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1

Yes, this is actually often used in a spectroscopic technique called REMPI -- see the image on this wikipedia page https://en.wikipedia.org/wiki/Resonance-enhanced_multiphoton_ionization There are some important physics techniques that rely on interaction with two photons -- two photon spectroscopy (http://cua.mit.edu/8.421_S06/Chapter9.pdf). Some other ...


1

(Moving this from my comment to an answer) Yes, the electric field simply penetrates the glass wall and charges (the electrons) placed in that field will feel a force and move. The glass does not really interact with the charges on either side, so you might as well remove it completely (theoretically).


4

You are probably thinking in terms of the shell model of the atom: (source) It is important to note that the shell model is ultimately inaccurate and is not really a good representation of how modern physics views atomic structure. It has a few advantages, and it is a nice tool to explain certain features in atomic structure (in particular, the fact that ...


1

If you have current flowing one way through a resistor, then the electrons flow through the other way. Since current flows from the high voltage end of a resistor to the low voltage end, then the electrons come in at the low voltage end and come out at the high voltage end. When electrons (which are negatively charged) go from low voltage to high voltage, ...


0

There may be a confusion between enery and power. While the current, which is the number of charges per second, and the energy of the ,electron do'nt change at the output of the resistor, the power does. Power is the amount of energy per unit time, and that does not affect the current which, again, is the amount of charge per unit time.


2

An electric current is the flow of electric charge. But electric charge is not an entity, it is a property that must be 'carried' by a charge carrier. An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. However, an electric current is not necessarily an electron current. ...


0

There are a number of ways to represent a 720 degree spin of a particle, but the particle has to be a little more complex then a spinning ball. Imagine this wiki image (from http://en.wikipedia.org/wiki/Spin-1/2) as the inside of a larger ball and you can see an example of 720 degree spin. My favorite representation is the idea of breaking the electron into ...


3

You have fallen prey to a popular simplification of spinors. The statement "you have to turn electron by 720 degrees in order to get the same spin state" does not refer to an actual rotation of an actual electron. In quantum mechanics, we describe the states of objects as elements of a Hilbert space $\mathcal{H}$. The crucial thing is that not all elements ...


0

The definition of work function that I am using is: the minimum thermodynamic work (i.e. energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface. This could be viewed in the context of the photoelectric effect, the minimum energy photon required, incident to a surface, to liberate an electron ...


0

The reason why they put equations in is because physics is best described in terms of equations. They tried to remove randomness with equation, ultimately believing that QM is not that random, that it is behaving according to certain parameters. They tried to set the boundaries in the randomness/unpredictability, but the equations still couldn't remove the ...


0

For what it's worth, I showed in my recent article http://link.springer.com/content/pdf/10.1140%2Fepjc%2Fs10052-013-2371-4.pdf (published in European Phys. J. C) that one can eliminate the Dirac field from the Dirac-Maxwell electrodynamics after introduction of a complex electromagnetic 4-potential (producing the same electromagnetic field as the real ...


5

In our modern understanding, every electron is thought to be a localized excitation of the electron (or Dirac) (spinor) field $\Psi(x^\mu)$, while every photon is considered to be an excitation of the photon (vector) field $A^\nu(x^\mu)$, which is the quantum field-theoretic counterpart of the classical four-potential. Thus, the answer to your questions ...


3

If the photon is massless, how can it make an electron change momentum? Because, relativistically, momentum isn't proportional to (invariant) mass? Thus, particles with zero invariant mass can have non-zero momentum.


0

Since the two slit experiment is a bit complicated for what I'm about to discuss, allow me to consider a simplified toy model. Consider an $N$ component state vector $| \alpha \rangle $, about to be acted on by some operation, and subsequently measured. This evolution can be represented by: $$| \beta \rangle = U |\alpha\rangle$$ The important bit here is ...


1

Assuming our only aim is to solve double slit experiment (or other problems that can be mapped into that). Actually the double slit experiment for electrons is a derivative/prediction from the quantum mechanical theory as it started with the Schrodinger equation ,its wavefunction solutions and the interpretation of differential operators with energy ...


3

You say that we are only interested in the probability distribution on the screen, $\rho(x,t) = \lvert \psi(x,t) \rvert^2$, which is essentially correct. So, why do we have $\psi(x,t) = \lvert\psi(x,t)\rvert\mathrm{e}^{\frac{\mathrm{i}}{\hbar}S(x,t)}$? Well, looking at the time evolution equation for the probability density, the continuity equation of ...


