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your answer is hidden in the self organized criticality (SOC) concept. there is many model that have applied in many branch of science that obey SOC. this concept for the first time was introduced by Bak et al with "Abelian Sand pile model". this model is the simplest model can receive to critical state (showing power law and infinite temporal-spatial ...

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One calculates the probability that the electron is inside the nucleus by integrating $\psi^*\psi$ over the volume of the nucleus. For example, the radial part of the hydrogen ground state wavefunction is $\psi=\frac{e^{-\frac{r}{a_0}}}{\sqrt{\pi a_0^3}}$, so the integral is $\frac{1}{\pi a_0^3}\int_0^b e^{-2r/a_0} 4\pi r^2 dr$. In the above, $a_0$ is ...

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In this link you will see the radial hydrogen wavefunctions. It is only the l=0 states, S states, that have a value different than zero at r=0. The other angular momentum states get a very small contribution to the probabilities from r>0 to r=1 fermi ( the charged radius of the proton) as 1 fermi is of order 10^-15meters, and the probability is the ...

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The type of interaction depends on the energy of the photon, based on the Klein-Nishina formula. From Wikipedia: Very low energy photons (visible light; as long as the photon energy is much less than the mass energy of the particle, i.e. Compton wavelength) yields Thompson scattering, which is elastic scattering with electrons. Low energy photon (a few ...

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Does a proton have a "bandgap"? If yes, what happens when a photon is absorbed by a proton? For single protons, as in a plasma , there exists Compton scattering . The photon transfers part of its energy to the proton and scatters off at a lower energy/frequency, the proton taking up the energy-momentum balance. This is a continuous spectrum, from very ...

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A system can absorb a photon if the energy of the photon matches an excitation in the system. So the hydrogen atom can absorb a photon if its energy matches one of the frequencies in the hydrogen spectral series. A proton is a composite object and it does have a spectral series. However the excited states of the proton involve rearrangements of the energy ...

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The mathematical formalism starts from a C*-algebra $A$, that of observables, and a set of physical states. For any observable $a\in A$, i.e. a self-adjoint element ($a^*=a$) and a state $\omega\in A^*$, the expectation value of $a$ on the state $\omega$ is simply $$\omega(a)$$ If you consider the C*-algebra generated by $a$ you can then apply the ...

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Probabilities of independent outcomes of the same measurement are always additive, which means that the expression $$P(U) = \sum_{n\in J}P(a_n),$$ is perfectly correct. If you want something that looks more formal, you can express $P(U)$ in as the expectation value of an appropriate operator, the projector  \Pi_U=\sum_{n\in ...

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Yes, your interpretation is correct, but only if the probabilistic nature of these events is caused by quantum mechanics. To give a counterexample. If you throw a die,the probability of throwing six is around 16.6 %. However in this case there is no splitting of worlds. The reason is that in this case the uncertainty in the outcome is a result of an ...

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Is it true that if $u_n(x)$ and $u_m(x)$ are orthogonal (which is true), then $u_n(x)$ and $u_m(x)f(x)$ will be also orthogonal? No. The simplest example of this is the case $f(x) = u_n(x)/u_m(x)$ for whatever $n$ and $m$ you're considering. More broadly, the result you're trying to prove is false. Consider the infinite square well between \$\pm ...

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Born calculated solution to Schroedinger's equation corresponding to electron scattering experiment and what he got was continuous function of scattering angles measured with respect to the original direction of propagation of electrons. However, in experiment electrons are always detected at definite points of a screen. Clearly, there is no direct match ...

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in a similar manner in which a fission reaction is moderated I think you are confusing ideas here, and it is leading you to ask a somewhat nonsensical question. The individual nuclear fission events don't change at all in a moderated nuclear pile. What does change is the odds of a neutron from one causing another and how long it take for that ...

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If you think in plain simple terms, probability arise from the discrete nature of interactions between entities. This is the real deal which turns everything so strange and interesting. Suppose you have 3 entities A, B and C, where A is the source of some perturbation to be sent to B and C at the 'same time'. Let's think of the perturbation in practical ...

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Within current quantum field theory, it does not make sense to ask "how long" a particular process takes to occur. There is a certain probability that a particle and its antiparticle annihilate. But there is no concept of a "process of annihilation". There's the in-state (particle and anti-particle) and the out-state (products of the annihilation, usually ...

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