# How do I show that there is no supersymmetry breaking in two-dimensional $U(1)$ gauge theory

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My question is regarding supersymmetric $U(1)$ gauge theory in two dimensions. In particular, I am interested in the supersymmetry which is preserved by Witten's topological A-twisting. It seems to me that quantum corrections result in an anomaly that breaks this supersymmetry. However, I doubt that this is accurate because I have not seen this in the literature, and am trying to show that there is in fact no anomaly.

On a worldsheet parametrized by $x^1$ and $x^2$, the action is \begin{aligned} S=\frac{1}{2\pi}{1 \over 2{e}^2}\int d^2x\Big(( {F}_{12})^2 +\partial_{\mu} {\sigma} \partial^{\mu} \overline{ {\sigma}} -( {D} )^2 - {i } \overline{ {\lambda}}_{+} ({\partial_-}) {\lambda}_{+} -{i } \overline{ {\lambda}}_{-} ({\partial_+}) {\lambda}_{-} \Big), \end{aligned} (where $\partial_{\pm}=i\partial_2\pm \partial_1$, and $d^2x=dx^1dx^2$ with $x^2$ understood to be the Euclidean time direction) and it is invariant under the following supersymmetry transformations, generated by the supercharge $Q_A$ \begin{align} \delta_{ {Q}_A}& {A}_{1} \ =\ \frac{i\epsilon( \overline{ {\lambda}}_{+}-{ {\lambda}}_{-})}{2} & \delta_{ {Q}_A}& {\lambda}_{-} \ =\ \, \epsilon\partial_- {\sigma} \nonumber\\ \delta_{ {Q}_A}& {A}_{2 } \ =\ \frac{\epsilon( \overline{ {\lambda}}_{+}+{ {\lambda}}_{-})}{2} & \delta_{ {Q}_A}& \overline{ {\lambda}}_{+} \ =\ \, \epsilon\partial_{+} {\sigma} \nonumber \\ \delta_{ {Q}_A}& {\sigma} \ = \ 0 & \delta_{ {Q}_A}& \overline{ {\lambda}}_{-} \ = \ -i\epsilon( {F}_{12}+ {D} ) \nonumber \\ \delta_{ {Q}_A}& \overline{ {\sigma}} \ = \ - i\epsilon (\overline{ {\lambda}}_{-} + {\lambda}_{+}) & \delta_{ {Q}_A}& {\lambda}_{+} \ = \ i\epsilon( {F}_{12}+ {D} )\nonumber \\ \delta_{ {Q}_A}& {D} =\frac{1}{2}\epsilon(\partial_-\overline{ {\lambda}}_{+}-\partial_+{ {\lambda}}_{-}). \end{align} We need to fix a gauge in order to quantize the theory, so we shall choose the Lorentz gauge $$\langle \psi '| \partial_{\mu} {A} ^{\mu}|\psi \rangle=0$$ (where $|\psi \rangle$ and $|\psi' \rangle$ are physical states) by including the following BRST gauge fixing action $$\label{gfaction} S_{BRST}=\frac{1}{2\pi}\frac{1}{2 {e} ^2}\int d^2x(-iB \partial_{\mu} {A}^{\mu} -(B )^2+\partial_{\mu} {b} \partial^{\mu} {c} -\frac{i\partial_2 {b} }{2}({ {\lambda}}_{-}+\overline{ {\lambda}}_{+})-\frac{\partial_1 {b} }{2}({ {\lambda}}_{-}-\overline{ {\lambda}}_{+})),$$ where $b$ and $c$ are fermionic ghost fields, while $B$ is an auxiliary bosonic field. Now, both $S$ and $S_{BRST}$ are invariant under the corrected supersymmetry transformations generated by $\hat{Q}_A=Q_A+ Q_{BRST}$: \begin{align} \delta_{\hat{Q}_A}& {A}_{1} \ =\ \frac{i\epsilon( \overline{ {\lambda}}_{+}-{ {\lambda}}_{-})}{2} +i\epsilon\partial_{1} {c} & \delta_{\hat{Q}_A}& {\lambda}_{-} \ =\ \, \epsilon\partial_- {\sigma} \nonumber\\ \delta_{\hat{Q}_A}& {A}_{2 } \ =\ \frac{\epsilon( \overline{ {\lambda}}_{+}+{ {\lambda}}_{-})}{2} +i\epsilon\partial_{2} {c} & \delta_{\hat{Q}_A}& \overline{ {\lambda}}_{+} \ =\ \, \epsilon\partial_{+} {\sigma} \nonumber \\ \delta_{\hat{Q}_A}& {\sigma} \ = \ 0 & \delta_{\hat{Q}_A}& \overline{ {\lambda}}_{-} \ = \ -i\epsilon( {F}_{12}+ {D} ) \nonumber \\ \delta_{\hat{Q}_A}& \overline{ {\sigma}} \ = \ - i\epsilon (\overline{ {\lambda}}_{-} + {\lambda}_{+}) & \delta_{\hat{Q}_A}& {\lambda}_{+} \ = \ i\epsilon( {F}_{12}+ {D} )\nonumber \\ \delta_{\hat{Q}_A}& {D} =\frac{1}{2}\epsilon(\partial_-\overline{ {\lambda}}_{+}-\partial_+{ {\lambda}}_{-}) \end{align} \begin{aligned} \delta_{\hat{Q}_A} b &=\epsilon B \\ \delta_{\hat{Q}_A} c &=-\epsilon {\sigma} \\ \delta_{\hat{Q}_A} B &=0. \end{aligned} In particular, $\delta_{\hat{Q}_A}^2=0$ on all fields. The Noether charge $\hat{Q}_A$ is computed to be \begin{aligned} \hat{Q}_A=&\frac{1}{2\pi}\int dx^1\Big(- \frac{i}{{e} ^2 } {F}_{12}\big(\frac{(\overline{ {\lambda}}_{+}-{ {\lambda}}_{-})}{2} +\partial_1 {c}\big) \\& - \frac{1}{2 {e} ^2}(\overline{ {\lambda}}_{-}+{ {\lambda}}_{+})i\partial_2 {\sigma} + \frac{1}{2 {e} ^2}(\overline{ {\lambda}}_{-}-{ {\lambda}}_{+c})\partial_1 {\sigma} %-\frac{i}{2}\sum_j^N\sum ^k\sum_d^k {Q}_{jc}v^j_d%\partial_1\tic (\theta^d-\overline{\theta}^d) \\& +\frac{1}{2 {e} ^2}B\big(\frac{-i(\overline{ {\lambda}}_{+}+{ {\lambda}}_{-})}{2}+\partial_2 {c} \big)+\frac{1}{2 {e} ^2}\partial_2b {\sigma} \Big) \end{aligned}

