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Quantum Hamiltonian II

  • Thread starter 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:

[tex](<\phi_{n}|H|\phi_{n}>)*=<\phi_{k}|H|\phi_{k}> [/tex]

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

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

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

Answers and Replies

Galileo
Science Advisor
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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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