(a) Generalize Problem 8.2, using the trial wave function
ψ(x)=(x2+b2)nA,
for arbitrary n. *Partial answer:* The best value of b is given by
b2=mωℏ[2(2n+1)n(4n−1)(4n−3)]1/2.
(b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form
ψ(x)=(x2+b2)nBx.
*Partial answer:* The best value of b is given by
b2=mωℏ[2(2n+1)n(4n−5)(4n−3)]1/2.
(c) Notice that the bounds approach the exact energies as n→∞. Why is that? *Hint:* Plot the trial wave functions for n=2, n=3, and n=4, and compare them with the true wave functions (Equations 2.60 and 2.63). To do it analytically, start with the identity
ez=n→∞lim(1+nz)n.