# Tag Info

0

Perhaps the most enlightening is just to show how it goes in OP's example. If the Lagrangian reads $${\cal L}_1(A,\phi)~:= ~{\cal F}(A)- Ay(\phi),\qquad F~=~{\cal F}^{\prime}(A),\tag{1}$$ then the eom for the "auxiliary" variable $A$ reads $$F(A)~\approx~ y(\phi) \qquad \Leftrightarrow\qquad A~\approx~ F^{-1}(y(\phi)),\tag{2}$$ where we have assume that ...

0

Suppose $\partial_\alpha$ is a derivative w.r.t. Grassmannian coordinate $\theta^\alpha$. We often define a super-derivative $${\cal D}_\alpha = \partial_\alpha + \cdots$$ where the $\cdots$ is chosen such that $$\{ {\cal D}_\alpha , {\cal D}_\beta \} \sim \gamma^\mu_{\alpha\beta} \partial_\mu$$ Now, let's do some dimensional analysis, since ...

0

Below is a summary of my very limited understanding of what the elliptic genus is. I'll first give you the mathematical definition, followed by an explanation of how it appears naturally in physics. As a first exposure to the subject, it's perhaps best to not consider the most general definitions. In what follows, I will only explain what the elliptic genus ...

1

The bracket you have written is of the form $$[\delta,\delta] A_\mu = v^\nu \partial_\nu A_\mu + \partial_\mu (v^\nu A_\nu).$$ As you pointed out, the first term corresponds to a translation by $v^\mu$. The second term corresponds to a gauge transformation $\delta A_\mu = \partial_\mu \lambda$ with $\lambda = v^\nu A_\nu$. So the algebra closes up to a ...

3

There are obviously differeing genus types according to which partition function in d-dimensional QFT. At the very outset of $0$-d QFT the index is the push forward in ordinary de Rahm Cohomology; in other words, the integration of differential forms. The genus as you put it, is non-existant here. In $1$-d QFT the index of the Dirac Operator is ...

1

Not only must supersymmetry be broken, it must be broken in a way that doesn't lead to phenomenological disasters. In the MSSM, $D$-term breaking with a Fayet-Iliopoulos term achieves neither: it doesn't break supersymmetry but it does lead to phenomenological disasters! Consider again your potential \begin{align} V&=\sum_i |m_i|^2 |\phi_i|^2 +1/2 ... 2 The variation \delta F for any field (or degree of freedom) F, given an infinitesimal transformation, is always calculated as the commutator \delta F = [ \bar\epsilon Q, F ]  where $\bar \epsilon$ is a parameter ("angle" or "shift" or some generalization) of the transformation and $Q$ is the generator. (Those may be replaced by other letters.) ...

0

One way to think about this is the following. In general, the partition function (which is the integrand of the vacuum amplitude and not the vacuum amplitude itself) will be of the form $Z_\psi^{\pm}\propto\sum_{a,b}C[^a_b]Z^a_b(\tau)$ where $a$ and $b$ sum over the different sectors as given in the text and the $C$s are some phases. Many of these phases ...

Top 50 recent answers are included