Recent content by landpt

Yes, you're right... It should be zero (therefore there is no first order correction of those energies). I found out today that there was a mistake in the equation of the potential - it should be only \lambda x y instead \lambda x y z. Now there is actually a correction. Thanks. About...

Homework Statement A particle with mass m moves in the potential: V(x,y,z) = \frac{1}{2} k(x^{2}+y^{2}+z^{2}+ \lambda x y z) considering that lambda is low. a) Calculate the ground state energy accordingly to Pertubations Theory of the second order. b) Calculate the energies of...