- #1
eljose
- 492
- 0
let be a particle in a potential well with mass m=1/2 so we have the equation:
[tex](p^{2}+V(x))\phi=E_{n}\phi [/tex]
we don,t know if V is real or complex but we have that if En is an energy,its complex conjugate En^*=Ek is also another energy of the system,my question is if the potential is real...
Proof?:taking normalized Eigenfunctions of the Hamiltonian...with [tex]<\phi|\phi>=1 [/tex] then we would have:
[tex](<\phi_{n}|T+V|\phi_{n}>)^{*}=(<\phi_{k}|T+V|\phi_{k}>)[/tex]
so in the end separating and knowing that [tex]<\phi|p^{2}|\phi> [/tex] is always real then we would have that:
[tex]\int_{-\infty}^{\infty}|\phi_{n}|^{2}V^*(x)-int_{-\infty}^{\infty}|\phi_{k}|^{2}V(x)=r [/tex] with r a real number...
so we would have for every k and n and complex part of the potential b(x) that:
[tex]} (|\phi_{n}|^{2}+|\phi_{k}|^{2})b(x)dx=0 [/tex]
so the complex part of the potential is 0...is that true?
[tex](p^{2}+V(x))\phi=E_{n}\phi [/tex]
we don,t know if V is real or complex but we have that if En is an energy,its complex conjugate En^*=Ek is also another energy of the system,my question is if the potential is real...
Proof?:taking normalized Eigenfunctions of the Hamiltonian...with [tex]<\phi|\phi>=1 [/tex] then we would have:
[tex](<\phi_{n}|T+V|\phi_{n}>)^{*}=(<\phi_{k}|T+V|\phi_{k}>)[/tex]
so in the end separating and knowing that [tex]<\phi|p^{2}|\phi> [/tex] is always real then we would have that:
[tex]\int_{-\infty}^{\infty}|\phi_{n}|^{2}V^*(x)-int_{-\infty}^{\infty}|\phi_{k}|^{2}V(x)=r [/tex] with r a real number...
so we would have for every k and n and complex part of the potential b(x) that:
[tex]} (|\phi_{n}|^{2}+|\phi_{k}|^{2})b(x)dx=0 [/tex]
so the complex part of the potential is 0...is that true?