show where this is done? If this is supposed to be the usual Passarino-Veltman stuff, this is not how it works. What you do there is to write the whole integral as combination of all lorentz vectors it depends on, but which are not integrated over,

$\int \frac{\mathop{d^dl}}{(2\pi)^d} \frac{l^\mu l^\sigma}{D(p,n)} = A g^{\mu\nu} + B p^\mu p^\nu+ C n^\mu n^\nu + D p^{\mu}n^\nu + E n^\mu p^\nu$

which you can do because there are no other vectors or tensors this integral can depend on. You then might solve for the coefficients. Why do you think you could write a general D dimensional vector as a sum of just two vectors?

Last edited: Mar 4, 2017
3. Mar 4, 2017

### CAF123

Hi @Dr.AbeNikIanEdL
I am reading it in a thesis and the external momenta are all written in terms of Sudakov basis vectors p and n in the Bjorken limit. They are two linearly independent vectors satisfying p.p=n.n=0 and p.n=1. So essentially by writing $l^{\mu} = p^{\mu} (l \cdot n) + n^{\mu} (l \cdot p)$ I have made a Sudakov parametrisation of my loop momenta which is the natural choice of frame in the high energy limit with which I am considering (working in the framework of collinear factorisation).

But I am not sure if that clarifies anything?
Could you give a reference for this way of doing things? It makes sense to me.

4. Mar 4, 2017

### Dr.AbeNikIanEdL

But Sudakov decomposition is something like

$k^\mu = Ap^\mu + Bn^\mu + Ck_\mathrm{T}$

where $k_\mathrm{T}$ is some object with $D-2$ degrees of freedom and $p$,$n$ are light like vectors which satisfy $pn=0$ as you said. This $k_\mathrm{T}$ part seems to be missing. Note that it might be valid to neglect it for external momenta, but not for the loop momenta you have to integrate over.

Just google Passarino Veltman integrals, I guess it is contained in lecture notes and so on.

5. Mar 4, 2017

### CAF123

I see. So Passarino-Veltmann is essentially a tensor reduction in that it reduces tensor integrals to a linear combination of scalar ones. So all the coefficients in your decomposition, i.e A, B,C etc.. should be understood to be a scalar integral with lorentz structure factored out in front of A,B,C etc..?

I just read the following in the thesis: 'It should be understood that all appearances of the loop momentum $l^{\mu}$ appear under an integral, thus each of the tensor coefficients and $l^{\mu} = p^{\mu} (l \cdot n) + n^{\mu} (l \cdot p)$ should be understood to be the numerator of an integral over the loop momentum.'
However, is this perhaps an ambiguous statement? It could mean

1) that all instances of $l^{\mu}$ actually mean an integral of the form $\int \frac{d^dl}{(2\pi)^d} \frac{l^{\mu}}{D}$ in which case the method in the thesis does indeed coincide with the Passarino-Veltmann prescription (i.e he didn't write the integral explicitly for quickness I can only guess)

or

2) he means the actual loop momentum vector is written as this decomposition (which given the comments you made, I see this is probably not the way the statement should be interpreted)

For 1) it would then be the case that e.g $l^{\mu} = p^{\mu} (l \cdot n) + n^{\mu} (l \cdot p)$ should be understood as $$\int \frac{d^dl}{(2\pi)^d} \frac{l^{\mu}}{D} = p^{\mu} \int \frac{d^dl}{(2\pi)^d} \frac{ (l \cdot n)}{D} + n^{\mu} \int \frac{d^dl}{(2\pi)^d} \frac{(l \cdot p)}{D}?$$

Last edited: Mar 4, 2017
6. Mar 4, 2017

### Dr.AbeNikIanEdL

Essentially yes.

I am obviously not familiar with that kind of notation, but it sounds reasonable to me. The last line looks correct to me, if $p$ and $n$ are the external (massless) momenta of the loop amplitude.

7. Mar 5, 2017

### CAF123

Thanks. This is perhaps a naive question but given the formulae at the back of e.g Peskin and Schroeder such as $$\int \frac{d^dl}{(2\pi)^d} \frac{l^{\mu} l^{\nu}}{(l^2- \Delta)^n} = \frac{(-1)^{n-1}}{(4 \pi)^{d/2}} \frac{i}{2} g^{\mu \nu} \frac{\Gamma(n-d/2-1)}{\Gamma(n)} \left(\frac{1}{\Delta}\right)^{n-d/2-1}$$ why would one have to bother with all this reduction prescription?

