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24

One cannot tell by the light spectra. Hydrogen and antihydrogen would give the same lines in the spectrum. The prevalence of matter over antimatter from other evidence indicates matter is predominant in the observable universe, and here is a nice review. How do we really know that the universe is not matter-antimatter symmetric? The Moon: Neil ...


10

In a neutral hydrogen atom the ground state has the electron and proton spins anti-parallel i.e. lined up with each other but pointing in opposite directions. The state with the spins parallel and pointing in the same direction has a slightly higher energy, and transitions between these two states produce the notorious 21cm hydrogen line. Since the ...


9

The probability density of the ground state is time independent, so there is no motion in this sense. Yet the expectation value of the kinetic energy is non-zero, so there is movement in this sense. How are these notions of movement reconciled? First off, classically, if we had a particle in a $1/r$ potential and released it from rest, it would indeed bob ...


8

The degeneracy of energy-levels can be traced to the fact that the hydrogen atom possesses an enhanced $SO(4)$ symmetry caused by (among other things) the conservation of the Laplace-Runge-Lenz vector operator, see e.g. this Phys.SE post and Ref. 1. References: G. 't Hooft, Introduction to Lie Groups in Physics, lecture notes, chapter 9. The pdf file is ...


8

$\newcommand{\d}[1]{\,\mathrm{d}#1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}\newcommand{\p}{\psi_{100}}\newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$The hydrogen ground state is the following: $$\psi_{100}=Y_{00}\frac{2}{a_0^{3/2}}e^{-r/a_0}$$ The expectation value of the position operator on this state is the following: ...


6

The orbitals, which recently have been observed for the hydrogen atom, are probability distributions. These probability orbital distributions have been calculated using quantum mechanical solutions of the Schrodinger equation which give the wave function, and the square of the wave function is the probability distribution for finding the electron at that ...


5

Is the Coulomb potential also used to solve the hydrogen atom in relativistic quantum mechanics? Yes, the Coulomb potential is there in the solution of the hydrogen atom with the Dirac equation, which is formulated in the relativistic framework. Now it is time to specialize to the hydrogen atom for which $$\frac{V}{\hbar c}=-\frac{Z\alpha}{r}$$ ...


5

You say: The Coulomb potential comes from classical electrodynamics but actually the Coulomb potential is predicted by quantum electrodynamics as a low energy limit. Quantum field theory describes the interactions between charged particles as the exchange of virtual particles, and it's not immediately obvious that it would lead to an inverse square ...


5

The graph shows the probability of finding the electron between the distances $r$ and $r + dr$. This probability is given by: $$ P = \psi^* \psi dV $$ where $dV$ is the volume element: $$ dV = 4\pi r^2 dr $$ So we get the probability: $$ P(r,r+dr) = \psi^* \psi 4\pi r^2 dr $$ and therefore when $r = 0$ the probability $P = 0$. It isn't that the ...


5

The answer is that the premise is wrong. There can't be a hydrogen wave function with the coefficients you have written. Even if there was no $| 1 0 0 \rangle$ state present, the state isn't normalized. That means that it isn't physical. However, remember that the coefficients are somewhat arbitrary, that is, we're allowed to multiply the whole wavefunction ...


4

The time-independent Schrödinger equation for the hydrogen atom is $$-\frac{\hbar^2}{2m}\vec \nabla^2\psi-\frac{e^2}{4\pi \epsilon_0r}\psi=E\psi $$ If your aim is just to verify that the $1s$-wave function $$\psi_{100}=\frac{1}{\sqrt{\pi a^3}}e^{-r/a}\hspace{2cm} a\equiv \frac{4\pi\epsilon_0\hbar^2}{me^2} $$ is indeed an eigenfunction, then your task ...


4

From the Virial Theorem we can say the total energy of the atom is propotional to the potential energy of the atom. The potential energy is given by a coulomb potential and so is it will be roughly proportional to $\frac{1}{\langle r \rangle}$ where $\langle r \rangle$ is the mean radius of the electron's orbital. For a hydrogen atom the energy $E\propto ...


3

$r=n^2a_0$ for the Bohr model. Similarly, in the Schrodinger model, the most probable radius is $r=n^2a_0$, when $n=l+1$


3

Yes, the mean radius grows as $n^2$ asymptotically, but to have $1-10^{-20}$ for sure you have to add some more ;-). Highly excited atoms are called Rydberg atoms and are dealt with in experiment.


