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I have a problem to reproduce the following identity:

\begin{equation} \Pi_{\mu\nu}(q^2) = i \int d^Dx e^{iqx} \langle 0 | T \{j_\mu(x) j_\nu(0) \} | 0 \rangle = (q_\mu q_\nu - g_{\mu\nu} q^2 ) \Pi(q^2) \end{equation} with $j_\mu(x) = :\bar q (x) \mu_\mu q(x):$ as the normal ordered vector current.

How can I derive the second identity?

In the script I am reading the explanation is simple given by: 'The second identity holds since the vector current is conserved, i.e. $\partial^\mu j_\mu (x) = 0$, and thus $q^\mu \Pi_{\mu\nu}(q^2)$ has to vanish, which implies the Lorentz structure on the right.'

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  • $\begingroup$ Couldn't have said it better than what the book already says. What's bothering you? $\endgroup$ – Prahar Oct 16 '15 at 6:51
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The proposition expressed by the last $=$ is not an "identity". It is a simplified Ansatz. The "equals to" sentence doesn't say that two pre-existing expressions are identical. It says that one may rewrite the first expression, $\Pi_{\mu\nu}(q^2)$, using a newly defined $\Pi(q^2)$.

The left hand side is a collection of $D^2$ functions of the variable $q^2$ because the indices $\mu,\nu$ attached to the "general" variable $\Pi_{\mu\nu}$ take $D$ values each and distinguish which of these functions we talk about. The right hand side is only one function $\Pi$ of the same variable $q^2$.

It is possible to rewrite these $D^2$ functions in terms of one, with the tensor dependence that is indicated, because the integral on the left hand side depends on the free indices $\mu,\nu$ but the structure of the integral depends on covariant tensors and vectors such as $j_\mu$ only. And the theory is Lorentz-covariant. It follows that the result of the integral has to depend on such tensors only, too. "Nice expressions for integrals have to produce nice results," if you wish.

But on the left hand side, we could use vectors such as $j_\mu(x)$ that also depend on the position $x$. However, the result of the integral has no $x$-dependence, it is a set of "global" functions, and there are no global dynamical variables such as $j_\mu$. The only covariant tensors (or vectors) that the integral may depend on are therefore $g_{\mu\nu}$ and $q_\mu$; the $\epsilon_{\kappa\lambda\mu\nu}$ could be possible for a larger number of indices but not here.

The indices $\mu,\nu$ have to arise covariantly from $q_\mu$ and $g_{\mu\nu}$. There are only two independent ways how the pair of free indices $\mu,\nu$ may arise from this tensor plus vector, so it follows that the integral must have the form of $$ g_{\mu\nu}\cdot A(q^2) + q_\mu q_\nu \cdot B(q^2) $$ That's why the $D^2$ functions $\Pi_{\mu\nu}$ expressed by the integral may be expressed in terms of the two functions $A,B$. However, because $q^\mu \Pi_{\mu\nu}(q^2)=0$, which we may express using $A,B$, these functions $A,B$ are not independent of each other. They may be calculated from one another because the vanishing is equivalent to $$ 0 = q^\mu \cdot g_{\mu\nu}\cdot A(q^2) + q^\mu\cdot q_\mu q_\nu \cdot B(q^2) $$ This equation is equivalent to $$q_\nu ( A(q^2) + B(q^2)q^2 ) = 0$$ which implies that whenever the vector $q_\nu\neq 0$, it must be true that $A=-q^2 B$. In the result above, they use the symbol $\Pi(q^2)$ for the function $B(q^2)$ and rewrite the result $$ g_{\mu\nu}\cdot A(q^2) + q_\mu q_\nu \cdot B(q^2)=\dots $$ in terms of $\Pi(q^2)$ as $$ (q_\mu q_\nu - q^2 g_{\mu\nu}) \Pi(q^2) $$

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  • $\begingroup$ Just to be clear - ansatz is usually meant to be a "guess" for the answer. Here we have actually derived what the form of the answer must be - so it's not really an ansatz. $\endgroup$ – Prahar Oct 16 '15 at 6:55
  • $\begingroup$ Dear Prahar, the "form" and "Ansatz" are exactly the same if your sentence is understood properly, so no tension like that may exist. Ansatz doesn't mean a particular guess where nothing can be adjusted anymore. An Ansatz means a guessed result that depends on some parameters that are left unspecified and that still have to be calculated or chosen, in this case on $\Pi(q^2)$. An Ansatz is a step towards the full solution, not the full solution per se. $\endgroup$ – Luboš Motl Oct 16 '15 at 6:56

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