13 votes

What makes energy "the" conserved quantity associated with temporal translation symmetry?

The OP's question is basically stating that in a system with time-translation invariant dynamics, we can define a conserved quantity by arbitrarily assigning a real number to each orbit; when the ...
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  • 4,585
12 votes

What makes energy "the" conserved quantity associated with temporal translation symmetry?

What is special about the conserved quantity $Q(x, p) := \frac{1}{2} (x^2 + p^2)$, when also the quantity $Q_2(x, p) = \sin(x^2 + p^2)$ is conserved too, by the same temporal translational symmetry? ...
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11 votes

How are anomalies possible?

It's true as the others have said that the path integral measure may not be invariant. However I don't really like that quote as a general description of anomalies, since it sounds like they rely on a ...
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9 votes

How are anomalies possible?

This is such a short answer I'm tempted to just leave it as a comment, but besides specifying the Lagrangian you need to also specify your regularization, and the regularization is what introduces the ...
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  • 7,481
8 votes

How are anomalies possible?

The reason for this is that in the classical theory the Lagrangian fully specifies the dynamics of the system, however, in the Quantum system this is not true. Rather the Quantum theory is given by (...
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7 votes

What makes energy "the" conserved quantity associated with temporal translation symmetry?

For what it's worth, OP seems to be mischaracterizing Noether's theorem by seemingly ignoring one of its main assumptions: That the physical system is equipped with an action formulation $$S~=~\int\! ...
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3 votes
Accepted

Symmetry implies Ward identity

$U = e^{i\epsilon Q}$ is a symmetry if : $$\langle\alpha | e^{-i\epsilon Q}Se^{i\epsilon Q}|\beta\rangle = \langle \alpha|S|\beta\rangle \tag{1}$$ You can see this as saying that the $S$-matrix is ...
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  • 3,424
3 votes
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Finding $\Sigma$ wavefunctions from proton wavefunction. Any operator which can achieve this?

U and V spins move you diagonally in SU(3), so that $U_- ~d= s$ and $V_- ~u= s$. It then follows that $$U_- ~p = \Sigma^+ ~;\\ I_- U_- ~p=V_- ~p = \Sigma^0 ~; \\ I_- V_- ~p =V_- I_- ~p = \Sigma^- ~. ...
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3 votes
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I'm confused about the number of Killing vectors in Schwarzschild metric

A brute force (and ugly) derivation of the Killing fields of Schwarzschild metric The Schwarzschild metric is \begin{equation} ds^2 = -\left(1-\frac{R_{\text{S}}}{r}\right) \text{d} t^2 + \left(1-\...
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  • 491
3 votes

Lagrangian first integral

Hint: Noether's theorem yields that $$\begin{align} L\text{ has no }&x\text{-dependence} \cr \quad& \Downarrow&\quad\cr \text{momentum } &p_x \text{ is conserved}, \end{align} $$ and $...
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  • 167k
2 votes
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Are there non time-symmetric systems that increase total energy over time?

Noether's theorem implies that if $L$ does not depend on time, then $H$ (Hamiltonian) is conserved. But $H$ is not always energy in the physics sense. See When is the Hamiltonian of a system not equal ...
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2 votes
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Why can't spherical nuclei rotate?

This is colloquial to say that spherical nuclei do not have a rotational spectra. The nucleus could have in principle rotational energy but we’d have no way of measuring energy differences through ...
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2 votes

Preservation of symmetries of Tensors under lowering and raising indices

If $$T^{ij} = T^{ji},$$ Then $$T^{ij}g_{jk} = T^{ji}g_{jk}$$ $$T^{i}_{\ \ k} = T_{k}^{\ i} = (T^{i}_{\ \ k})^T$$ And $$T^{i}_{\ k}g_{li} = T_{k}^{\ i}g_{li}$$ $$T_{lk} = T_{kl},$$
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2 votes
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Translation invariance for scalar field

This is a straightforward application of Lagrange's shift operator, namely the formal rewriting of Taylor's expansion around 0. ($\phi$ being translationally invariant means its form does not change ...
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2 votes

What makes energy "the" conserved quantity associated with temporal translation symmetry?

If I understand the question correctly, then a flow $\Phi^t$ uniquely determines a Hamiltonian vector field $\mathbf X := \mathrm d\Phi^t\big|_{t=0}$. Such a vector field can be written as $$\mathbf ...
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2 votes

What makes energy "the" conserved quantity associated with temporal translation symmetry?

