# Connection of Sugawara construction to the regular energy-momentum tensor

Update:

As pointed out by @ConnorBehan this problem is related to the rearrangement lemma in the 'Yellow Book'. In fact this problem is already mentioned page 649 in the book in the discussion about free fermion representation of WZW model.

In that specific derivation however, I still have trouble understanding the steps. In particular there is a normal-ordered term $$$$(\psi_i ( \psi_j (\psi_k \psi_l)))$$$$ which I couldn't figure out how to get rid of.

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I am reading about bosonization and specifically constructing the energy-momentum tensor from currents using Sugawara constructions, and this question occurs:

Suppose we are looking at a free fermion system with one flavor. We then have a left-moving component $$\psi_R$$ and a right-moving one $$\psi_L$$. The Sugawara construction demands that the holomorphic energy-momentum tensor T(z) given by $$$$\label{eq:1} T(z) \propto \; :J(z)J(z):$$$$ with $$$$J(z) =\; :\psi_L^\dagger \psi_L:$$$$ However, this is a fermion bilinear and when substituting into the Sugawara form we get a term containing four fermions, which is not manifestly equivalent to the free fermion energy-momentum tensor, being proportional to: $$$$\psi_L^\dagger i\partial \psi_L^\dagger$$$$ I am wondering how one could resolve this puzzle. My current thought is that it has something to do with the anomalous commutator and what one should do is to do a point splitting, for example $$$$: \psi_L^\dagger (x+\epsilon) \psi_L^\dagger (x-\epsilon) \psi_L^\dagger (x+\epsilon') \psi_L(x-\epsilon') :$$$$ and expand out the normal ordering, and carefully take the limit of $$\epsilon, \epsilon' \rightarrow 0$$. But I do not know how to go further from here. Any clues and suggestions would be helpful.

• Point splitting is exactly right. You need to show that $((\psi\psi)(\psi\psi))$ rearranges to $(\psi \partial \psi)$. This is exercise 6.9 in the yellow book. See also physics.stackexchange.com/q/562566 which is for bosons. Nov 23 '21 at 0:43
• @ConnorBehan Thanks for your comment. I tried the method you mentioned and reproduced the derivative terms. However I still got the term $(\psi^\dagger (\psi (\psi^\dagger\psi)))$ and I don't know how to make them disappear. Nov 28 '21 at 0:43
• That term is just zero because $Re(\psi)$ and $Im(\psi)$ at the same point anti-commute right? Nov 28 '21 at 11:02
• I can see why $Re(\psi)$ and $Im(\psi)$ anti-commute once we view them as Majorana fermions. However I don’t see how this renders the term zero :( Can you be a little more specific? Nov 28 '21 at 17:08
• I think it's enough to use the formula $(\psi (\psi^\dagger \psi))(0) \left | 0 \right > = \psi_{-1/2} \psi^\dagger_{-1/2} \psi_{-1/2} \left | 0 \right >$ which you can show to be minus itself by moving the modes around. Dec 1 '21 at 12:21