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Studying dimensional reugularization one often encounters the following identity:

$$ \int d^d q\, \, q^\mu q^\nu f(q^2) = \frac{1}{d}g^{\mu\nu}\int d^d q\,\, q^2 f(q^2) $$

often justified by some symmetry arguments which I admit i never really understood, and since I didn't understand them I tried to prove the identity mathematically, the proof below is valid in general, so I ask you

  • Is it really valid outside of the integral sign? If it isn't, why?
  • Independently of the answer to the first question can someone explain me with clarity and details the reasons why that identity is true, without doing the proof, like basically every textbook does?

My proof basically relies on index gymnastics:

$$ q^\mu q^\nu = g^{\mu\rho} g^{\sigma\nu} q_\rho q_\sigma $$

multiplying by $g^{\rho\sigma}$ on both sides

$$ g^{\rho\sigma}q^\mu q^\nu = g^{\mu\rho} g^{\sigma\nu} q^2. $$

Multiplying now by $g_{\rho\sigma}$

$$ g_{\rho\sigma} g^{\rho\sigma}q^\mu q^\nu = g_{\rho\sigma} g^{\mu\rho} g^{\sigma\nu} q^2 $$

which is equal to

$$ d \,\,q^\mu q^\nu = \delta^\mu_\sigma g^{\sigma\nu} q^2 = g^{\mu\nu} q^2 $$

which is equivalent to

$$ q^\mu q^\nu = \frac{1}{d} g^{\mu\nu} q^2 $$

Which is the required identity.

On the other end i don't see why terms like $q^0 q^1$ would be zero in general, if $q$ is momentum for example, $q^0 q^1$ is just the product of energy and momentum along the x direction, why would it be zero?

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    $\begingroup$ Check Peskin's QFT book, the discussion around equations 6.45 and 6.46. In general the expression is only valid under integration and relays also on the fact that $f$ depends only on $q^2$. $\endgroup$
    – secavara
    Commented Oct 27, 2019 at 14:26
  • $\begingroup$ Indeed, why would it be zero? Your identity is only true under the integral sign, as is easily demonstrable by choosing a $q^{\mu}$ that doesn't satisfy $q^{\mu} q^{\nu} \sim g^{\mu \nu}$. $\endgroup$
    – Prof. Legolasov
    Commented Oct 27, 2019 at 14:50
  • $\begingroup$ @AnOrAn, mind you that the identity is correct conditional on the fact that integration is finite or logarithmically divergent. Symmetry arguments DO NOT work for integrals which are more than logarithmically divergent: ABJ anomaly is exhibit A. $\endgroup$
    – MadMax
    Commented Oct 28, 2019 at 14:42
  • $\begingroup$ @AnOrAn, it boils down to the surface terms which spoil the symmetry argument for non-logarithmically divergent integrals. You could check out any text book on ABJ anomalies. $\endgroup$
    – MadMax
    Commented Oct 29, 2019 at 14:57
  • $\begingroup$ @AnOrAn, e.g. the integral may NOT equal to zero for $\mu \neq \nu$, if the integral is quadratically divergent. It depends on the non-trivial cutoff/boundary condition, which results in the surface term. $\endgroup$
    – MadMax
    Commented Oct 29, 2019 at 15:07

4 Answers 4

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Proposition. If the integrals $$\begin{align} I^{\mu\nu}~:=~&\int_{\mathbb{R}^d}\! \frac{d^dq}{(2\pi)^d} f(q^2) q^{\mu}q^{\nu},\cr I~:=~&\int_{\mathbb{R}^d}\! \frac{d^dq}{(2\pi)^d} f(q^2) q^2, \end{align}\tag{1} $$ are convergent, then $$ I^{\mu\nu}~=~g^{\mu\nu}I/d.\tag{2}$$

Sketched proof of eq. (2): First Wick-rotate to Euclidean signature $$g^{\mu\nu}~=~\delta^{\mu\nu}.\tag{3}$$ Secondly, off-diagonal $I^{\mu\nu}$ must be zero, because the integrand is then odd. Thirdly, the diagonal elements $$ I^{11}~=~I^{22}~=~\ldots~=~I^{dd} \tag{4}$$ must all be equal. Fourthly, by definition $$ I~\stackrel{(1)}{=}~I^{11}+I^{22}+\ldots+I^{dd}.\tag{5}$$ Combine the above observations to derive the sought-for eq. (2). $\Box$

It is not hard to see that the sought-for identity does not hold without the integral sign.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT; eq. (6.46).
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Your proof is not valid. The $\rho$ and $\sigma$ indices are dummy indices in a summation, attempting to multiply by $g^{\rho\sigma}$ on both sides amounts to multiplying each term in a sum by different factors (which results in an inequivalent expression).