There are nice answers already. I just want to add one small observation. If $H$ is the energy, ordinary time is defined by the Hamiltonian flow with $H$, so observables evolve according to $$\frac{dO}...
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2 votes
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The mathematics of different particle rotations

Consider a point in space, and the effect a rotation would have on that point. Because the point has no internal structure (in other words, it is characterized entirely by its spatial location), a ...
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2 votes

Why is the gravitational potential inside a hollow sphere same as that of the gravitational potential on the surface of the hollow sphere?

Suppose one point on the shell is of potential V0. Due to spherical symmetry(with constant mass density throughtout the shell), every other point on the shell will be of that same potential. Now, the ...
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2 votes
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Why is the gravitational potential inside a hollow sphere same as that of the gravitational potential on the surface of the hollow sphere?

The 'Shell theorem' states that inside a hollow sphere there is no net gravitational pull. This is because the pull of all the parts of the surface cancel each other out perfectly. This is not the ...
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2 votes
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Noether current associated with transformation $\delta \psi=i\alpha \psi$

Mindful of @Robbie's comment, let's first rewrite your calculations as $$\begin{align}0&=\partial_\nu\frac{\partial\mathcal{L}}{\partial\partial_\nu\psi}-\frac{\partial\mathcal{L}}{\partial\psi}\\&...
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2 votes
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Given a representation of $su(3)$ labelled by $(p, q)$, is there a way to construct its state of greatest weight?

I don't have the formal Lie-algebraic statement you are seeking, but the practical seat-of-the-pants answer is evident: The origin of your Y here (in your normalization it is strangeness), that is, ...
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2 votes

How does parity work for the electric field and electric dipole and electric quadrapole transitions?

The electric quadrupole operator is quadratic in position: $$ Q_{jk}^{(i \rightarrow f)} = \langle i|r_jr_k-\frac 1 3 r^2\delta_{jk}|f\rangle$$ so it's even.
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  • 23k
2 votes

What is parity useful for in physics?

The parities of two atomic states (or nuclear states) factor into the selection rules for transitions between them mediated by electromagnetism, which in turn influences decay rates. Consider a ...
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1 vote

What is a rigorous and general definition of the parity operator?

Parity is a discrete spacetime symmetry under which the spatial coordinates $x^i$ transform as $x^i \rightarrow -x^i$ and the time coordinate transforms as $t \rightarrow t$. Various fields also ...
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  • 33.2k
1 vote

Problem 6 of Sheet 1 - Quantum field theory David Tong - Variation of Lagrangian density

I will try to complement the answer by Thomas, since some things were not that clear to me. We start by varying the Lagrangian $\mathcal{L}(\phi(x),\partial_{\mu}\phi(x),x^{\mu}):$ $$ \begin{split}\...
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1 vote

From the point of view of physics, why is it useful to know the irreps of rotation group?

In a nutshell: If we know the tensor/spinor/vector components in one coordinate system, and we know how it transforms, then we know it in every coordinate system, which is pretty useful. This sort of ...
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1 vote
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Chiral symmetry of the Dirac Lagrangian

The already used identity for $\gamma^0$ can be generalized to arbitrary $\mu$, i.e. $[\gamma^\mu, \gamma^5] =0$, hence one picks up a sign in $$\gamma^\mu e^{i\alpha \gamma^5} = e^{-i\alpha \gamma^5}...
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  • 407
1 vote
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Quantum Time Crystals

He is referring to a more precise definition of order parameter. The idea is to capture spontaneous symmetry breaking: the ground states of the system are not individually invariant under the ...
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  • 1,145
1 vote

How to unfold electronic bands? How do the reduced/extended schemes work? Confusion regarding the quasi-momentum

This is not a complete answer, but it was too long to fit into a comment. I'm having troubles with the same question. I haven't found a good explanation about this anywhere yet. So far what I managed ...
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  • 1,352
1 vote

Lagrangian first integral

with $$L=\frac{\sqrt{\dot x^2+\dot y^2}}{y}$$ and because L is not a function of x you obtain that $$\frac{\partial L}{\partial \dot x}=\frac{\dot x}{\sqrt{\dot x^2+\dot y^2}\,y}=\text{constant}$$ ...
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