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4

The basic idea is the following. For the shake of simplicity, I henceforth assume that every function does not depend explicitly on time (with a little effort, everything could be generalized dealing with a suitable fiber bundle over the axis of time whose fibers are spaces of phases at time $t$). On a symplectic 2n dimensional manifold (a space of ...


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Great question! One way to look at what is going on is to use the Hamiltonian version of Noether's theorem. The Noether procedure generates a conserved charge $Q$ associated with the symmetry with parameter $\theta$. It turns out that $Q$ is the generator of that symmetry, in the sense that for some function $A$ of phase space variables \begin{equation} ...


1

Your question seems to contain two parts. First, you're asking how to set up the equations of motion for this coupled system. Second, you are asking how to use symmetry considerations to find the normal modes and frequencies. Let's first answer the bit about symmetry first Normal modes - symmetry Your observation about the reflection symmetry is spot on. ...


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Consider an element $g$ of the symmetry group. Say $g$ is represented by a unitary operator on the Hilbertspace $$ T_g = \exp(tX) $$ with generator $X$ and some parameter $t$. It acts on an operator $\phi(y)$ by conjugation $$ (g\cdot\phi)(y) = T_g^{-1}\phi(y) T_g = e^{-tX}\phi(y) e^{tX} = \big[ 1 + t[X,\cdot]+\mathcal{O}(t^2)\big]\phi(y)$$ On the other ...


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"Derivation" of Baryon Number Conservation - Consider the QCD Lagrangian (density) $$\mathcal{L} = \bar{\psi}(i\gamma^\mu D_\mu - m)\psi - \frac{1}{4}G^a_{\mu\nu}G_a^{\mu\nu}$$ where the symbols have their usual meaning. This is invariant under $U(1)$, which is nothing but a multiplication of $\psi$ by a global phase factor $e^{i \theta}$. This is ...


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The definitions are equivalent because the action is invariant in each case, that is, $\delta S = 0$. Let us take case 2, in which $\delta \mathcal{L} = \partial^\mu F_\mu$. By Stokes' theorem, the total divergence results in a surface integral at infinity, $$ \delta S = \int d^4x\delta\mathcal{L} = \int d^4 x \partial^\mu F_\mu= \int d\Sigma^\mu F_\mu = 0 ...


7

I) We interpret OP's question (v2) as essentially asking about the following. What happens L1) if the Lagrangian density $\delta {\cal L}= 0$ does not transform? L2) if the Lagrangian density $\delta {\cal L}=\varepsilon~ d_{\mu} f^{\mu}$ transforms with a total space-time divergence? Here $\delta$ denotes an infinitesimal transformation ...


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It's because magnetic field has zero divergence combined with the symmetry of the problem. At each point on our Ampere's loop we define three orthogonal unit vectors: $\hat{t}$ which is tangential to the loop, $\hat{r}$ which points radially outward from the center of the loop, and $\hat{z}$ which is parallel to the wire. Using these direction, we write the ...


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You seem to be confusing some things. Noether's theorem tells us that continuous symmetries of the Lagrangian of a system are in one-to-one correspondence with conserved quantities (you can find many references and explanations of this on the web, including on this site. The canonical example is $$\text{spacetime translation invariance}\Leftrightarrow ...


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Taking a step back, I'd suggest a look at Ashcroft and Mermin's "Solid State Physics" where they treat the harmonic crystal modes (chapter 22 in my edition). Nowhere do they suggest that LO-TO splitting occurs only in ionic solids. Instead, they make it clear that a Bravais lattice with a mono-atomic basis has acoustic modes only. Once a poly-atomic ...


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So I think you should check out this problem for some useful commentary. Problem understanding sign of volume integral in Minkowski space Essentially if you look at the integral on the LHS we see it takes the form explained in the problem above $$ I^{\mu\nu}[f] = \int d^4k\, f(k^2)k^\mu k^\nu $$ Where f is just 1 The integral is clearly Lorentz invariant ...


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I don't think that you should think of particle conservation as a conservation law in the context of classical physics. As Qmechanic says in a classical system, the particle number is the number of degrees of freedom and is fixed as a definition of the problem. Once we have decided how many particles will be there we can write a Lagrangian, investigate its ...


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I think this calculation is the answer for your question: $$\sum_{\alpha,\mu}\frac 1 2 \omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)=\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)+\frac 1 2 \sum_{\mu<\alpha}\omega_{\alpha \mu} \left(x^\mu ...


1

The concept of number operator is most natural in the context of second quantization. I assume you are familiar with it, and denote by $a(x)$ and $a^*(x)$ the creation and annihilation operators(-valued distributions). They satisfy the commutation relation $[a(x),a^*(y)]=\delta(x-y)$. The number operator $N$ is the second quantization of the identity, and ...


2

In single particle picture (with wave functions) it is clear that probability distribution $|\phi(x)|^2$ won't change under the $U(1)$ transformation $\phi(x) \rightarrow \phi(x) e^{i\phi},\quad \phi^* (x) \rightarrow \phi^* (x) e^{-i\phi}$ In many-particle picture, using field operators, simplest $U(1)$ symmetric Hamiltonian can be rewritten as: $$ ...


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This is one of the few times when using the Einstein convention isn't helpful: recall that in Noether's theorem you need to sum over the independent parameters of your transformation, which is usually trivial but, for Lorentz transformations, you need to be careful because of the antisymmetry of $\omega_{\mu\nu}$, which implies that you have only 6 ...


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Your guess is correct. That classification of order $n$ multipole moments can be viewed as a symmetry classification of rank $n$ tensors, so how many of them there are is a matter of what symmetries exist in spacetime. There are two types of "inversions" that can be performed on spacetime's two types of physical dimensions: a spatial inversion, also called ...



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