# Quantum problem

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 $$E_{n}$$ is an eigenvalue of the Hamiltonian also its conjugate $$E^*_{n}=E_{k}$$ will also be an eigenvalue we have necesarily that V is real.. ..my proof is that

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

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

$$<\phi]p^{2}+V[\phi>=E_{k}$$
(3) ow equating down 2 and 3 we would have that

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

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

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

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

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Are you asking if your work is right? If so, I got lost at the third line. Is that a new wavefunction, or is it the same as the one from line 1? Your notation suggests it's the same, which doesn't follow from your work. And it's easy to prove p^2 is hermitian if p is.

Another question how i would prove that $$<\phi|p^{2}|\phi>=<p\phi|p\phi>$$ ?..we know p is an observable so p=p+...

eljose said:
Another question how i would prove that $$<\phi|p^{2}|\phi>=<p\phi|p\phi>$$ ?..we know p is an observable so p=p+...
$$<\phi|a\phi>=<a*\phi|\phi>$$ and since p*=p $$<\phi|p^{2}|\phi>=<p\phi|p\phi>$$

another question,let,s suppose we have the identity:

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

valid for all the k and n eigenfunctions of a given Hamiltonian, then necessarily we must have that f=0

Galileo