Some Hermitian Hamiltonians like H = p2 + x 2 + x2m+1
do not have boundstate behaviour . If this is so ,why we consider all hermitian operators yield real eigenvalues ? The basic formulation of quantum mechanics.
@Osamah
I know this . My question shows finger on postulates of QM . Further argument of Bender using PT symmetry also fails in many more systems.
I dont see the problem. For a bound system there are just
real discrete eigenvalues. For an unbound system you have a continuous real function E(k) describing the spectrum.
There are proofs that Hermitian operators do have real eigenvalues.
Actually, your question is quite complicated. Part of the answer lies in the difference between symmetric operators and Hermitian operators. The latter always have real eigenvalues, the former may have complex eigenvalues.
An operator like the one you mention, say for m=1, is in fact only symmetric when defined straightforwardly. To make it Hermitian, one must define a boundary condition at infinity. If you do that, then the spectrum of the Hamiltonian will in fact turn out to be discrete, even though the potential is unbounded from below.
https://en.wikipedia.org/wiki/Extensions_of_symmetric_operators
@Leyvraz
Thank you for your understanding of hermitian operator relating to spectrum. In fact this was in my mind .Hermitian matrix will not give real eigenvalue always but hermitian operator with bounded wave function will yield the result .
Regards
B.Rath
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