Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 307 (Supersymmetry)

Complex conjugation reverses the order of the Grassmann variables. Thus the conjugate of $i\bar{\psi }\dot{\psi }$ differs from itself only by integration by parts, while the remaining terms are manifestly real. The action is therefore real up to a boundary term.

Substituting the stated transformations into the Lagrangian, using anticommutation of $\psi ,\bar{\psi },ϵ,\overline{ϵ}$, and integrating the terms containing $\ddot{x}$ and $\dot{\psi }$ by parts leaves a total derivative. A convenient convention for the resulting Noether charges isOverall signs can be moved between the charges and the Grassmann transformation parameters. These charges generate the displayed transformations and obey the classical supersymmetry algebra.

Canonical quantization giveswith all other elementary graded commutators zero. Represent $p=-id/dx$, let $\psi$ act by exterior multiplication by $dx$, and let $\bar{\psi }$ act by contraction with ${\partial }_x$. The Hilbert space is thenthe square-integrable complex differential forms on the line. Up to an inessential factor of $-i$, $Q$ is the twisted de Rham differentialand $\bar{Q}$ is its Hilbert-space adjoint. The Hamiltonian is $H=\{Q,\bar{Q}\}/2$, so a zero-energy state must be annihilated by both charges.

On zero-forms the zero-mode equation is $(d/dx+h^{\prime })u=0$, giving $u=C{e}^{-h}$. On one-forms it is $(-d/dx+h^{\prime })v=0$, giving $v=C{e}^h$. For $h=-x^4$, only the one-form is square integrable, so the unique ground state isFor a generic cubic polynomial, $h(x)$ tends to opposite infinities at the two ends of the real line. Each of $e^h$ and $e^{-h}$ therefore diverges at one end, so neither candidate is square integrable. There is consequently no normalizable zero-energy state.

Let $\Lambda$ be a chiral gauge parameter. One consistent Abelian supergauge transformation convention isfor which ${\bar{\Phi }}_i{e}^{q_{i}V}{\Phi }_i$ is invariant. Wess-Zumino gauge uses the nonordinary components of $\Lambda$ to remove the superfluous scalar and spinor components of $V$, leaving the photon, gauginos, complex scalar, and real auxiliary field of the two-dimensional vector multiplet. Ordinary $U(1)$ gauge transformations remain.

Define the field-strength multiplet, up to conventional normalization, byGauge invariance follows because chirality, antichirality, ${\bar{D}}_{+}^2=D_{-}^2=0$, and $\{{\bar{D}}_{+},D_{-}\}=0$ annihilate the variation of $V$. The same identities giveso $\Sigma$ is a twisted chiral superfield. The Fayet–Iliopoulos term is a twisted F-term. In the standard axial convention its measure $d{\theta }^{+}d{\bar{\theta }}^{-}$ has axial charge $-2$, so invariance requiresReversing all axial-charge conventions reverses both signs but leaves this statement unchanged: the field and measure have opposite charges.

The charged matter fermions are chiral with respect to the axial symmetry. In a background with gauge flux, their functional measure has the two-dimensional axial anomalyup to orientation and current normalization. Equivalently, a Fujikawa transformation multiplies the torus path integral by a phase proportional to $\alpha ({\sum }_i{q}_i)\int F$. Since the background flux may be nonzero, the continuous axial $U(1)$ survives quantum mechanically precisely when

Put $\theta ={\theta }^{+}$, $\bar{\theta }={\bar{\theta }}^{+}$ and $y^{+}=x^{+}+i\theta \bar{\theta }$. The general two-dimensional N=(0,2) supersymmetry chiral superfield isIn ordinary coordinates this is $\Phi =\varphi +\sqrt{2}\theta {\psi }_{+}+i\theta \bar{\theta }{\partial }_{+}\varphi$. A supersymmetric kinetic action iswhose component form, up to light-cone conventions and total derivatives, is

A Fermi superfield satisfying ${\bar{D}}_{+}{\Lambda }_{-}=f(\Phi )$ has expansionHere $G$ is a complex bosonic auxiliary field. With the normalization in the question, the component action isup to equivalent sign conventions for the fermions.

The chiral integral in $S_3$ is supersymmetric only if its integrand is chiral. Applying ${\bar{D}}_{+}$ gives the necessary and sufficient conditionwith every $J^a$ holomorphic and with gauge charges chosen so that each product ${\Lambda }_{-a}{J}^a$ is gauge invariant. Its component expansion couples $G_a$ linearly to $J^a$ and supplies the corresponding Yukawa term ${\lambda }_{-a}{\psi }_{+}^i{\partial }_i{J}^a$.

Eliminating each $G_a$ by its algebraic field equation gives the nonnegative scalar potential. In the normalization displayed above and in the question,If the conventional definitions ${\bar{D}}_{+}{\Lambda }_{-}^a=\sqrt{2}{E}_a$ and $S_J=-\int d\theta {\Lambda }_{-a}{J}^a/\sqrt{2}+h.c.$ are used instead, the same result is written $V={\sum }_a(|E_a{|}^2+|J^a{|}^2)$; the two forms differ only by the normalization of $f$ and $J$.