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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.

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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.

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  • $\begingroup$ 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

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