# Tag Info

7

The Hamiltonian for the hydrogen atom $$H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r}$$ describes an electron in a central $1/r$ potential. This has the same form as the Kepler problem, and the symmetries are similar. There is an obvious $SO(3)$ generated by the angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. In other words, the components of ...

4

"Total spin conservation" means global $SU(2)$ spin-rotation symmetry (a continuous symmetry) of the Heisenberg model, and "spin wave" indicates an ordered ground state that spontaneously breaks the spin-rotation symmetry. Thus, according to Goldstone theorem, there must be a gapless mode for spin wave.

4

Yes, there is a structural reason for the existence of flat band on the Kagome lattice. This is related to the wave function localization due to the destructive interference on the lattice. The flat band has many physical interpretations. In the momentum space, looking at the dispersion relation, a flat band means the effective mass of the particle is ...

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 ...

2

Axial charge'' refers to the (isovector) axial coupling constant $g_A$ of the nucleon $$\langle p|A_\mu^a|p\rangle = g_A \bar{u}(p)\gamma_\mu\gamma_5\tau^a u(p)$$ where $A_\mu^a=\bar{\psi}\gamma_\mu\gamma_5\tau^a\psi$ is the QCD axial current, $|p\rangle$ is a nucleon state with momentum $p$, $u(p)$ is a free nucleon spinor, and $\tau^a$ is an isospin ...

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 ...

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 & ...

2

It's because there is another vector quantity $A_i$ conserved in addition to the angular momentum $L_i$. Furthermore, the commutation relations of $A_i$'s and $L_i$'s are those of $SO(4)$. See for instance this reference : http://hep.uchicago.edu/~rosner/p342/projs/weinberg.pdf

2

For reference, Weinberg p. 378: A metric space is said to be homogeneous if there exist infinitesimal isometries (13.1.3) that carry any given point $X$ into any other point in its immediate neighborhood. Equation (13.1.3) defines an infinitesimal transformation and (13.1.5) concludes the Killing equation $\xi_{\sigma;\rho} + \xi_{\rho;\sigma} = 0$ ...

2

One way to understand it is to recognize that for the spherical harmonic $|l,m\rangle$ with $l=0$ (and obviously $m=0$), we have $\hat L_i|0,0\rangle=0$, where $\hat L_i$ is the angular momentum operator in the direction $i=x,y,z$. It is obvious for $\hat L_z$, which eigenvalue is $m=0$, and can be verified for the other two. Then, the rotation operator ...

1

Suppose that there existed a spherically symmetrical wavefunction $\psi({\bf r})=f(r)$ for which $l\neq0$. This cannot be, for if we calculate $\langle \psi | L^2 | \psi \rangle$ we will always get zero, as each term in $L^2$ has derivatives with respect to $\theta$ and $\phi$. Conceptually speaking, a spherically symmetric state gives the electron the ...

1

I'm not entirely sure I understand what you're asking but here is how I've interpreted your question: It seems like all of the laws of physics are reversible in time. That is, given the state of a physical system, it's possible to both go forward in time or backwards in time from that state. Assuming this is the case for physical laws and the equations ...

1

The short answer is yes. One can convince oneself this is indeed the case by doing the dimensional counting as it was done by Everett You. However, it is by no means a proof. The problem is that the valence bond states are not linearly independent. Even though there are much more valence bond states than the number of singlets made from $N$ spin-one-half ...

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 ...

1

Sorry I found David Z' answer a bit confused just when discussing the crucial point. Since the two functions ψ(x) and ψ(−x) satisfy the same equation, you should get the same solutions for them, except for an overall multiplicative constant; in other words, ψ(x)=aψ(−x) Normalizing ψ requires that |a|=1, which leaves two possibilities: a=+1 ...

Only top voted, non community-wiki answers of a minimum length are eligible