Therefore,We should have the inequality (see Problem5.33)where is the true ground state energy.b) To find the exact ground state of the system, notice that odd wave
functions of a symmetric oscillatorproblem(from to )will also be
solutions for these boundary conditions since they tend to zero at
Therefore, the ground state wave function of this halved oscillator will corre-
spond to the first excited state wave function of the symmetricaloscillator.
The wave function can easily be obtained if you take the ground state
and act on it by the creation operator (see Problem 5.16):
The ground state energy of our halved oscillator will in turn correspond to
the first excited state energy of the symmetricaloscillator:Comparing this result with that of (a), we see that the inequality (S.5.36.13)
holds and that our trial function is a fairly good approximation,since it
gives the ground state energy to within 15% accuracy.
QUANTUM MECHANICS 285