Is the Potential in Quantum Hamiltonian II Always Real?

AI Thread Summary
The discussion centers on whether the potential V in the Hamiltonian H = T + V is necessarily real if the energies E_n and E_k are related as E*_n = E_k. The argument presented uses normalized eigenfunctions to show that the expected value of the kinetic energy T is real, leading to the conclusion that the complex part of the potential, b(x), must equal zero. This implies that V is real, resulting in all eigenfunctions being real as well. Additionally, it is noted that if V were non-real, H would not be Hermitian, contradicting the requirement for energy eigenvalues to be real. The conclusion emphasizes that the nature of the Hamiltonian ensures real eigenvalues when V is real.
eljose
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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.
 
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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.
 
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