I could for example, reduce $D$ down to the form $(l^2 - \Delta)^n$ using Feynman parameters and then simply use these known formulae provided by P&S. The formulae provided by them (shown above) has only the metric because I guess all other lorentz vectors that would be included in $\Delta$ are involved in a contraction. But then this same argument could be applied to $D(p,n)$ - it only involves contractions of p and n so I don't understand the reconcilation of the fomulae provided by P&S and the tensor reduction formulae in post #2 $$\int \frac{d^dl}{(2\pi)^d} \frac{l^{\mu} l^{\nu}}{D(p,n)} = g^{\mu \nu} A + p^{\mu} p^{\nu} B + \dots,$$ where we can make the tensor structure of p and n explicit.

Thanks!

Edit: Ooops, is it simply because in my choice of frame I have Sudakov decomposed my loop momenta in terms of linearly independent $p, n$ and transverse $l_T$? I was too focused on $D(p,n)$ earlier hehe. If that is the case, then does it then follow that the formulae in P&S cannot be used (ie they are not valid)?

Last edited: Mar 5, 2017
8. Mar 5, 2017

### Dr.AbeNikIanEdL

So if you do the actual working out of the coefficients, you would get the same result, only the coefficient of $g^{\mu\nu}$ is non-zero, and you should get factors consistent with the ones in Peskin-Schroeder. You can do the decomposition first, or first introduce Feynman parameters, and of course you can tabulate some general intermediate results like it is done in P&S. The integrals needed for usual one loop calculations are of course all known, so you can of course just look them up (or probably just get the right mathematica package). But I guess that is true for almost everything that is not your original research.

9. Mar 5, 2017

### CAF123

I worked out the coefficients A,B,C, D (=C by symmetry) and E and they weren't zero. Is it not simply the case that in P&S they didn't make a Sudakov decomposition of their momentum $l^{\mu}$ so the only available lorentz structure was the metric in their case? For A I was getting $$A = \int \frac{d^dl}{(2\pi)^d} \frac{1}{D} \left( \frac{-2}{d-2} (l \cdot n) (l \cdot p) + \frac{1}{d-2} l^2\right),$$ C was $$C = \int \frac{d^dl}{(2\pi)^d} \frac{1}{D} \frac{1}{d-2} \left( d(l \cdot n) (l \cdot p) - l^2 \right)$$ and similar equations for the other coefficients. Now from the tensor decomposition in #2, by contracting with the metric on both sides we have that $\int \frac{d^dl}{(2\pi)^d} \frac{l^2}{D} = dA + 2C$ and using the explicit expression for A and C given above, we see this holds. I think this had to be the case because the integral with l^2 in the numerator is lorentz invariant so independent of the basis of l (while the integral with the 2 tensor $l^{\mu} l^{\nu}$ in the numerator is frame dependent, and so in my frame using Sudakov vectors I get a different answer from P&S but such that any lorentz scalar integrals obtained from it are the same? Makes sense to me but would you agree?)

Since e.g A involves a product $(l \cdot n) (l \cdot p)$ necessarily I will have numerator factors in my integrals. These numerators are however scalars so I will end up with terms like $$\int \frac{d^dl}{(2\pi)^d} \frac{l^2}{D'},$$ which can be evaluated using the P&S formulae after feynman parameters.

10. Mar 5, 2017

### Dr.AbeNikIanEdL

Probably that was not precise, my point was that you either first simplify your denominator by using Feynman parameters, and then solve the somewhat simpler tensor integrals, or you reduce the tensor integrals to scalar ones and then solve the scalar ones by introducing Feynman parameters and so on. Either way, you should get the same result.

11. Mar 5, 2017

### CAF123

Ok thanks! Can you say whether you agree with this statement of mine above? Basically I am saying the 2 tensor integral is different from P&S's result because of my choice of basis for l but is such that the coefficients of the tensor decomposition conspire to yield the same results for lorentz scalar integrals (as I demonstrated for the case where numerator is $l^2$). E.g with sudakov basis for l, we have the decomposition in #2 while in P&S they don't have such $p^{\mu} p^{\nu}$ terms for example and the coefficients are not zero hence I think the results should be different for tensor integrals.

Thanks again!