3

The non-negative real probability distribution can't interfere like a complex wave function can. To produce interference phenomena it is necessary for quantum mechanics to deal with probability amplitudes, not just probabilities.


1

The answer to your question is yes and there are experiments which use multiple excitations. A very famous one is the Lamb-Rutherford-Experiment where they could prove the existence of the lamb shift. First they excited a beam of hydrogen atoms which were in the $1S_{1/2}$ groundstate into the $2S_{1/2}$ state by bombarding them with electrons. This has a ...


2

Multielectron atom has much more complex energy spectrum than hydrogen atom. As the electrons interact with each other, the hydrogenic energy levels get shifted, and much of the hydrogen-specific degeneracy, as well as degeneracy resulting from electrons mass&charge equality, is lifted. Moreover, since the electrons do interact with each other, we can't, ...


2

The simplest example I can think of to illustrate this is the spectrum of the hydrogen atom. The excitation of the electron from the ground state, n = 1 produces a series of absorptions known as the Lyman series. The first line is excitation of the electron from n = 1 to n = 2, the second line is n = 1 to n = 3 and so on. But there are also absorptions due ...


0

I think it's pretty typical in nature. Some elements have a a very large number of electrons, too. I would be interested if somebody could answer what the lifetime of an excited electron is. In classical QM the excited electron would stay excited forever. QED is needed to explain the electron dropping back to the ground state and releasing a photon before ...


0

The electron density is proportional to the square of the wavefunction. More precisely it's: $$ D \propto \Psi\Psi^* $$ where $\Psi^*$ is the complex conjugate of the wavefunction. The wavefunction can be complex, but the product $\Psi\Psi^*$ is always real, which is just as well since the electrons are real too. So a diagram of the electron density looks ...


0

Well, according to the wild ER=EPR conjecture by Maldecena and Susskind, two entangled electrons are connected by a quantum wormhole. The mechanism and details of this quantum wormhole are left unspecified by these authors, though.


0

Electrons are very close to the energy of self-capacitance of a quantum of charge. The size of the electron is very close to $r_e$, the energy supposed if one tries to charge a sphere of that radius with a single electronic charge, ie $mc^2 = e^2/4\pi\epsilon r_e$.


-3

In light of the fact, documented in the paper by Carron, N. J., that the simplicity of the system in question allows for a gauge invariant formulation of Maxwell's Laws (not requiring Lorentz invariance -- since everything is motionless), the answers thus far given are inadequate due to their appeal to the arbitrary choice of gauge. It is for this reason ...


6

There is no universally accepted quantum theory of gravity. Quantumly, the "shape" of a fundamental particle is a very fuzzy notion - we know that states are often not localized, so it is wholly unclear what it means to say "the electron is pointlike". The proper formal interpretation of a "pointlike particle" is simply a particle that is not composite - ...


1

The electron gun produces electrons by heating the cathode; this shakes out electrons from the metal ("boils them off"), and as soon as they're out, they are repelled and accelerate away to do your bidding. This is called Thermionic Emission. When an electron is emitted, another one comes in from the cable connecting the cathode to the power source. The ...


0

You would need to supply an infinite amount of energy to an electron (or any massive particle) to accelerate it to the speed of light. The (theoretical) absence of a decelerating force simply implies that the electrons won't slow down once they have been accelerated to a certain velocity.


0

You are correct that Aharanov and Bohm's effect means that the potential $A^\mu$ cannot be blithely disregarded. However, that doesn't mean it's physically meaningful or well defined in the way you seem to imagine. It's still a gauge variant quantity meaning it can be locally set to zero by a gauge transformation. Its physical significance (independently of ...


0

The diagram for A resembles that of a field due to a dipole that varies inversely as r cube which is not the case here though directions of A are OK. However the situation here is that of a current sheet of a cylinder that has been deformed so that the ends meet. There is no B outside. By redefining A all over the space you end up with zero A outside and ...


6

First, there is nothing wrong with our charge polarity conventions. They are predict electrical phenomena just as accurately as the opposite convention would have done. Did we have a problem if since the begin of their discovery we called them positive particles and negative to protons? We could predict the behavior of electrical phenomena equally ...


1

There are two kinds of momentum. One kind is simply a frame dependant portion of a larger tensor. It is exemplified in the total stress-energy tensor, which is a symmetric rank two tensor that is divergence free. The divergence free part means that is is conserved locally in in the sense that the momentum (or energy) in a region at one time, is equal ...


