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2

I) OP is asking about the case where the infinitesimal symmetry transformations $$ \delta\phi^a~=~t^a{}_b \phi^a \tag{A}$$ are linear in the fields in the path integral$^1$ $$Z[J] ~=~\exp\left[\frac{i}{\hbar}W_c[J]\right]~=~\int \! {\cal D}\phi~\exp\left[\frac{i}{\hbar}\left( S[\phi]+J_a\phi^a \right)\right]. \tag{B}$$ Recall the Legendre transformation ...


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From my readings; the key to conservation of momentum appears to be based on defining a closed system to see if any mass crosses the boundaries of the system.


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I think a much more natural way to come to $\hbar$ is the one used by Dirac in his principles of quantum mechanics. In it you start by stating that you want a Poisson Bracket which has the same algebraic structure as that of classical mechanics. Then, since quantum operators don't commute (unlike what happens in classical physics) you quickly find out that ...


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Time reversal essentially means a system looks the same if you reverse the flow of time. The only difference beeing that things like velocity go in the opposite direction. In condensed matter systems it is represented as a unitary matrix times complex conjugation $\mathcal{T} = U\mathcal{K}$. A simple system that follows T-symmetry would be a system ...


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Ref. 1 defines a local translation on spacetime $M$ as a diffeomorphism. The word local means here point-dependent ($x$-dependent). An infinitesimal diffeomorphisms $\xi^{\mu}(x)$ can be identified with a vector field. Note that in flat Minkowski spacetime we called affine transformations for global Poincare transformations. The word global means here ...


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All quarks have baryon number 1/3, so that the nucleons can be built up. Baryon number is a conserved quantity in the standard model . In models where the proton can decay, it is not conserved, but no proton decays have been detected up to now. One has to realize that the quantum numbers S,C,B,T are attributes for all quarks and have a value, even if that ...


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They do. It's the third component of isospin, which came about before Murray Gell-Mann's quark model. \begin{equation} I_3 = \frac{(N_u-N_\bar{u})-(N_d-N_\bar{d})}{2} \end{equation}


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Let's check that parity is violated by the weak interaction lagrangian: $$\mathcal{L}(x) = \bar{\psi}(x) \gamma^\mu \frac{(1-\gamma^5)}{2} \psi(x) W_\mu(x)$$ Saying that parity is violated means that the transformed lagrangian $\mathcal{L}'(x)$ is not equal to the old lagrangian resulting from new coordinates $\mathcal{L}(x')$ where $x'^0 = x^0$ and ...


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Let $\xi^\alpha$ be a Killing vector of a metric $g_{\mu\nu}$, i.e. it satisfies $$ \nabla_\mu \xi_\nu + \nabla_\nu \xi_\mu = g_{\mu\alpha} \partial_\nu \xi^\alpha + g_{\nu\alpha} \partial_\mu \xi^\alpha + \xi^\alpha \partial_\alpha g_{\mu\nu} $$ Then the quantity $$ Q = \xi^\alpha u_\alpha $$ is conserved along any geodesic. To see this, we can compute $$ ...


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But why is it the component of the dual vector $p_\phi$ that is constant rather than $p^\phi$? From the bottom of page 189: The geodesic equation can thus, in complete generality, be written $$m \frac{dp_\beta}{d\tau} = \frac{1}{2}g_{\nu \alpha,\beta}\;p^\nu p^\alpha$$ We therefore have the following important result: if all of the ...


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A short answer to questions 2 and 3: In Mermin-Wagner's paper the short-range condition is stated as $\sum_{\bf R} {\bf R}^2 |J_{\bf R}|<+\infty$. For interactions with (or more precisely majorized by a) power law decay $|J_{\bf R}| \sim R^{-\alpha}$, this requires $\alpha > D+2$, where $D$ is the space dimensionality (i.e., $\alpha >4$ for $D=2$ ...


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Just wanted to add that it's not totally true that the drawn orbitals are "the regions where an electron can be found". But my answer grew and grew... Let's take a neutral boron atom where it has filled 1s and 2s shells and one electron in the 2p shell. Suppose it is floating in space, far away from messy interactions with other things. So, you wonder, ...


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My answer has a lot of overlap with the earlier answer by NoEigenvalue, but grew too long for a comment. The orbitals are segregated by the amount of angular momentum that they carry. No angular momentum, $\ell=0$, are called $s$-orbitals, and are spherically symmetric. The first two examples in your figure are $s$-orbitals with $n=1$ and $n=2$. Angular ...


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The orbitals are only spherically symmetric for $S$-states, for which the angular momentum number $l$ is zero. So in you picture the first two orbitals are the $s=1$ and the $s=2$ state, while the other three pictures correspond to the $n=2,l=1$ with the three different possibilities of $m_l$. There are several ways to see that the wavefunctions with $l=0$ ...


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Comments to the question (v2): The fact that the TISE is invariant under a symmetry group $G$ (in this case the Lie group $G=SO(3)$ of 3D rotations) does not imply that the orbital/wave-function solutions $\psi$ must be $G$-invariant as well. (Think e.g. on spontaneous symmetry breaking where the governing laws of a physical system are invariant under a ...


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All these mathematical objects (scalars, vectors, tensors) are carefully selected to represent real world phenomena and their observable properties. For example, invariance under transformation, means that the variable (or physical object) represented via the transformed, let's say vector or tensor, does not change ,i.e. it is conserved. In general, theories ...


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Just as a quick example: say that the dot product were not invariant under transformations. Then let's say that we have two reference frames, A and B, where reference frame B is rotated and displaced with respect to A and which moves at a constant speed w.r.t. A (where $v\ll c$). Then the researcher in A wants to calculate the gravitational attraction ...


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The 4-vector $A^{\mu}$ is an operator which acts on a Hilbert space with states $|a\rangle$. These things are called tensor operators - see chapter 4 of Howard Georgi's book Lie Algebras in Particle Physics. So, they have a matrix representation $\langle a|A^{\mu}|b\rangle$ which I'll write as $A^{\mu a}_{\ \ \ b}=\langle a|A^{\mu}|b\rangle$ to emphasize the ...


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The consideration of ground states is an equilibrium assumption. Imagine that instead of developing a theory around a ground state we were doing it around any general state. Actually, for simplicity I'll only consider states with constant $\varphi$, but the same remarks will be valid with the appropriate modifications -- modulo some topological ...



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