Heisenberg picture describes emission, Schroedinger picture does not
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Jul24-11, 02:55 PM
Anyway, the way I see it there is no distinction between whether H or S is more real. Neither picture, as you pointed out, makes any difference when one considers only observable quantities, i.e. the matrix elements of the operator under question.
When we use classical terminology such as energy to describe quantum concepts it is easy to forget that we are describing reality as predicted by a mathematical model something that human languages weren't designed to do.