The argument is fundamentally one of symmetry of the integration region, hence why there needs to be an integral. It is basically the same flavor of logic as one uses to simplify the integration of odd/even functions over symmetric intervals in one dimension. If you specialize to the Minkowski metric and integrate over all space (as I suspect will generally be the case for your application), it will be easier to see why the result is valid.

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  • $\begingroup$ Ok, thank you very much for clarifying where exactly the proof is wrong. I'm not really sure I understood your point on why i can't multiply by the metric with dummy indices, could you please explain further, rephrase or propese an example? About your second point: I understood that the intergral is zero unless $\mu=\nu$ and that it has to be a tensor cause of Lorentz symmetry, but why the metric tensor? Can't it be any other diagonal tensor that is different from zero only for $\mu=\nu$? $\endgroup$
    – Erontado
    Commented Oct 29, 2019 at 14:23
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    $\begingroup$ Sure, a simple example would be a two-dimensional space with diagonal metric $g^{00}=1$ and $g^{11}=2$. Consider a vector with $q_0=1$ and $q_1=1$. Then $3=g^{\mu\nu}q_\mu q_\nu$, but $g^{\mu\nu}g^{\mu\nu}q_\mu q_\nu=5\neq 3g^{\mu\nu}$. $\endgroup$
    – ProfJMiller
    Commented Oct 30, 2019 at 15:25
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    $\begingroup$ This result is more general than just for spaces with Lorentz symmetry, and generally the metric is non-diagonal even for flat spacetimes (just pick a non-orthogonal basis for the Minkowski metric). Consider a flat two-dimensional Euclidean space with a non-orthogonal $(q^0,q^1)$ with an angle $\theta$ between them. Convert $\int d^2q\,q^0q^1\,f(q^2)$ back to orthogonal Cartesian coordinates and you'll see the result is non-zero and proportional to the off-diagonal metric component. $\endgroup$
    – ProfJMiller
    Commented Oct 30, 2019 at 16:20
  • $\begingroup$ ok, thanks that's even clearer now. I'm just struggling to see it in general, I proved that special case but how could be this done in general $\endgroup$
    – Erontado
    Commented Oct 31, 2019 at 9:18
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Since the integrand is a Lorentz (2,0) tensor so must be the answer. Your integrand does not depend on any other momenta or position vectors or tensors so the only two tensor that (Always) exists in the problem is the metric tensor. Hence the result must be proportional to it. The result is definitely not valid outside of the integral (there is, however a tensor decomposition into antisymmetric, trace and symmetric traceless parts.

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  • $\begingroup$ you write: the only two tensor that (Always) exists in the problem is the metric tensor. What does that mean? that that tensor always exists in that problem? why can't it be an energy momentum tensor multiplied by some appropriate constant in order to have the correct dimensions? $\endgroup$
    – Erontado
    Commented Oct 29, 2019 at 14:18
  • $\begingroup$ The result can only depend upon objects that enter the integrand but aside from the integration variable there is nothing else if could depend upon aside from the single tensor that exists due to geometry. $\endgroup$
    – nox
    Commented Oct 29, 2019 at 19:18
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This identity is true even outside the integral sign. It is just a consequence of $g^{\mu\nu}g_{\mu\nu}=d$, as you observed in your proof.

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    $\begingroup$ This is not true, given that the RHS is diagonal and the LHS is not. $\endgroup$
    – Javier
    Commented Oct 27, 2019 at 14:49
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    $\begingroup$ $g^{\mu \nu} g_{\mu \nu} = d$ is true, but $q^{\mu} q^{\nu} \sim g^{\mu \nu}$ is not. The latter does not follow from the former. $\endgroup$
    – Prof. Legolasov
    Commented Oct 27, 2019 at 14:51