Question 10.3: The potential function of a one-dimensional oscillator of ma......
The potential function of a one-dimensional oscillator of mass m and angular frequency ω is V(x)=\kappa x^{2} /2+\xi x^{4}, where κ is the spring constant for a harmonic potential and the second term is small compared with the first.
(a) Show that, to first order, the effect of the anharmonic term is to change the energy of the ground state by 3\xi (\hbar /2m\omega )^{2}.
(b) What would be the first-order effect of an additional x³ term in the potential?
(a) Using the position operator \hat{x} =(\hbar /2m\omega )^{1/2} (\hat{b}^{\dagger } +\hat{b}), one may express the perturbation as
\hat{W}=\xi x^{4}=\xi \left(\frac{\hbar }{2m\omega } \right)^{2} (\hat{b}^{\dagger } +\hat{b})^{4}\hat{W}=\xi \left(\frac{\hbar }{2m\omega } \right)^{2} (\hat{b}^{\dagger } +\hat{b})(b^{\dagger 3}+\hat{b} \hat{b}^{\dagger }\hat{b}^{\dagger }+\hat{b}^{\dagger }\hat{b}\hat{b}^{\dagger }+\hat{b}\hat{b}\hat{b}^{\dagger }+\hat{b}^{\dagger }\hat{b}^{\dagger }\hat{b}+\hat{b}\hat{b}^{\dagger }\hat{b}+\hat{b}^{\dagger }\hat{b}\hat{b}+\hat{b}^{3 })
\hat{W}=\xi \left(\frac{\hbar }{2m\omega } \right)^{2}(b^{\dagger 4} +\hat{b}^{\dagger }\hat{b}\hat{b}^{\dagger }\hat{b}^{\dagger }+\hat{b}^{\dagger }\hat{b}^{\dagger }\hat{b} \hat{b}^{\dagger }+\hat{b}^{\dagger }\hat{b}\hat{b}\hat{b}^{\dagger }+\hat{b}^{\dagger }\hat{b}^{\dagger }\hat{b}^{\dagger }\hat{b}+\hat{b}^{\dagger }\hat{b}\hat{b}^{\dagger }\hat{b}+\hat{b}^{\dagger }\hat{b}^{\dagger }\hat{b}\hat{b}+\hat{b}^{\dagger }\hat{b}^{3}+\hat{b}\hat{b}^{\dagger3}+\hat{b}\hat{b}\hat{b}^{\dagger }\hat{b}^{\dagger }+\hat{b}\hat{b}^{\dagger }\hat{b}\hat{b}^{\dagger }+\hat{b}\hat{b}\hat{b} \hat{b}^{\dagger }+\hat{b}\hat{b}^{\dagger }\hat{b}^{\dagger }\hat{b}+\hat{b}\hat{b}\hat{b}^{\dagger }\hat{b}+\hat{b}\hat{b}^{\dagger }\hat{b}\hat{b} +\hat{b}^{4} )
Energy eigenvalues in first-order perturbation theory couple the same state, so only symmetric terms with two \hat{b}^{\dagger } s and two \hat{b}s will contribute.
W_{nn} =\xi \left(\frac{\hbar }{2m\omega } \right)^{2}((n+1)(n+2)+(n+1)^{2}+n^{2}+n(n-1)+2n(n+1) )W_{nn} =\xi \left(\frac{\hbar }{2m\omega } \right)^{2}(6n^{2}+6n+3 )
so, to first order,
E_{n} =\hbar \omega \left(n+\frac{1}{2} \right)+\xi \left(\frac{\hbar }{2m\omega } \right)^{2} (6n^{2}+6n+3 )and the new ground state is
E_{n} =\hbar \omega \left(n+\frac{1}{2} \right)+3\xi \left(\frac{\hbar }{2m\omega } \right)^{2}(b) There is no first-order correction for a perturbation in x³, because it cannot couple to the same state (there are always an odd number of operators \hat{b}^{\dagger } or \hat{b} ).