12. Mar 5, 2017

### Dr.AbeNikIanEdL

I don't think so. Lets for simplicity only take the integral with one external momentum, so $D$ looks something like

$D = l^2(l+p)^2~.$

and introducing Feynman parameters yields something like

$\int_0^1 \mathop{dx}\int_l \frac{l^\mu l^\nu}{l^2 +2l.px + m^2 x}$

where $p^2 = m^2$ and I use the shorthand $\int_l = \int \frac{d^d l}{(2\pi)^d}$. Now to get rid of the extra factor $l.p$ we would need to replace $l\rightarrow l-xp$ which will make

$l^2 +2l.px + m^2 x \rightarrow l^2 + m^2x(1+x) =: l^2 + \Delta~.$

However we will also need to make the same shift in the numerator, and get something like

$\int_0^1 \mathop{dx} \left[\int_l \frac{l^\mu l^\nu}{l^2+\Delta} + x^2p^\mu p^\nu \int_l \frac{1}{l^2+\Delta} \right]~.$

These integrals can now be evaluated like in the formulas from P&S you found. As said, we could have anticipated that the result is something like

$Ag^{\mu\nu} + Bp^\mu p^\nu ~,$

with some coefficients $A$, $B$ and could have exploited this to make the integrals easier.

Now if you are considering an integral with two external momenta, i.e. your denominator looks like

$D = l^2(l+p_1)^2(l+p_2)^2$

you can do the same thing. But you know that the result will look like

$Ag^{\mu\nu} + Bp_1^\mu p_1^\nu + Cp_2^\mu p_2^\nu + Dp_1^\mu p_2^\nu + Ep_2^\mu p_1^\nu ~,$

for some $A$, $B$, $C$, $D$ and $E$ simply because the integral only depends on these vectors, not because of any specific choice of frame.

Last edited: Mar 5, 2017
13. Mar 5, 2017

### CAF123

Ok. I suppose you neglected the cross terms in the product $(l-xp)^{\mu} (l-xp)^{\nu}$ because they contribute a single power of loop momenta so vanish under symmetric integration? If this is the case, then when I talked about the integral $\int_l \frac{l^{\mu}}{D}$ isn't this just zero for the same reason? So the only interesting case is the 2 tensor and there is no need for consideration of a 1 tensor integral decomposition.

14. Mar 6, 2017

### Dr.AbeNikIanEdL

It holdes that

$\int_l \frac{l^\mu}{l^2-\Delta} = 0$

where $\Delta$ does not depend on $l$, which I used above, yes. However, this is not how a general loop integral will look, there the denominator will be something like

$D = l^2(l+p_1)^2(l+p_2)^2(l+p_3)^2\dots$

with an extra factor for every external momentum (note that all the time we also neglect the masses of the particles in the loop, so the most general denominator would also include these). This is not just symmetric in $l^\mu$, and for example

$\int_l \frac{l^\mu}{l^2(l+p)^2} \neq 0 ~.$

15. Mar 6, 2017

### CAF123

Yes, I was being silly, somehow I forgot I had originally defined D in terms of a product of propagators and not just a single one after feynman parametrisation, It's clear thanks. I guess my last question on this is to do with recovering the A and B coefficients in the ansatz $A g^{\mu \nu} + B p^{\mu} p^{\nu}$. Using that $l^{\mu} l^{\nu} = (1/d) g^{\mu \nu} l^2$, using what you wrote I can write $$\int_l \frac{l^{\mu} l^{\nu}}{l^2 (l+p)^2} = \int_{x,l} \frac{1}{d} g^{\mu \nu} \frac{l^2}{(l^2 - \Delta)^2} + p^{\mu} p^{\nu} \int_{x,l} x^2 \frac{1}{(l^2 - \Delta)^2}$$ and then from the ansatz I should expect to recover say $$A = \frac{1}{d} \int_{x,l} \frac{l^2}{(l^2- \Delta)^2}$$
So, from $$\int_l \frac{l^{\mu} l^{\nu}}{l^2 (l+p)^2} \overset{!}{=} Ag^{\mu \nu} + B p^{\mu} p^{\nu}$$ I contract with $g_{\mu \nu}$ and $p_{\mu}$ in turn to get $$\int_l \frac{l^2}{l^2 (l+p)^2} = dA + m^2 B\,\,\,\,\text{and}\,\,\,\,\int_l \frac{ (p \cdot l) l^{\nu}}{l^2 (l+p)^2} = Ap^{\nu} + Bm^2 p^{\nu}$$ Multiplying first one by $p^{\nu}$ and subtracting from the second I get $$\int_l \frac{l^2 p^{\nu} - (p \cdot l) l^{\nu}}{l^2 (l+p)^2} = Ap^{\nu}(d-1)$$ Contracting again with $p_{\nu}$ gives me $$A = \frac{1}{d-1}\frac{1}{m^2} \int_l \frac{l^2 m^2 - (p \cdot l)^2}{l^2 (l+p)^2} = \frac{1}{d-1} \frac{1}{m^2} \int_{x,l} \frac{l^2 m^2 - (p \cdot l)^2}{(l^2 - \Delta)^2}$$
I'm not seeing at the moment how this concides with the correct A I wrote above. Can you see if and how it does?

