- #1
PeteSampras
- 44
- 2
In a text a exercice says that for the Hamiltonian
##H_0 = \frac{p^2}{2m}+V(x)## the eigenfunction and eigen energy are ##\phi_n, E_n##. If we add the perturbation ## \frac{\lambda}{m}p## ¿what is the new eigenfunction?
The solution is
## \frac{p^2}{2m} + \frac{\lambda}{m}p+V= \frac{(p+\lambda)^2}{2m}- \frac{\lambda^2}{2m}+V##
but, the solution says ##\psi= exp(-i x \lambda/ \hbar) \phi_n##
I understand that ##H_0 \phi_n = E \phi_n## but,
¿why the solution ##\psi## has it form?,
##H_0 = \frac{p^2}{2m}+V(x)## the eigenfunction and eigen energy are ##\phi_n, E_n##. If we add the perturbation ## \frac{\lambda}{m}p## ¿what is the new eigenfunction?
The solution is
## \frac{p^2}{2m} + \frac{\lambda}{m}p+V= \frac{(p+\lambda)^2}{2m}- \frac{\lambda^2}{2m}+V##
but, the solution says ##\psi= exp(-i x \lambda/ \hbar) \phi_n##
I understand that ##H_0 \phi_n = E \phi_n## but,
¿why the solution ##\psi## has it form?,