# Superpartner for the stress-energy tensor

I would like to understand what is meant when one introduces a generator $G(z)$ as the superpartner of the energy-momentum tensor $T(z)$.

• How does one decide that this $G(z)$ should have a "conformal weight" of $\frac{3}{2}$ ?

• How are the following OPEs for it derived ?

$T(z)G(w) = \frac{\frac{3}{2}}{(z-w)^2}G(w) + \frac{ \partial_w G(w) } {z-w} + ...$

$G(z)G(w) = \frac{ \frac{2c}{3} }{(z-w)^3} + \frac{2T(w)}{z-w} + ...$

• If one wants to here create a $\cal{N}=2$ superconformal algebra then one apparently needs to introduce two such conformal weight $\frac{3}{2}$ supercurrents say $G^1(z)$ and $G^2(z)$ (with OPEs as above) and another "$U(1)$ current" $J(z)$ such that,

$G^1(z) G^2(z) = \frac{ \frac{2c}{3}}{(z-w)^3} + \frac{2T(w)}{(z-w)} + i( \frac{2J(w)}{(z-w)^2} + \frac{\partial _w J(w)} {(z-w)} ) +...$

$T(z)J(w) = \frac{J(w)}{(z-w)^2} + \frac{ \partial _w J(w)} {(z-w)} + ...$

$J(z)G^{1/2}(w) = \pm \frac{ iG^{2/1}(w)}{(z-w)} + ...$

$J(z) J(w) = \frac{ \frac{c}{3} }{ (z-w)^3 } + ...$

I would be grateful if someone can give some explanations about how the above construction is made.

Especially regarding the need and motivation to introduce the field $J$ (..and its OPEs..)

-

I don't know much about supersymmetry but in absent of any other answers, maybe you will benefit infinitesimally from my guesses.

Lets think in terms of a non-interacting SUSY theory with a bosonic and a fermionic field $S \sim \int\text d^2z\left(\partial X\bar{\partial}X - \left(\psi\bar{\partial}\psi + \bar{\psi}\partial\bar{\psi}\right)\right)$. Then from space-time symmetries we have a conserved energy-momentum tensor $T(z) = T_{boson}(z) + T_{fermion}(z)$ (no coupling, since they don't interact), and similarity for the anti-holormorphic part. It is then very natural to guess that your $G(z)$ is nothing but the conserved charge coming from supersymmetry (which could be to derive from Noether's theorem). Now, we know that we have a primary field $\psi(z)$ (the fermion) with $h_{\psi}=\frac 12$ and $j(z)=i\partial X(z)$ (the boson) with $h_j=1$. It is natural to guess that $G(z)$ is a combination of $\psi(z)$ and $j(z)$, which potentially could make it a primary field with conformal dimension $h_G = h_{\psi} + h_j = \frac 32$. The $T(z)G(w)$ OPE is nothing but the statement that $G(z)$ is an primary field with $h_G=\frac 32$. The $G(z)G(w)$ OPE should then be derived by using the OPE's of $\psi(z)$ and $j(z)$ together with Wicks theorem, as usual.

For the $\mathcal N = 2$ case, I have no idea. If you are less lazy than me and actually carry out this calculation, please write a comment on whether it works or not.

WARNING: the content of this answer is not thought through and could be completely wrong.

EDIT: Regarding why $G(z)$ is called a superpartner to $T(z)$. I vaguely remember that one usually do supersymmetry in terms of superspaces and superfields. My guess would be that one can construct a super (field) energy-momentum tensor, probably of the form $\sim G(z) + \theta T(z)$, and then they are "superpartners" in this sort of way.

EDIT2: @Arnirbit, in my answer I assumed knowledge of some basic aspects of non-supersymmetric 2d CFT. In particular the free boson and fermion (I must declare myself as newcomer to CFT and therefore not an expert).

