(a) Calculate the expectation value of the operator \hat{X}^{4} in the N-representation with respect to the state |n〉 (i.e., 〈n|\hat{X}^{4}|n〉).
(b) Use the result of (a) to calculate the energy E_{n} for a particle whose Hamiltonian is \hat {H}=\hat{P}^{2}/(2m)+\frac{1}{2}m\omega ^{2}\hat{X}^{2}-\lambda \hat{X}^{4}.
(a) Since \Sigma ^{\infty }_{m=0}|m〉〈m|=1 we can write the expectation value of \hat{X}^{4} as
〈n|\hat{X}^{4}|n〉=\sum\limits_{m=0}^{\infty }〈n|\hat {X}^{2}|m〉〈m|\hat{X}^{2}|n〉=\sum\limits_{m=0}^{\infty }\left|〈m|\hat{X}^{2}|n〉\right| ^{2}. (4.318)
Now since
\hat{X}^{2}=\frac{\hbar }{2m\omega }\left(\hat{a}^{2} +\hat {a}^{\dagger 2}+\hat{a}\hat{a}^{\dagger }+\hat{a}^{\dagger }\hat {a}\right) =\frac{\hbar }{2m\omega }\left(\hat{a}^{2}+ \hat {a}^{\dagger 2}+2\hat{a}^{\dagger }\hat{a}+1\right), (4.319)
the only terms 〈m|\hat{X}^{2}|n〉 that survive are
〈n|\hat{X}^{2}|n〉=\frac{\hbar }{2m\omega }〈n|2\hat {a}^{\dagger }\hat{a}+1|n 〉=\frac{\hbar }{2m\omega }(2n+1), (4.320)
〈n-2|\hat{X}^{2}|n〉=\frac{\hbar }{2m\omega }〈n-2|\hat {a}^{2}|n 〉=\frac{\hbar }{2m\omega }\sqrt{n(n-1)}, (4.321)
〈n+2|\hat{X}^{2}|n〉=\frac{\hbar }{2m\omega }〈n+2|\hat {a}^{\dagger 2}|n 〉=\frac{\hbar }{2m\omega }\sqrt{(n+1)(n+2)}. (4.322)
Thus
〈n|\hat{X}^{4}|n〉=\left|〈n|\hat{X}^{2}|n〉\right| ^{2} +\left|〈n-2|\hat{X}^{2}|n〉\right| ^{2}+\left|〈n+2|\hat{X}^{2}|n〉\right| ^{2}=\frac{\hbar ^{2} }{4m^{2}\omega ^{2}} \left[(2n+1)^{2}+ n(n-1)+(n+1)(n+2) \right]
=\frac{\hbar ^{2} }{4m^{2}\omega ^{2}}\left(6n^{2}+6n+3 \right). (4.323)
(b) Using (4.323), and since the Hamiltonian can be expressed in terms of the harmonic oscillator, \hat{H}=\hat{H}_{HO} -\lambda \hat{X}^{4}, we immediately obtain the particle energy:
E_{n}=〈n|\hat{H}_{HO}|n〉-\lambda 〈n|\hat{X}^{4}|n〉 =\hbar \omega \left(n+\frac{1}{2} \right) -\frac{\lambda \hbar ^{2} }{4m^{2}\omega ^{2}}\left(6n^{2}+6n+3 \right). (4.324)