True. But then, what does it means that r in U?
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Replying to @VergaraLautaro
you could just as well ask what does the x,y,z mean in *any* potential V(r) mean.. They are just the coordinates or the electron--whatever they may be..
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Replying to @InertialObservr
Perhaps is just the language. My fault that I didn't thought in these terms for a while. But such questions came to me few days ago. You just gave me the opportunity to ask you. :)
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Replying to @VergaraLautaro @InertialObservr
If one says U = e^2/r, one thinks on 1) a given position 2) a localized charge. which is usually not always the case the case in QM, not even in a bound state like an atom.
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Replying to @VergaraLautaro
but all of that is covered from the fact that its wave function must obey the wave equation.. if you ask where the particle is you have to integrate it against the probability distribution defined by the schrodinger wave equation, where the coordinates are treated as random vars
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Replying to @InertialObservr
Of course, but one could say that that is a posteriori.
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Replying to @VergaraLautaro
i'm not sure what you mean.. the coordinates are treated as random variables of a distribution that need to be integrated over for meaningful results.. this isn't just physics.. just because a distribution f(x,y) has coordinates x,y doesn't mean x,y definitely happen
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Replying to @InertialObservr
I mean, one first establishes the Hamiltonian, which is a function, not a distribution and afterwards one writes down the Schrödinger eq. and voilá QM and so on. I'm thinking about the Hamitonian and the meaning of its terms, with QM. Perhaps my way of thinking is not well posed
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Replying to @VergaraLautaro
you can just think of the terms of the Hamiltonian in position space as H(x,y,z) such that H(x,y,z)|ψ|^2 d^3x is the average energy in a box of volume d^3x located at (x,y,z)
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Replying to @InertialObservr
Well, in this sense, the Hamiltonian in QM only has a meaning within a distribution.
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that's right, at least in position space
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