https://physics.stackexchange.com/q/149004/ was put on hold as too broad, but edited the same day. Now it's been on hold for a few days. Is it still too broad?
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$\begingroup$ For those with <10k rep, the question can be viewed here. A newer version of the question can be found here. $\endgroup$ – Kyle Kanos Dec 3 '14 at 14:41
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Questions that are edited after they are closed enter the reopen queue. There has not been a single reopen queue cast on that question, indicating that the reviewers still thought it too broad.
My personal view is that this is a mixture between too broad, unclear what you're asking and non-mainstream:
You just throw the matrix exponential in there and essentially ask "Is this a wavefunction?". I bet one can write down something looking like a wave equation to which that exponential would be the solution, but what do you mean by "[...]is there a natural way to extend the previous wavefunction interpretation[...]"? What was the previous wavefunction interpretation? (Just writing down an exponential doesn't make it a "wavefunction"!) Why would you expect the matrix exponential to have the same interpretation? What system are you trying to model? It's really not a very focused question.
answered Dec 1 '14 at 15:29
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$\begingroup$ In the scalar-variable case, k is the wavenumber. In the vector-variable scalar function, k is the wave-vector. Since we can define the same function with matrix variables, it seems valuable to ask if some properties of a matrix k (norm, eigenvalues, etc.) determine the behavior of the system in an analogous way as the scalar and vector k's. You say: "Just writing down an exponential doesn't make it a "wavefunction"!. Very good! Please don't stop there but answer the question. A good answer to the question would illustrate why this particular function is or isn't a wavefunction. $\endgroup$ – Andrew Dec 1 '14 at 21:39