Now to your questions. A primary field is a field which transforms in a particularly "nice" way under conformal transformations $(z,\bar z)\rightarrow (f(z), \bar f(\bar z))$,

$$\phi(z,\bar z) \rightarrow \left(\frac{\partial f}{\partial z}\right)^h\left(\frac{\partial \bar f}{\partial \bar z}\right)^{\bar h} \phi\left(f(z),\bar f(z)\right),$$ where $\phi(z,\bar z)$ is a primary field with conformal weight $(h,\bar h)$. There are a connection between transformation properties of fields and OPE's, one can for example show that a primary field of weight $(h,\bar h)$ have the following OPE with the energy-momentum tensor $$T(z)\phi(w,\bar w) = \frac h{(z-w)^2}\phi(w,\bar w) + \frac 1{z-w}\partial_w\phi(w,\bar w) + \text{regular terms},$$ and similar for the anti-holormorphic part (one way to derive this is by using Ward-identities). This equality must of course be interpreted as valid under a time-ordered correlation function. Another property of primary fields is that their two-point function is by symmetry fixed to be $$\left<\phi(z,\bar z)\phi(w,\bar w)\right> = \frac C{(z-w)^{2h}(\bar z-\bar w)^{2\bar h}}.$$

Now, lets concentrate on the free Bosonic theory $$S[X] = T\int\text d^2z\;\partial X(z,\bar z)\bar{\partial}X(z,\bar z).$$ The equations of motion is $\partial\bar{\partial}X(z,\bar z) = 0$, meaning that $j(z)\equiv i\partial X(z,\bar z)$ is holormorphic (depends only on $z$, not $\bar z$) and similarly for $\bar j(\bar z)\equiv i\bar{\partial}X(z,\bar z)$. One can calculate the two-point function $\left<X(z,\bar z)X(w, \bar w)\right> \propto \log\left|z-w\right|^2$. This does not have the above form, and therefore $X(z,\bar z)$ is not a primary field! However $$\left<j(z)j(w)\right> = -\left<\partial_zX(z,\bar z)\partial_wX(w,\bar w)\right>\propto -\partial_z\partial_w\log\left|z-w\right|^2 = \frac 1{(z-w)^2}.$$ This says that $j(z)$ is a (chiral) primary field with conformal dimension $(1,0)$, with the OPE $j(z)j(w) = \frac 1{(z-w)^2} + \dots$ (if you like fancy words, this OPE gives you a $\mathfrak{u}(1)$ affine Kac-Moody algebra). You can calculate the energy-momentum tensor which will be of the form $T(z) = -\gamma :\partial X\partial X: = \gamma :jj:(z)$, for some appropriate constant $\gamma$ and normal ordering $:\; :$ is necessary on the quantum level. Using Wicks theorem one finds $$T(z)j(w) = \frac 1{(z-w)^2}j(w) + \frac 1{z-w}\partial_wj(w) + \text{regular terms},$$ which again says that $j(z)$ has conformal dimensions $(h,\bar h) = (1,0)$. A similar analysis for the free fermion will tell you that $\psi$ is a primary field with $(h,\bar h) = (\frac 12, 0)$ and give you the OPE $\psi(z)\psi(w) = \frac 1{z-w} + \dots$ (and similarly for $\bar{\psi}$).

If you are not already familiar with all this, then my quick sketch will not be sufficient and I recommend you to consult a textbook. Finally, I think that when you combine the free bosonic and free fermionic theory in a supersymmetric way, then you can find out why $G(z)$ with conformal weight $h = \frac 32$ appears.

-
Thanks for your reply. Can you elaborate as to what you are defining as "primary field" of which you consider $\psi(z)$ and $j(z)=i\partial X(z)$ to be examples? (..i can hardly connect between the various things that I see being called as "primary" fields!..) Also kindly explain as to what OPEs you have in mind among the $\psi$s and $j$s. –  user6818 Jun 20 '11 at 10:54