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1

I don't know that this qualifies as a full answer, but the “intuitive” answer would be that on large enough scales, the details of the lattice won't matter: If you look from far enough away (conversely, the lattice parameter is sufficiently small), it will look like a continuum. In this picture, you would expect large-scale structures to be fully ...


3

You are correct in your interpretation that Weisner's method is geometric in nature: it is a method for finding generating functions for special functions using representation theory of Lie groups and Lie algebras. And as you know, Lie groups play an enormous role in modern geometry, on several different levels. Lie groups are smooth differentiable ...


0

The error is that $\gamma_5$ doesn't intrinsically change sign under parity. Also, don't forget that under parity the spatial components of $W_\mu$ change sign. And also $\gamma^0 \gamma^\mu \gamma^0 \neq \gamma^\mu$.


0

We know from translations symmetry that the expectation value of the wavevector operator is constant -- that is, $$ \langle \mathbf{k} \rangle = \langle \mathbf{k}_1 + \mathbf{k}_2 + \ldots \rangle = \text{const.} $$ In other words, wavevector is a conserved quantity. If the constant of proportionality between $\mathbf{k}$ and momentum were different for ...


0

Weak interactions with $W$ and $Z$ gauge bosons violate parity simply because the righ-handed and the left-handed fermions coupled differently to $W$ and $Z$. For example, the $W$'s couple only to the left-handed fields. A parity inviariant dynamics would require that both left- and right- fields couple in the same way to the gauge vector since they get ...


0

I believe that the simplest explanation, in the plainest language, is that the Wigner-Eckhart theorem is a quantum-mechanical expression of conservation of angular momentum. This may not be self evident, but it's hard to get simpler.


6

What happens if the twin in the spaceship doesn't return? Would he still be younger than his other twin? It's really a moot point, because you can't compare clocks. There is no absolute time! You can't say, "What's each twin's age at this instant?" because "this instant" depends on the observer. Is the symmetry broken simply by accelerating out of earth? ...


0

I guess it is because you first of all change sign of $\vec x$ to $- \vec x$ in physical space(this is parity transformation in a nutshell). All this peculiar algebra concerning left and right chirality fields comes from $J = 1/2$ Lorenz group representation, so transformation rules are defined as representatives of parity transformation of physical space.


-1

Okay, I think I have an idea why the terminology is used, but I think this argument makes little sense: The Lagrangian term describing weak interactions is of the form $$ \bar \Psi \gamma_\mu P_L \Psi W^\mu $$ Under parity transformations $ \Psi \rightarrow \gamma_0 \Psi$ and $ \bar \Psi \rightarrow \bar \Psi \gamma_0 $, therefore $$ \bar \Psi ...


2

The laws of physics are discovered through a mixture of heuristics, modelling and inference. In case of momentum, the story goes like this: It is possible to 'transfer motion' from one body to another. However, experiment shows that it is not velocity that is conserved during such transfers, but another 'quantity of motion'. We give that quantity the name ...


0

the Dirac Point on Graphene is protected by hidden symmetry. And it is explained very well in the paper arXiv:1406.3800. It is not that easy to understand the hidden symmetry. Personally speaking, I thought it is combination of inversion, time reversal and reflection symmetry, though the hidden symmetry in that paper has a totally different form with my ...


3

Stack all the $\phi^i$s into a column "vector" $\vec\phi$. The mass term $m^2\vec\phi\cdot\vec\phi$ is obviously invariant by $R^{-1}=R^T$. The same with the kinetic term $(\partial_\mu\vec\phi)\cdot(\partial^\mu\vec\phi)$ because $\partial_\mu R=0$. It is $SO(n)$ invariant because I take it $i$ runs over $n$ values. Thus your $r^i_j$ generates $SO(n)$. ...


1

The Lagrangian is what is integrated over spacetime in the action, i.e. has to be a 4-form. As such, it is necessarily a (pseudo-)scalar under Lorentz transformations. When wondering about Lorentz transformations and such, the Hamiltonian is, as a non-Lorentz-covariant object, not a good starting point, by the way. It is often better to start with the ...


0

The hexagonal Graphene lattice can be considered as a superposition of two identical sub-lattices set off by one one carbon-carbon bond length. As a result, it has two sets of wavevectors k,that are picked out by the lattice, inequivalent (since the two sublattices really are distinct) but otherwise identical (since it's semantics to say which sublattice is ...



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