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eljose

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**Quantum problem...:)**

let be the next quantum problem we have a Hamiltonian with a potential V so H=T+V we don,t know if V is real or complex the only thing we now is that if [tex]E_{n} [/tex] is an eigenvalue of the Hamiltonian also its conjugate [tex]E^*_{n}=E_{k} [/tex] will also be an eigenvalue we have necesarily that V is real.. ..my proof is that

[tex] <\phi]p^{2}+V[\phi>=E_{n} [/tex] taking the conjugate:

[tex] <\phi^*]p^{2}+V^{*}[\phi^*>=E^*_{n}=E_{k} [/tex] (2)but E_k is also an energy so we would have the equation:

[tex] <\phi]p^{2}+V[\phi>=E_{k} [/tex]

(3) ow equating down 2 and 3 we would have that

[tex]<\phi^*]p^{2*}+V^{*}[\phi^*>= <\phi]p^{2}+V[\phi> [/tex]

so we would have in the end that [tex]<\phi]V(x)[\phi>=<\phi^*]V^*(x)[\phi^*> [/tex] so the potential would be real...

we have that [tex] <\phi][x>=\phi^*(x) [/tex] and that [tex] <x][\phi>=\phi(x) [/tex] we would also have that [tex]\phi^*(x,E_{n})=\phi^(x,E^*_{n})=\phi(x,E_{k}) [/tex]

Another question is suppsed that p^2*=p^2 and that [tex]<p\phi][p\phi>=<\phi]p^{2}[\phi> [/tex]

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