Okay, I have a very basic question about the SUSY gauged linear sigma model which is driving me crazy. I am following Chapter $15$ of Mirror Symmetry by Hori et al. I am considering the SUSY gauged linear sigma model with gauge group $U(1)$ and $N$ chiral multiplets with the following Lagrangian: $$ L=\sum_{j=1}^{N} \left[ -|D_{\mu}\phi_{j}|^{2}+i\bar{\psi}_{j-}(D_{0}+D_{1})\psi_{j-}+i\bar{\psi}_{j+}(D_{0}-D_{1})\psi_{j+} - |\sigma|^{2}|\phi_{j}|^{2} - \bar{\psi}_{j-}\sigma\psi_{j+}-\bar{\psi}_{j+}\bar{\sigma}\psi_{j-} + i (\bar{\phi}_{j}\lambda_{+}\psi_{j-}-\bar{\phi}_{j}\lambda_{-}\psi_{j+}-\bar{\psi}_{j-}\bar{\lambda}_{+}\phi_{j}+\bar{\psi}_{j+}\bar{\lambda}_{-}\phi_{j}) + \frac{1}{2e^{2}}\left( -|\partial_{\mu}\sigma|^{2}+i\bar{\lambda}_{-}(\partial_{0}+\partial_{1})\lambda_{-}+i\bar{\lambda}_{+}(\partial_{0}-\partial_{1})\lambda_{+}+v_{01}^{2} \right) +\theta v_{01} -\frac{e^{2}}{2}\left(\sum_{i=1}^{N}|\phi_{i}|^{2}-r\right)^{2} \right] $$ where the terms $(\phi_{j},\psi_{j\pm})$ form chiral supermultiplets and the terms $(\sigma,v_{\mu},\lambda_{\pm})$ form a vector supermultiplet.

For $r>0$, the scalar potential is clearly minimised by configurations: $$ M_{vac}:=\{\phi_{i}:\sum_{i=1}^{N}|\phi_{i}|^{2}=r\}/U(1)\cong\mathbb{C}P^{N-1} $$ My problem centres around trying to figure out the (perturbative), masses of the various fields in this theory. The book claims that the massive modes constitute a massive supermultiplet of mass $e\sqrt{2r}$, in particular $v_{\mu},\sigma$, and $\lambda_{\pm}$ are claimed to have this mass, as are transverse modes of $\phi_{i}$ and certain modes of the $\psi_{j}$.

Clearly to find these masses, we want to pick a classical vacuum and expand around it. We expect the potential term: $$ U= |\sigma|^{2}\sum_{i=1}^{N}|\phi_{i}|^{2} + \frac{e^{2}}{2}\left(\sum_{i=1}^{N}|\phi_{i}|^{2}-r\right)^{2} $$ to give mass to $\sigma$ and some modes of $\phi_{j}$. We expect the gauge kinetic term: $$ \sum_{j=1}^{N} |D_{\mu}\phi_{j}|^{2} $$ to give mass to the gauge field (via the Higgs mechanism), and the Yukawa-like terms: $$ L_{F}:= i (\bar{\phi}_{j}\lambda_{+}\psi_{j-}-\bar{\phi}_{j}\lambda_{-}\psi_{j+}-\bar{\psi}_{j-}\bar{\lambda}_{+}\phi_{j}+\bar{\psi}_{j+}\bar{\lambda}_{-}\phi_{j}) $$ to give masses to the appropriate fermions.

We pick the vacuum $\sigma=\phi_{1}=\ldots=\phi_{N-1}=0$, $\phi_{N}=\sqrt{r}$. We let $\phi_{N}=\phi_{N}'-\sqrt{r}$, and begin to expand the Lagrangian around this vacuum.

Already, we see: $$ U=|\sigma|^{2}(\phi_{N}'^{2}+\sqrt{r}\phi_{N}'+r)+\cdots=r|\sigma|^{2}+\cdots $$ Suggesting a mass of $\sqrt{r}$, not a mass of $e\sqrt{2r}$.

We now set $\phi_{N}=\rho e^{i\theta}$, i.e. switch to polar variables. Here our vaccuum is $\rho=\sqrt{r}$, $\theta=0$. I will discuss the gauge and fermion terms separately. Expanding $-|D_{\mu}\phi_{N}|^{2}-U$, and writing $\rho=\sqrt{r}+\epsilon$, we have: $$ -(\partial_{\mu}\epsilon)^{2}+rv_{\mu}'^{2}-r|\sigma|^{2}-2e^{2}r\epsilon^{2}+\cdots $$ where $\cdots$ represents interaction terms, and where we have defined a new gauge field $v_{\mu}'=v_{\mu}+\partial_{\mu}\theta$. This shows that $\epsilon$ has the claimed mass, but $v_{\mu}'$ has an incorect mass of $\sqrt{2r}$.

What happens to $L_{F}$ is even worse. In the only term involving $\phi_{N}$, The Goldstone boson $\theta$ still appears (rather than having been eaten entirely), to yield: $$ i\rho \left( e^{i\theta}(\bar{\psi}_{N+}\bar{\lambda}_{-}-\bar{\psi}_{N-}\bar{\lambda}_{+}) + e^{-i\theta}(\lambda_{+}\psi_{N-}-\lambda_{-}\psi_{N+}) \right) $$ I have no idea how to extract nice terms from this mess.

I know this should be basic stuff, but I have managed to thoroughly confuse myself, and any help would be much appreciated.


First, wouldn't hurt to specify spacetime dimension and number of supersymmetries. I guess $D=2$ ${\cal N}=(2,2)$.

  1. Notice that the kinetic terms of the vector multiplet have the factor $1/(2e^2)$. For canonical kinetic terms you need to multiply the vector multiplet components by $\sqrt{2}e$.

  2. You can absorb the Goldstone by a phase rotation of the chiral fermion. Or just set $\theta=0$ as a gauge choice.

  • $\begingroup$ Thanks, I knew it had to be something really simple! I also figured out that you can see this even with the funny kinetic terms by making a free-field approximation and finding the corresponding equations of motion. $\endgroup$
    – CoffeeCrow
    Apr 28 at 1:54

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.