($\int_{x,l}$ means integration over x and l, x the feynman parameter)

16. Mar 6, 2017

### Dr.AbeNikIanEdL

Contracting with $g^{\mu\nu}$ and $p$ is the right starting point, but I don't get what you do afterwards. So first, notice that the integral you get after contracting with $g^{\mu\nu}$ is zero (in the massless case at least), which is a property of dim. regularisation for integrals not involving a mass scale, i.e.

$\int_l \frac{1}{l^n} = 0 ~,$

so you get

$A = -\frac{p^2}{d} B$

from that. The second one is a bit more messy, but contracting with $p^\nu$ as well as $p^\mu$, and writing $l.p$ as

$l.p = \frac{1}{2}((l+p)^2 - l^2 - p^2)$

should finally give

$Ap^2 + B p^4 = p^4 \frac{d-1}{d}B = -p^2 p^\mu \int_l \frac{l_ \mu}{l^2(l+p)^2} = \frac{p^4}{4}\int_l \frac{1}{l^2(l+p)^2}$

where the first equality is using the result for $A$ from above, the second follows from simplifying what you get by contracting the original integral with $p^\mu p^\nu$, and the third by doing everything again for $\int_l \frac{l^\mu}{D}$. This last integral can now be evaluated using Feynman parameters to finally get the coefficient $B$.

Edit:
That is not true.

Last edited: Mar 6, 2017
17. Mar 6, 2017

### CAF123

I understand the first equality and how the third follows from the second (although I think the factor of 1/4 should be 1/2 here?) but I am not seeing how the first to the second is obtained. Using that $p^4 = m^4$ I believe we get $$m^4 \frac{d-1}{d} B = \int_l \frac{(p \cdot l)^2}{l^2 (l+p)^2}$$ then I guess you expanded out the product $(p \cdot l)^2$. That gives $$\frac{1}{4} \left( (l+p)^4 - 2l^2 (l+p)^2 - 2m^2 (l+p)^2 + l^4 + 2 l^2 m^2 + m^4\right)$$ The second, third, and fifth terms here give rise to scaleless integrals I think so I put them to zero but out of the remaining terms I don't see how it is just $-m^2 \int_l \frac{p \cdot l}{l^2 (l+p)^2}$?

Oh, why is that? $l^2 = l^{\mu}l_{\mu} = g_{\mu \nu} l^{\nu} l^{\mu}$ Then invert the equation means $g^{\mu \nu} l^2 = d l^{\mu} l^{\nu}$ (I checked this by contracting the lhs with $g_{\mu \nu}$ to give $g_{\mu \nu} g^{\mu \nu}l^2 = dl^2 = d g_{\mu \nu} l^{\mu} l^{\nu}$ which is consistent no?)

Thanks!

18. Mar 6, 2017

### Dr.AbeNikIanEdL

The step with "inverting" the equation you do does not make sense at all. To see that is not consistent, take $l = (1,0,0,-1)$, it would follow $l^\mu l^\nu = 0$, but we know e.g. $l^1 l^1 = 1$.

Not quite, just write

$(p.l)(p.l) = \frac{p.l}{2}\left((l+p)^2 - l^2 -p^2\right)$

and the first and second term will vanish under the integral as they are antisymmetric (shift l in the second term by p to make this more visible), the last one is already of the claimed form.

19. Mar 6, 2017

### CAF123

Thanks! The derivation of A and B is now clear, except it would still be nice to see that the results obtained in your post #12 agree with what we got. In particular, in the fourth equation display, first integral in #12, what is the correct way to then write $l^{\mu} l^{\nu}$ in terms of the metric to make contact with our decomposition? I see for example the replacement $l^{\mu} l^{\nu} \rightarrow (1/d) g^{\mu \nu} l^2$ is used in P&S in their appendix and that the integral result for $\int_l \frac{l^2}{(l^2-\Delta)^2}$ may be derived from that of $\int_l \frac{l^{\mu} l^{\nu}}{(l^2-\Delta)^2}$ using precisely this replacement. So what warrants P&S's use of $l^{\mu} l^{\nu} \rightarrow (1/d) g^{\mu \nu} l^2$ here?

