I am struggling to understand Polchinski’s derivation (https://arxiv.org/abs/hep-th/9210046) of the conditions for marginality of the 4-fermi operator.

For a scattering process $(\mathbf{p}_1,\mathbf{p}_2) \to (\mathbf{p}_3,\mathbf{p}_4)$, the delta function Imposing momentum conservation can be written as

$\delta^d(\mathbf{p}_1 + \mathbf{p}_2 - \mathbf{p}_3 - \mathbf{p}_4) = \delta^d(\delta \mathbf{k}_3 + \delta \mathbf{k}_4 + \delta \mathbf{l}_3+\delta \mathbf{l}_4)$

where we have expressed the momenta as $\mathbf{p} = \mathbf{k} + \mathbf{l}$ where $\mathbf{k}$ is the orthogonal projection of $\mathbf{p}$ to the Fermi surface (unique assuming convex Fermi surface) and $\mathbf{l}$ is orthogonal to the Fermi surface.

For an inversion-symmetric Fermi surface, it is clear that the momentum changes $\delta\mathbf{k}_3$ and $\delta \mathbf{k}_4$ are parallel to each other for $\mathbf{p}_1 = \pm \mathbf{p}_2$. Polchinski seems to be claiming on page 19 that for these initial momentum configurations, in the scaling limit $\mathbf{l} \to s\mathbf{l}$, the delta function scales as $s^{-1}$. I do not follow his logic here. He claims that one component of $\delta\mathbf{k}_3+\delta\mathbf{k}_4$ vanishes, but in which coordinate system? Is there a coordinate-free explanation for this phenomenon?

I am even more baffled by this statement in footnote 7 on the bottom of page 20:

“One would seem to find the same enhancement for $\mathbf{p}_1= +\mathbf{p}_2$ In that case, however, the delta-function is degenerate only at one point on the Fermi surface, so that second order terms in $\delta \mathbf{k}$ are nonzero and the enhancement is only by $s^{-1/2}$.”

What second-order terms is he referring to?

Edit: Here is an attempt to rationalize the first statement. Clearly $\mathbf{p}=\mathbf{p}'$ implies $\mathbf{k}=\mathbf{k}'$ and $\mathbf{l}=\mathbf{l}'$ assuming convex Fermi surface. For an inversion-symmetric Fermi surface we further have that $\mathbf{p}=-\mathbf{p}'$ implies $\mathbf{k}=-\mathbf{k}'$ and $\mathbf{l}=-\mathbf{l}'$.

Now consider the following subsets of phase space

$$S = \{ (\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3,\mathbf{p}_4) \in \mathbb{R}^{4d} : \mathbf{p}_1 = -\mathbf{p}_2 \} $$ $$ S' = \{ (\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3,\mathbf{p}_4) \in \mathbb{R}^{4d} : \mathbf{p}_1 = -\mathbf{p}_2, \; \mathbf{k}_3 = -\mathbf{k}_4 \} $$ $$ S'' = \{ (\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3,\mathbf{p}_4) \in \mathbb{R}^{4d} : \mathbf{p}_1 = -\mathbf{p}_2, \; \mathbf{p}_3 = -\mathbf{p}_4 \} $$

By inversion symmetry $\mathbf{p}_3 = -\mathbf{p}_4 \implies \mathbf{k}_3 = -\mathbf{k}_4, \; \mathbf{l}_3 = -\mathbf{l}_4$ and thus $S'' \subseteq S' \subseteq S$.

Configurations in $S''$ conserve momentum so the delta function is divergent there. For a generic $(\mathbf{p}_1,\mathbf{p}_2,\mathbf{p}_3,\mathbf{p}_4) \in S'$, however, we have

$$ \delta\mathbf{k}_3 = \mathbf{k}_3 - \mathbf{k}_1 = -\mathbf{k}_4 + \mathbf{k}_2 = -\delta\mathbf{k}_4 $$ and thus $$\delta^d(\mathbf{p}_1 + \mathbf{p}_2 - \mathbf{p}_3 - \mathbf{p}_4) = \delta^d(\delta \mathbf{l}_3 + \delta \mathbf{l}_4) $$ so we recover the desired scaling of the delta function.

So it appears what is left to understand is the case $\mathbf{p}_1=\mathbf{p}_2$ in footnote 7 on page 20.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.