Then, using the nonzero canonical commutation relations \begin{aligned} \lbrack {\sigma} (x^1), \frac{1}{2 {e}^2}\partial_2\overline{ {\sigma}}(y^1) \rbrack&= 2\pi \delta(x^1-y^1)\\ \{ {\lambda}_{\pm}(x^1),\frac{1}{2 {e}^2} \overline{ {\lambda}}_{\pm }(y^1)\}&=2\pi \delta(x^1-y^1)\\ \lbrack {A}_{1}(x^1),-\Big(\frac{1}{ {e} ^2} {F}_{12}(y^1) \Big)\rbrack&=2\pi \delta(x^1-y^1)\\ %\lbrack {A}_{1c}(x^1),-\frac{1}{2 % {e}^2}\big(2 {F}_{12d}(y^1)-i\sum_j^N \sum_e^k % {Q}_{jd} v^j_e \textrm{Im}(\theta^e(y^1))\big)%\rbrack&=2\pi \delta(x^1-y^1)\\ \lbrack {A}_{2}(x^1),\frac{1}{2 {e}^2}(-iB(y^1))\rbrack&=2\pi \delta(x^1-y^1)\\ \{ {b} (x^1),\frac{1}{2 {e}^2}\big(\partial_2 {c}(y^1)\big)\}&=2\pi \delta(x^1-y^1)\\ \{ {c} (x^1),\frac{1}{2 {e}^2}\big(\partial_2 {b}(y^1)\big)\}&=2\pi \delta(x^1-y^1),\\ \end{aligned} we find \label{anomaly} \begin{aligned} \hat{Q}_A^2=\frac{1}{2}\{\hat{Q}_A,\hat{Q}_A\} %=&\frac{1}{2}\int dx^1\int dy^1\{\hat{J}^2(x^1),\hat{J}%^2(y^1)\} =&\frac{1}{2\pi}\int dx^1 \Big( \frac{i}{ {e} ^2 } {F}_{12}\partial_1 {\sigma} -\frac{i}{ {e} ^2 } {F}_{12}\partial_1 {\sigma} \\& - \big(\frac{1}{2 {e} ^2}\big)\partial_2 {\sigma} B \Big) \\ =&\frac{1}{2\pi}\int dx^1 \Big( - \big(\frac{1}{2 {e} ^2}\big)\partial_2 {\sigma} B \Big) \end{aligned} Now, if $\hat{Q}^2_A$ is nonzero, there can be no ground states annihilated by $\hat{Q}_A$, and therefore supersymmetry is broken. I suspect that this is not true in the case at hand, but there is still a term proportional to $B$ present. This term is actually the generator of time-dependent gauge symmetries, i.e., it generates the gauge transformations of $A_2$. Why should it annihilate physical states?

This post imported from StackExchange Physics at 2017-02-26 23:00 (UTC), posted by SE-user Mtheorist

asked Jan 5, 2017
edited Feb 26, 2017
Does your SUSY algebra close on-shell or off-shell? If it only closes on-shell, maybe you get to use the equations of motion in your last expression, e.g. B=0

This post imported from StackExchange Physics at 2017-02-26 23:00 (UTC), posted by SE-user user2309840
It closes off-shell, since the auxiliary field $D$ is present. However, since $B$ is proportional to $\partial_{\mu}A^{\mu}$ via equations of motion, and since the latter vanishes via Lorentz gauge, maybe we still can have $B=0$.

This post imported from StackExchange Physics at 2017-02-26 23:00 (UTC), posted by SE-user Mtheorist

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