- #1

Karlisbad

- 131

- 0

[tex] \bold H =0 [/tex]

then it's obivious that if you want to get its energies you would get [tex] E_{n}=0 [/tex] for every n, this is a non-sense since you must have positive energies (and a ground state) ...then if we "cheat" and make:

[tex] \bold H \Phi =i\hbar \frac{\partial \Phi}{\partial u}=\lambda_{n} \Phi [/tex] (2)

where u is a parameter that dictates the "evolution" of the system..of course we must have that [tex] \frac{\partial H}{\partial u}=0 [/tex]

then using (2) we get the energies of the system.. but are they "really" the energies of our initial system with H=0 ? i don't think so..perhaps we should put [tex] \lambda=0=\lambda (n_x ,n_y, n_z,n_t) [/tex] and from it recover the initial energies or consider that the factor..

[tex] \lambda_{n} \Phi [/tex] is an extra term inside our potential.. how you can quantize H=0 and get finite and positive energies?..