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The only thing that's really important is the differential eqaution. The situation is, outside the well, in both cases: $\dfrac{d^2 \psi}{dx^2}= - \frac{2mE}{\hbar^2} \psi$ Now it's foundamental notice that for bound states E<0 so we can write: $E=-|E|$ and Sc. equation become: $\dfrac{d^2 \psi}{dx^2}= + \frac{2m|E|}{\hbar^2} \psi$ So the usual way ...

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It is like the interesting proposition in simulation hypothesis which paraphrased goes, if we could create a computer simulation to understand reality it would be the approximate size and age of the current entire universe.

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The white colour of clouds is due to Mie scattering. This arises because the refractive index of water is different from air. If the particles are large enough all wavelengths are scattered equally so there is no change in the colour of the scattered light. This is the case in clouds where droplet sizes are typically 10 to 20 microns. The light scattering ...

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There are two possible Feynman graphs for Bhabha scattering at tree level. I have shown them below. Are you asking what will go wrong if these two are modified like shown below. If this is what you are asking then the only thing that we should be concerned about is the conservation of charge at each vertex point. We can ...

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The thing is, the electrons are known (even at the original time of this experiment) to be very light compared to a alpha particle very weakly bound compared to the kinetic energy of the alpha particle If an alpha hit one head on, the electron would be ejected at high speed but the alpha would be barely deflected. And at the energies used in the ...

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Short version: In the infinite potential well, $E \geq 0$ (because $V_{min}=0$, and $E \geq V_{min}$). In your finite potential well, it sounds like you are looking for bound states, in which case $E < 0$, so you absorb the negative into the square root. Long version: When you are tackling a QM problem, first you should figure out the admissibility of ...

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Followup to my comment: take a look at the table here http://hyperphysics.phy-astr.gsu.edu/hbase/starlog/staspe.html . Ignoring for the moment the spectral absorption lines, which shouldn't matter much to the Mark-I Human Eyeball, use the nominal temperatures of each star class in combination with the standard Planck distribution curves for a black body ...

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Good question; I remember spending hours trying to understand this when I first learned QFT. Let's address your two main points in turn. First, you say I don't understand how rhyme these two different pictures. Let's outline how to connect the two pictures in steps. It's a good exercise to try and work through all of the gory details yourself, so I ...

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