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Thinking of the photon itself as a single particle, and imagining the building up diffraction patterns with one photon in the apparatus at a time will give you an idea of the correspondence between classical wavelike behaviour and the probability waves of the wavefunction, at least for bosons. The following, I believe, is what Akrasia's answer means when ...


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Double slit experiment with electrons is a demonstration of a quantum behavior. When we say observe we mean expose to some kind of interaction. So when electron travels from its source towards the double slit and then passes through, and hits the detector, we see that it is a particle. But, when many electrons pass through the slit we observe a interference ...


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A free moving electron does not emit photons. If one want to see an electron you have to illuminate it. Sending photons, the electron reflect (absorb and re-emit) photons which one can detect then. By doing this the electron get disturbed and changes his direction. The fringes on the observation screen will be destroyed. Since one cannot synchronize the ...


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When you send the photons through the double-slited wall, they form a diffraction pattern on the other side, which is a wave-like phenomenon. It makes no sense to think of the photons as "particles" anymore, since they would need to cross both slits simoultaneously in order to create the diffraction pattern. So a way to find out what's happening is to place ...


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Your question is not silly. In the two slit expariment you measure the position if the photon, that is a particle-like quality. What happens is that a photodetector records when it absorbs a photon. That gets recorded in the memory. Some schools of physics interpret the interaction of the photon with the detector as a measurement. Others interpret that the ...


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We know from translations symmetry that the expectation value of the wavevector operator is constant -- that is, $$ \langle \mathbf{k} \rangle = \langle \mathbf{k}_1 + \mathbf{k}_2 + \ldots \rangle = \text{const.} $$ In other words, wavevector is a conserved quantity. If the constant of proportionality between $\mathbf{k}$ and momentum were different for ...


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The second answer is correct, some problems need wave-treatment, others may be treated with the particle approach. Let's make clear the specific of each case: in the quantum theory an object (which typically is microscopic) is described by a wave-packet. In some experiments, and for some particles, this wave-packet is small (we say in this case that the ...


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Each photon has an oscillating electric and a magnetic component. In vacuum the components are always perpendicular to each other and also perpendicular to the motion of the photon (see this sketch). When Ppotons are unpolarized, these oscillations are equally distributed in 360 °. A polarization grid can rotate about 50% of the photons so that they can pass ...


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In physics, complementarity is a fundamental principle of quantum mechanics, closely associated with the Copenhagen interpretation. It holds that objects have complementary properties which cannot be measured accurately at the same time. The more accurately one property is measured, the less accurately the complementary property is measured, according to the ...


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The photon as an elementary particle is special: the quantum mechanical wave equation whose solutions squared will describe the probability of finding the photon at a given phase space point is the Maxwell equation that describes the classical electromagnetic wave, except it is the potential form of it and it acts as an operator on the wave function of ...


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To answer the question in your title - no, the kinetic energy of an electron is a function of its velocity, same as for any other particle. The charge of an electron is always $1.6\cdot 10^{-19}C$, and so the electron will pick up $1.6\cdot 10^{-19}J$ of energy for every Volt of accelerating potential. The key to answering the question (part d in the ...


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0.16 aJ is the energy an electron acquires by going through 1 volt. If it goes through 500,000 volts, it gets an energy almost equal to its mass.


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No. Kinetic energy depends on how much energy you give to an electron. $Volt = Work$ $done$ / $ unit$ $ test$ $ charge$. $1.6 * 10 ^ -19 J$ is the amount of work done to accelerate an electron of charge $1.6 *10^-19$ to $1$ $volt$ potential.


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Not at all. Space is expanding, as in space is constantly being added. Although space might be added between the electron and nucleus, that does not effect the atom to significant degree. Its like having an atom in flatland and turning the flatland into a ball. the atom is not going to know much of a difference. If the sphere is returned to flat land the ...


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hv is the energy per photon, the power radiated is essentially this times the number of photons emitted per second, so a constant frequency antenna with variable power is emitting a variable number of photons per unit time.


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Light diffracts through any object but the wavelength must be greater when compared to the size of the object. In this case it is slits so.


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A particle does not need to have a hard edge. It can for example, be a density function, which sort of fades to zero. One might note that waves can intersect each other and come out as if the other wave was not there. Particles with hard edges are more an artefact of our minds rather than 'what's really there', until we get some real evidence otherwise. ...


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I like the way you're thinking, but I'd like to try to open your mind a bit further. A lot of our everyday concepts are a product of the net behaviour of billions of particles following the laws of physics. For example, consider a steel ball. It has a "hard edge". What does this mean? It means another ball can be anywhere outside of the hard edge, but ...


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This is one of the key results of quantum field theory: particles are not single points, they are disturbances in quantum fields that are spread out over space. Typically the disturbance is not spread out very much, otherwise it looks more like what we know as a wave than a particle. The technical term for what you're calling a "smear" is a wavepacket.


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In fact any object with non-zero physical extent is required to be deformable by relativity (but see below); otherwise pushing on it can transmit energy and information faster than light. And quantum mechanics rules at those scales so everything does have fuzzy edges. At the scale of the very small there are no sharply defined objects. A consequence of this ...


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This is misconception that light is some kind of 'mix' of waves and particles. Instead, It actually IS both waves and particles at the same time, you can't separate them from each other. So probably, the answer could be: you see particles as well as waves.


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It would be physically impossible to be able to "see" light as anything other than a particle (photon). The only time photons, or any other subatomic particle for that matter, can be described as a wave is when we are NOT looking at them.


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You are seeing particles. However there's more to this than meets the eye so I need to explain exactly what I mean by this. Light is neither a particle nor a wave. Instead it is a quantum field. As a general rule while light is travelling it appears as a wave, but when the light quantum field is exchanging energy with anything it does so in quanta that ...


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Comment to the question (v2): The question formulation conflates on one hand a non-relativistic particle in an infinite potential well/box, which has a quadratic dispersion relation $E \propto p^2$; and on another hand a massless relativistic dispersion relation $E \propto p$. See this Phys.SE question and links therein for a similar misunderstanding.


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Wave equations have a long history in physics, they are usually equations involving second derivativs, the solutions are sinusoidal ( sines and cosines) and have been used to model classical waves, starting from water, sound, pressure waves, and finally light classically, with Maxwell's equations. When Schrodinger's equation was able to reproduce the Bohr ...


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The quantum wavefunction is not a wave in the classical sense. In particular, it is not a wave obeying $E = \hbar \omega$, as electromagnetic waves do. The term wavefunction is, in this sense, just a bad naming decision. It is no wave, there is nothing oscillating, and it has no connection whatsoever except the mathematical form to physical waves.



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