# Tag Info

1

A quick answer, if I may. You need $\theta$ to be smooth since you want to derive it. So mathematics imposes you to choose $\theta$ smooth. Now the trick: choosing $\theta$ to be smooth means you can always impose $\mathbf{A}$ to be smooth, and use several patches related to each other by a gauge transform. Then you should always discuss smooth vector ...

2

Suppose in classical mechanics, I told you that $\ddot{x} = \frac{F}{m}$, where $F$ is a conservative force. You would have no trouble finding a conserved value along the trajectory (I hope). What you need, therefore, is to use the GR version of the work-kinetic-energy theorem. If I were in a hurry, I would start with the Lagrangian that gives the EOM and ...

2

Well, after symmetry breaking, all that remains is electromagnetic $U(1)$, so the only generator that is truly a symmetry generator is $Q$. The fermions couple to the "Higgs" via the Yukawa coupling: $\mathcal{L}_y = -y_e^{ij} \bar L_{L,i} \Phi e_{R,j} - y_u^{ij} \bar Q_{L,i} \tilde{\Phi} u_{R,j} - y_d^{ij} \bar Q_{L,i} \Phi d_{R,j} + h.c.\,$ which mixes ...

0

Your hamiltonian is still gauge invariant because the canonical momentum, that is the generator of translations, is $m \dot{\vec{x}} - q \vec A$. In other words, your Hamiltonian is still $H = \frac{1}{2m}\Pi^2 - q \Phi$, where $\vec \Pi = m \vec{ \dot{x}}$ is the mechanical momentum.

2

Again, thanks to the $SU(2)$ PSG proposed by prof.Wen, I can answer my question now, $THT^{-1}$ is in fact $SU(2)$ gauge equivalent to $H$, and the statement "$H$ is also not SU(2) gauge equivalent to the time-reversal transformed Hamiltonian $THT^{-1}$" in my question is wrong. Let's rewrite the Hamiltonian as ...

2

I don't know the article you refer to, but I believe the Hamiltonian you discuss should get a $\pi$-phase shift after one turn around a (2D) lattice cell. So I guess it should read $H=F^{\dagger}\cdot H_{\pi}\cdot F$ with $$H_{\pi}=t\left(\begin{array}{cccc} 0 & e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4}\\ e^{-\mathbf{i}\pi/4} & 0 & ... 3 Tarek (OP) e-mailed me to contribute to this thread. Here's the response that I gave him (slightly edited for clarity). I see why this was confusing, my apologies! I was perhaps too glib in the post. Iwas implicitly talking about a chiral rotation but wanted to present it somewhat more intuitively. Let me try to spell it out more carefully, and hopefully ... 0 Just to follow up on Trimok's answer, you might find the following references useful: Correlation functions in the CFT(d)/AdS(d+1) Correspondence (http://arxiv.org/abs/hep-th/9804058) Supersymmetric Gauge Theories and the AdS/CFT (Correspondence http://arxiv.org/abs/hep-th/0201253) The first paper contains a discussion of the discrete inversion isometry ... 3 Answer: There is none. The issue at hand is that the Kaehler invariance is just that - an invariance, not a continuous symmetry of the fields. Most prominently the superpotential must transform as$$ \mathcal W \to \mathcal W e^{-h} $$A general superpotential that leads to consistent theories is$$ \mathcal W =\frac{1}{2} m_{ab} \phi^a \phi^b + \frac{1}{3} ...

3

It is known that the Hubbard model possesses the global $SU(2)$ spin-rotation symmetry, which means that the Hamiltonian commutes with the total spin $\sum_i\mathbf{S}_i$(where $\mathbf{S}_i=\frac{1}{2}c_i^\dagger \mathbf{\sigma}c_i$), which is the generators of the global $SU(2)$ spin-rotation group, and it does not commute with $\sum_i\mathbf{S}_i^2$.

0

Please look at "On The Galilean and Lorentz Transformations" published by American Open Advanced Physics Journal, Vol. 1, No. 2, November 2013, PP: 11 - 32, Available online at http://rekpub.com/Journals.php. In this paper, detail distinction between the two transformations is given. But the answer to this question is completely different from Answer 1.

1

I just found a relative rigorous argument supporting my conjecture: The Chern number $N=\frac{1}{2\pi}\int _{BZ}b(\mathbf{k})$, where $b(\mathbf{k})$ is the Berry curvature. Since $H(-\mathbf{k})=H(\mathbf{k})$, it's easy to show that $b(-\mathbf{k})=b(\mathbf{k})$, accordingly, we can divide the $BZ$ into two halfs called $BZ_1$ and $BZ_2$, therefore, ...

1

What is in the same electroweak doublet are just left-handed components of electron and neutrino. Note that the mass of the particle can be thought of as the strength of the coupling between left-handed and right-handed components of particle's field. You have such coupling for electron (in a gauge-invariant way, via Higgs), and not for neutrino (which lacks ...

0

If you are looking for a book with an "elegant mathematical framework which accounts for all the patch up work that is needed" to build a realistic perturbatively renormalizable quantum field theory like the recently completed and validated Standard Model, its title could be : Noncommutative Geometry, Quantum Fields and Motives. It is written by Alain Connes ...

5

I think that Adam's and my original answer below are right (after having spent about 24 hours thinking they were wrong). The idea we were all missing is the notion of Berry phase. Hopefully an expert in anyons can confirm Adam's and my thinking. Your argument is "specious"; I say this word wholly with its technical meaning of "not relevant" and with no ...

8

The thing is that the operation "exchange of two particles" has to be defined properly. What is the meaning of $P$ ? We can imagine the operator $P$ is not physical (in the sense that is does not correspond to a physically possible operation). For instance, $P\psi(x_1,x_2)=\lambda\psi(x_2,x_1)$ in the sense that it only exchange the argument of the ...

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