20. Mar 6, 2017

### Dr.AbeNikIanEdL

I see, I think the replacement is valid under the integral, where it follows from partial integration arguments (i.e. it is true up to some surface term, which vanishes under the integral) if I remember correctly.

21. Mar 6, 2017

### CAF123

Ok, I only ever replace $l^{\mu} l^{\nu}$ with $(1/d)g^{\mu \nu} l^2$ under an integral so I think I can make this substitution. So, in the fourth display,first integral I would rewrite $$\int_0^1 dx \int_l \frac{l^{\mu} l^{\nu}}{(l^2- \Delta)^2} = \int_0^1 dx \frac{g^{\mu \nu}}{d} \int_l \frac{l^2}{(l^2-\Delta)^2}$$ and then read off $$A = \int_0^1 dx \frac{1}{d} \int_l \frac{l^2}{(l^2-\Delta)^2}$$ But this is different from the A we found, which was $$A = \frac{1}{1-d} \frac{m^2}{2} \int_l \frac{1}{l^2 (l+p)^2} = \frac{1}{1-d} \frac{m^2}{2} \int_0^1 dx \int_l \frac{1}{(l^2-\Delta)^2}?$$ so I am still a little confused.

22. Mar 6, 2017

### Dr.AbeNikIanEdL

I think your $A$ from the decomposition is wrong by a factor of $1/2$ still? Apart from that I get (using $\Delta = p^2x(1-x)$) from the decomposition

$A = -\frac{1}{d-1}\frac{p^2}{4}\int_0^1 \mathop{dx}\int_l \frac{1}{(l^2-\Delta)^2} \\= -\frac{1}{d-1}\frac{p^2}{4} \int_0^1 \mathop{dx} \frac{(-1)^ni}{(4\pi)^{d/2}} \frac{\Gamma(2-\frac{d}{2})}{\Gamma(2)}\left(\frac{1}{p^2x(1-x)}\right)^{2-d/2} = -\frac{(-1)^ni}{(4\pi)^{d/2}} \frac{\Gamma(1-\frac{d}{2})}{2\Gamma(2)}\left(\frac{1}{p^2}\right)^{1-d/2} \times \frac{1-d/2}{2(d-1)}\int_0^1 \mathop{dx} \left(\frac{1}{(1-x)x}\right)^{2-d/2}$

where the second equality is the P&S formula for our case $n=2$, and the third one uses $\Gamma(x+1) = x\Gamma(x)$. The other integral gives

$A' = \int_0^1 \mathop{dx}\frac{1}{d}\int_l \frac{l^2}{(l^2-\Delta)^2} = -\frac{(-1)^ni}{(4\pi)^{d/2}}\frac{\Gamma(1-\frac{d}{2})}{2\Gamma(2)}\left(\frac{1}{p^2}\right)^{1-d/2} \times \int_0^1 \mathop{dx} \left(\frac{1}{x(1-x)}\right)^{1-d/2} ~,$

which is again the P&S formula with the terms rearranged such that only the last bit differs. Using

$\int_0^1 \mathop{dx} \left(\frac{1}{x(1-x)}\right)^y = \frac{\Gamma(1-y)^2}{\Gamma(2-2y)}$

one arrives at expressions for these final bits, which can be simplified by again using the property of the gamma function, and I get in both cases

$\frac{\Gamma(d/2)^2}{\Gamma(d)}$

for the part after the "$\times$"-symbol, so that indeed $A=A'$.

Edit: corrected sign of first equation

Last edited: Mar 6, 2017
23. Mar 6, 2017

### CAF123

Thanks! Ok I got everything you said except i think my answer is off from yours by a (crucial) minus now. Given our starting equations $$\int_l \frac{(p \cdot l)^2}{l^2 (l+p)^2} = Am^2 + Bm^4$$ and $0 = dA + Bm^2$ we deduce $$Am^2(1-d) = \int_l \frac{(p \cdot l)^2}{l^2 (l+p)^2}$$ equal to $$Am^2(1-d) = - \frac{m^2}{2} \int_l \frac{p \cdot l}{l^2 (l+p)^2} = + \frac{m^4}{4} \int_l \frac{1}{l^2 (l+p)^2} = + \frac{m^4}{4} \int_{l,x} \frac{1}{(l^2 - \Delta)^2}$$ which is off from your answer for A by a minus but I just don't see where this minus came from.

24. Mar 6, 2017

### Dr.AbeNikIanEdL

Hm, I think there is a typo and I flipped $\frac{1}{d-1}$ from my notes to $\frac{1}{1-d}$... I edited the post, but I will have to check my notes when I get back to my office tomorrow.