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5

It can! The question isn't very specific, so I'll only answer broadly. Of course it all depends on what you mean by high and low energies, but many naive supersymmetric models you might write down will affect low-energy physics. If your model predicts that the proton will decay quickly, or that a new particle will be created if you collide two electrons ...

1

Thank you for your answer. I fortunately already found the mistakes in my computation: From the second to the third line I simply misinterpreted the identities for the $\theta$-functions. I forgot the $\sqrt{2}$ contribution, which would've also lead to the identity you used. As a remark: Regarding 1.: My central charges in the computation are chosen ...

2

First, I'm going to work out the calculation to help you be clear about getting the desired result and how similar computations are performed in the literature. The character of the evolution operator for a free fermion in the Ramond sector on the torus is: $$Tr_{R}[q^{L_{0}-c/24}].$$ Now recall that the central charge of the free fermion CFT is $c=1/2$ and ...

1

The Witten index in a supersymmetric theory counts the number of bosonic ground states minus the number of fermionic ground states. The reason for this is that the trace sums over all states and for each excited bosonic state, we also get an excited fermionic state of the same energy, due to supersymmetry, which results in a cancellation. The reason for $\... 0 I will give an answer to this old question since I just saw this question referenced in Reddit, and since Qmechanic's very comprehensive answer above did not really answer OP's question. I understand for some energies for E<0 that T=1. I also understand resonant scattering means at some values of E when E>0 that T=1. I do not understand how "... 0 Firstly, the final term that you have written does not have Lorentz indices so the expression is incorrect. Assuming that what you meant is $$D_{\mu \nu}(k) = \frac{-i g_{\mu \nu}}{k^2 - M^2_V},$$ this is still incorrect. An easy way of seeing this is, recognizing that computing a loop with such as $$\int d^4k \frac{(-i)^2 g_{\mu \nu} g^{\mu \nu}}{(k^2 -M^... 2 Properties of R-symmetry group depends on spinor structure: Here M means Maiorana spinors, MW - Maiorana-Weyl, S- symplectic. Spinor structure depends on dimension and signature of space. For more details one can consult Tools for supersymmetry. Put supersymmetric theory on curved space is not simple task. To do such procedure, one must find generalized ... 1 I think yes, it is a correct way to think about it. R-symmetry is a global symmetry of the the theory. For the N=1 it is typically U(1) symmetry, which acts by multiplication by phase. And as for extended supersymmetry, when we consider action on fields, it is a flavor symmetry in some sense. For instance, the N=4 gauge multiplet has a following ... 3 Supersymmetric theories are based on the introduction of an extra symmetry between fermions and bosons. Standard Model is very non-supersymmetric, so people have been considering "minimal" supersymmetric extensions of the standard model (MSSM). So, first of all, your question title is not entirely correct - we are talking about minimal versions of ... 1 There are different formulations of SUGRA in superspace. Here I will describe only \mathcal{N}=1 D=4 Einstein supergravity: General coordinate transformations + super Lorentz transformations in real superspace Z^M =(x, \theta, \bar{\theta}):$$ Z^M \to Z^M + \lambda^M(Z) $$Physical fields are in supervielbeins E_M^{\;A}(x, \theta, \bar{\theta}) ... 0 The usual tracing rules of DeWitt do not apply to the connection, and in general, they do not apply to any object that is not tensorial. The tracing of indices of nontensorial objects is carried out by contracting and adjoining a parity factor that reflects the parity of the contracted index:$$\Gamma^a_{\,\,bc}\rightarrow (-1)^d\Gamma^d_{\,\,bd}.$\$ Note ...

1

Let's invent some new quantum number, call it, ah, whatever, why not R-parity. Now let's postulate that this quantum number is a conserved property in reactions. Further, let's assign our known particles a "0" in this quantum number, and any dark matter or other beyond-the-standard-model-particle a "1". In that case you have guaranteed ...

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