2

Nice question. This is connected to gauge invariance. The interaction term in the Lagrangian (interaction between charge and field) is not gauge invariant. Thus, a curl-free vector potential, which corresponds to zero magnetic field, appears to have a non-zero momentum. Nevertheless the equation of motion is not changed in a new gauge, i.e., you will get ...


0

The system of equations belong to the ESU, and presumably gaussian, where charge^2 = force*length^2. Putting this in, you get $F^2 * L^4 / F * L * L^3$ gives F = E/L. Hint, the correcting factor for SI is to replace $e^2$ with $e^2/4\pi$. If you really are going to read a lot of older physics, it is best to be versed in the gaussian units: their units ...


0

Without having access to the book, the best guess I can come up with is that the formula is written in a natural unit system where $c=1$, $\hbar=1$, and $e=1$. In this system, charge is unitless, and energy has the same unit as inverse length. So, using $\equiv$ to mean "congruent to" (i.e. having the same units), ...


0

UPDATE Thank you for posting back your thought experiment. My calculations apply again. I think I found the source of your confusion. Your analysis is missing the displacement current. You seem to think that the linear momentum of the electron could only go back to the current in the torus: Hence there is at least one scenario in which the electron can ...


2

We do consider that energy. It reduces the amount of energy needed for the electron to be freed from the surface. An analogy: If a satellite is already in orbit, you need less energy to make it escape earth's gravity than if you started with the satellite on a launch pad on earth. The energy of the satellite in orbit is like the energy of the electron ...


2

Exactly what is this "circuitous route"? Does the thing I touch also have to be touching the carpet? Though I'm not a native English speaker I am pretty sure that a circuitous route is a path that combines you shoes and the carpet as were they a part of a circuit. The thing you touch has to be connected to the carpet (by touching the carpet itself or ...


7

First, the electron is not a point particle. The abstraction you are thinking of is what we would call a naked electron. In an experiment, you do not see the naked particle ever. It is always surrounded by virtual pairs. Hence, what you measure as the electron is really a many-body system. Second, you might want to read this. The take-home message is "the ...


0

It sounds like you want an electron, a statically charged torus, and then to consider the two cases of the torus with and without a current. You also bring up the canonical momentum conjugate to position: $\vec{p}+e\vec{A}$, which evolves according to: $$ ...


0

Ferromagnetic materials exhibit a long-range ordering phenomenon at the atomic level which causes the unpaired electron spins to line up parallel with each other in a region called a domain. Within the domain, the magnetic field is intense, but in a bulk sample the material will usually be unmagnetized because the many domains will themselves be randomly ...


0

When you cool it the domains so formed align to get minimum energy and that is not the energy in which all are pointing in the same direction.


1

Since you don't have velocity information on the electron, then it doesn't have momentum, so any equation involving p can't be used. Just convert the electron's rest mass to energy directly, using E=mc²


1

The micro world of atoms and molecules is ruled by quantum mechanics. The forces controlling the interactions between them at the level affecting everyday situations, as touching, are electromagnetic. In quantum mechanics the electrons are in bound states around the atoms, and at most can be mobile in bands when in a solid, i.e a bound state due to the ...


0

There are bound electrons and there are free electrons - the former interact with radiation and electric fields and the like by bouncing around energy levels and emitting photons (dissipating energy, as in a resistive element), the latter do actually move around and are strictly not localized around the nucleus of a few atoms, and largely preserve energy ...


0

I need to marry the classical model of the colliding electrons and atoms of the medium with the Quantum principles involving the excitation of charges to higher energy bands and emission of photons. My answer to this question explains something about the concept of the conduction band. Think of it this way. In all the filled shells (including shells ...


0

The Fermi function or Fermi-Dirac distribution was developed as a quantum mechanical expansion of the Boltzmann distribution. With it, you can describe the free electron gas appropriately, whereas the Boltzmann distribution fails by predicting zero energy for all particles at absolute zero (see Pauli exclusion principle). The fermi-dirac distributiom arises ...


1

There are two important contributions: First the density of states, which tells you the states, which can potentially be occupied. Then there is the Fermi-Dirac distribution, which tells you, which energies are occupied. The Fermi-Dirac distribution does not include allowed and forbidden states. You must fold it with the density of states, which is then ...


1

In a (perfect) semiconductor, there are no electronic states available in the band gap. Therefore, despite the fact that the Fermi-Dirac distribution "predicts" (or better: accounts for) electrons with energies lying in the band gap, they cannot exist there because no states are available. If the Fermi level lies in the band gap, the undoped semiconductor ...



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