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The definition of the expectation value of an operator A is \begin{equation} \langle A\rangle=\int{\psi^* (x) A(x) \psi (x) dx} \end{equation} (because it represents "the value of the variable" $A(x)$ times "the probability of being in that configuration" $P(x)=\psi^* (x) \psi (x)$) and for the particular case of the expectation value of the position ...


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You are imagining the particle in the well as a classical system i.e. a point particle moving to and fro in the well. However this is not a good description of the system. A quantum particle does not have a position. By this I mean that it is meaningless to ask what the position of the particle is because position, in the sense we normally use the term, is ...


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So the principle that systems find themselves at an energy minimum needs to be heavily qualified: it is a consequence of ever-present friction/drag forces having a sign opposite to your velocity, so that the power wasted due to friction is terminally negative. The principle therefore states, "when a system only has two forces, one due to drag and the other ...


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A particle subjected to only a conservative force field, without other restrictions, will move in a direction which reduces the potential energy of the system. The particle itself does not have potential energy; the particle-field system has potential energy (some may say the field-field system, but let's not nit-pick that yet). So what's important is not ...


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You have to integrate (i.e. find the primitive) in $dy$ as well, and then compare derivatives of both equations: $$ f(x,y) = \int 1 d x = x + g(y) $$ $$ f(x,y) = \int 2 d y = 2 y + h(x) $$ now compare derivative terms $$ \frac{\partial f}{\partial x} = h'(x) = 1 $$ and $$ \frac{\partial g}{\partial y} = g'(y) = 2 $$ therefore $$h(x) = x + c_x $$ ...


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Both the zeroes are "real" to answer the question. There is absolutely no problem with multiple points in a space having the same electric potential. Note that we have ASSUMED the potential to be zero at infinity and based upon that assumption we have found out the potential to be again zero at the equidistant point between two opposite charges. So there ...


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However it did pass within Δx of the electron. The Δx is not the difference in space with the electron, as the electron is bound to a nucleus with a potential simulated by "an infinite potential well" . The Δx is related to the whole system, from the center of its mass as a possible location to start with. So the problem is : "photon + atom" as a ...



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