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I'm studying this paper on supersymmetric ground state wavefunctions. In section 5 "quantum mechanical gauge theories", it says:

"We begin with the ${\cal N} = 2$ gauge theory which may be obtained by dimensionally reducing the ${\cal N} = 1$ gauge theory in 1 + 1 dimensions. We reduce the gauge field as $A^{\mu}_A = (A_A, \phi_A)$, and write the theory in 1 + 0 dimensions in terms of a complex fermion $\lambda_A$, where the index $A$ labels the generators in the gauge group. The resulting Lagrangian is \begin{align*} L &= \frac{1}{2}D\phi_A D\phi_A + i \bar{\lambda_A} D \lambda_A - igf_{ABC}\bar{\lambda_A}\phi_B \lambda_C \end{align*} where \begin{align*} D\phi_A &= \dot{\phi} - g f_{ABC}A_B\phi_C \\ D\lambda_A &= \dot{\phi} - g f_{ABC}A_B\lambda_C \end{align*} are the covariant time derivatives.

Quantisation leads to the algebra and Hamiltonian: \begin{align*} [\phi_A,\pi_A] &= i \delta_{AB} \\ \{\lambda_A, \bar{\lambda}_B\} &= \delta_{AB} \\ H &= \frac{1}{2} \pi_A \pi_A + igf_{ABC}\bar{\lambda}_A \phi_B \lambda_C. \end{align*} In the Hamiltonian formulation, the variable $A_A$ is eliminated and the generator of gauge transformations \begin{align*} G_A &= f_{ABC}(\phi_B \pi_C - i \bar{\lambda}_B \lambda_C) \end{align*} is constrained to vanish on physical states by Gauss' law.

The model is invariant under the supersymmetry transformations \begin{align*} \delta\phi_A &= - \delta A_A = \bar{\lambda}_A\xi + \bar{\xi}\lambda_A \\ \delta \lambda_A &= -i \xi D \phi_A \\ \delta \bar{\lambda}_A &= i \bar{\xi}D \phi_A." \end{align*}

  1. I don't follow the gauge reduction part, nor where the fermions are coming from in this process. Any clarifying comments would be appreciated.

  2. If I do a Legendre transformation on the given Lagrangian, the resulting Hamiltonian I get is the Hamiltonian they have plus a term $g A_A f_{ABC}[\phi_B \pi_C - i \bar{\lambda}_B \lambda_C] = g A_A G_A$. Why are they neglecting this extra term in the Hamiltonian? And where does the requirement that $G_A$ vanish on physical states come from? And how does one see it is the "generator of gauge transformations"?

  3. I'd like to confirm the Lagrangian is invariant under the given transformations. But when calculaitng $\delta L$, having to expand out all the various leibniz rules and covariant derivatives is quite cumbersome. Is there any better way to see how these are symmetries? Or how certain parts are obviously going to cancel?

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1. Gauge Reduction and Fermions

Dimensional Reduction involves reducing the number of spatial dimensions. When reducing to 1 + 0 dimensions, the spatial component of $A_\mu$ is reinterpreted as a scalar field, $\phi_A$. This leads to the gauge field in 1 + 0 dimensions being expressed as $A^\mu_A = (A_A, \phi_A)$, with $A_A$ being the temporal component.

Upon reduction, the fermions are reinterpreted in the lower-dimensional theory. In your specific case, the complex fermion $\lambda_A$ is introduced, which corresponds to the fermionic degrees of freedom in the reduced theory. The index $A$ labels these fermions in the context of the gauge group.

2. Hamiltonian and Gauge Transformations

When performing a Legendre transformation on the given Lagrangian to obtain the Hamiltonian, the term $g A_A G_A$ appears due to the transformation of the gauge field $A_A$. In many gauge theories, certain components of the gauge field do not have corresponding kinetic terms (they are non-dynamical), and as such, they do not contribute independent degrees of freedom. These components are often eliminated using constraint equations, such as Gauss' law in electromagnetism. The term you identified, $g A_A G_A$, can be related to the constraints that are imposed on the system. In the Hamiltonian formulation of gauge theories, some variables (like $A_A$ in your case) can be eliminated in favor of others using constraint equations.

Gauss' law in gauge theories serves as a constraint that physical states must satisfy. It originates from the gauge invariance of the theory and essentially ensures that the physical states are gauge invariant. In your case, $G_A$ is identified as the generator of gauge transformations, which means it generates infinitesimal changes in the gauge field that leave the physical content of the theory unchanged. The requirement that $G_A$ vanishes on physical states ensures that these states are invariant under gauge transformations.

3. Lagrangian Invariance

You can try to look for terms in the Lagrangian that transform into total derivatives under the given transformations, as these do not affect the action. But overall it is likely that ugly calculations will be required.

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  • $\begingroup$ Thanks, but 1. where are the fermionic degrees of freedom coming from? And 2. how is this particular constraint deduced in this example? And how do you see it generates the gauge transformations? $\endgroup$
    – Gleeson
    Dec 2, 2023 at 12:44
  • $\begingroup$ @Gleeson In supersymmetric theories, fermionic and bosonic degrees of freedom are paired, reflecting the symmetry between fermions (particles with half-integer spin) and bosons (particles with integer spin). When you dimensionally reduce a supersymmetric gauge theory, you effectively compactify one or more dimensions. In doing so, certain components of the higher-dimensional fields become independent fields in the lower-dimensional theory. $\endgroup$
    – MrDBrane
    Dec 8, 2023 at 1:46
  • $\begingroup$ For the case of reducing a N=1 gauge theory in 1+1 dimensions to 1+0 d: The gauge field A in 1+1 dimensions has a spatial component that, upon reduction, is reinterpreted as a scalar field. This scalar field is a bosonic degree of freedom in the reduced 1+0 dimension theory. $\endgroup$
    – MrDBrane
    Dec 8, 2023 at 1:47
  • $\begingroup$ Correspondingly, the fermionic partners of these gauge fields in the higher-dimensional theory (which may be part of a supermultiplet in the supersymmetric framework) become the fermionic degrees of freedom $\lambda_A $ in the reduced theory. This maintains the supersymmetry, as each bosonic degree of freedom, has a fermionic counterpart $ \lambda_A $ $\endgroup$
    – MrDBrane
    Dec 8, 2023 at 1:48

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