3

The potential energy function is the same for both. The energy level solutions are the same for both. The key difference is that in (most modern interpretations of) the Schrodinger model the electron of a one-electron atom, rather than traveling in fixed orbits about the nucleus, has a probablity distribution permitting the electron to be at almost all ...


3

The shortest and correct answer: this degeneracy is determined by the symmetry of the system. The case of degeneracy in hydrogen atom is so-called "accidental degeneracy", when eigenfunctions belonging to different irreducible representations of the symmetry group of a Hamiltonian correspond to the same energy. This type of degeneracy can also occurs in ...


3

What you're trying to do is called radiometric calibration. The problem with doing it with an ordinary incandescent light bulb is that the bulb itself would have to be radiometrically calibrated to get a precise spectral calibration of the detector. That's because the bulb's spectral emissivity deviates from an ideal blackbody source with an emissivity ...


3

The electron is the lightest lepton and the proton is the lightest baryon, so it's hard to see what reaction could occur without violating lepton number or baryon number. I suppose if proton decay (to a pion and positron) occurs then there could be a reaction to give a pion and two photons.


3

The symmetries that you're missing are conservation of baryon number $B$ and lepton number $L$. We strongly suspect that baryon number is not an exact symmetry, because the universe appears to contain very many baryons and very few antibaryons. Actually, a better metric for the baryon asymmetry of the universe is to compare the baryon density to the density ...


3

These relations are based on the fact that both the position and the momentum distributions are centred around zero, which is in turn due to the symmetry of the atom. Given that, the width of the position and momentum distributions ($\Delta x$ and $\Delta p$) is of the same order as a typical position or momentum within those distributions ($r$ and $p$).


3

It is a misconception to say that In QM [...] we know that the the electron does not radiate EM-Waves because it is not actually circling around the nucleus. It is sometimes here and there. In QM the notion of "circling round the nucleus" does indeed fail to make sense, but this is not why electrons don't radiate EM waves. Instead, an atom in its ...


3

Hydrogenic wavefunctions (as well as anything with well-defined angular momentum about a given axis) come in two flavours. The first set is 'cylindrical', and has wavefunctions $\psi\sim e^{\pm i|m|\phi}$. The second set is 'cartesian', and has wavefunctions $\psi_\text{even}\sim\cos(m\phi)$ and $\psi_\text{odd}\sim\sin(m\phi)$. Both sets are perfectly ...


2

The maximum $s_z$ eigenvalue an electron can have is $\hbar/2$. Therefore, the only way that a quantum state can have $\langle s_z \rangle = \hbar/2$ is if the state lies completely within the $s_z = \hbar/2$ eigenspace. Thus your answer may only include pure states with $s_z = \hbar /2$.


2

In nonrelativistic quantum mechanics you set up a wavefunction that is a function of configuration space, $\Psi(x_1,y_1,z_1, \dots , x_n,y_n,z_n,t)$. Where $(x_i,y_i,z_i)$ is the position of the i-th particle. If any particles are identical then you make sure that the wavefunction is (anti)symmetric in those coordinates. Then if you are doing a simple ...


2

The main fusion reaction in the sun is the proton-proton chain reaction, which takes six protons and produces two protons, one alpha particle, two anti-electrons, and two electron neutrinos. The deuterium nucleus is only barely bound and can be destroyed — dissociated into a proton and neutron — by absorbing a gamma ray with energy more than 2 MeV. This ...


2

The state you have given is not normalisable as a consequence of the results of the calculations you have done. Even if the first state (with coefficient $A$) was not present it would not be normalised. To normalise what you have given, another constant needs to multiply everything through (so that the relative proportions are unchanged


2

Your main question was "Is there an example where the overlap between a bound state and the scattering states makes a measurable contribution to the energy in the perturbative regime?" Actually, I disagree with the statement of the other answer that the scattering states must be included in the perturbative calculations only if the result is to be highly ...


2

I pulled most of this from Wikipedia here. A stationary state is called ''stationary'' because the system remains in the same state as time elapses, in every observable way. For a single-particle Hamiltonian, this means that the particle has a constant probability distribution for its position, its velocity, its spin, etc. A stationary state is not ...


2

This is because each spherically invariant state $\psi$ must have zero angular momentum. Indeed, by hypotheses, the state $\psi$ verifies $$\psi(R_{\vec n}(\theta)\vec{x} ) = (e^{i \theta \vec{n}\cdot \vec{\hat{J}}} \psi)(\vec{x}) = \psi(\vec{x})\tag{1}$$ where $\vec{n}\cdot \vec{\hat{J}}$ is the self-adjoint generator of rotations $R_{\vec n}(\theta)$ ...



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