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6

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 ...


3

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, ...


3

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 $l=0$ ...


3

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 $$ ...


3

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 ...


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 ...


2

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}


2

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 ...


1

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 $D=2$ ...


1

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 ...


1

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 ...


1

The following situation is not uncommon: classically a symmetry may be (spontaneously) broken, but, quantum mechanically, the symmetry is restored. Put differently, quantum fluctuations can, under certain, well understood conditions, destroy the classical asymmetry ("order"). The simplest example is probably the one-dimensional double well potential, ...


1

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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