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eljose
Sep6-05, 11:46 AM
Let be a Hamiltonian in the form H=T+V we don,t know if V is real or complex..all we know is that if E_n is an energy also E*_n=E_k will be another energy, my question is if this would imply V is real...

my proof is taking normalized Eigenfunctions we would have that:

(<\phi_{n}|H|\phi_{n}>)*=<\phi_{k}|H|\phi_{k}>

so the expected value of T is always real,then we would have the identity with the complex part b(x) of the potential:

\int_{-\infty}^{\infty}dx(|\phi_{n}|^{2}+|\phi_{k}|^{2})b (x)=0

for every k,and n so necessarily b=0 so the potential is real and all the eigenfunctions would be real.

Galileo
Sep7-05, 02:00 AM
I find the question a bit weird. The energy of a system is an observable and its corresponding operator is the Hamiltonian and is thus hermitian which implies having real eigenvalues. If you take V nonreal H isn't Hermitian anymore.