Question 7 of 7: Perturbed Harmonic Oscillator — Quartic Anharmonicity

Question 7: Perturbed Harmonic Oscillator — Quartic Anharmonicity (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Unperturbed SHO $H_0=p_x^2/2m+\frac12m\omega^2x^2$ with normalized eigenstates $\psi_0,\psi_1$; perturbation $H'=bx^4$ ($b>0$).

Find. First-order energy shifts $\Delta E_0^{(1)}$ and $\Delta E_1^{(1)}$, and which is larger.

Approach. First-order perturbation theory gives $\Delta E_n^{(1)}=\langle\psi_n|H'|\psi_n\rangle=b\langle x^4\rangle_n$; evaluate each $\langle x^4\rangle_n$ with the supplied even-power Gaussian integral.

1. Ground-state shift. $$ \Delta E_0^{(1)} = b\langle x^4\rangle_0 = b\frac{\alpha}{\sqrt\pi}\int_{-\infty}^{\infty}x^4e^{-\alpha^2x^2}dx = \frac{2b\alpha}{\sqrt\pi}\int_0^\infty x^4e^{-x^2/(1/\alpha)^2}dx. $$ Using the supplied formula $\int_0^\infty x^{2n}e^{-x^2/A^2}dx=\sqrt\pi\frac{(2n)!}{n!}\left(\frac{A}{2}\right)^{2n+1}$ with $2n=4$ ($n=2$) and $A=1/\alpha$: $$ \int_0^\infty x^4e^{-\alpha^2x^2}dx = \sqrt\pi\cdot\frac{4!}{2!}\left(\frac{1}{2\alpha}\right)^5 = \frac{3\sqrt\pi}{8\alpha^5}. $$ So $$ \langle x^4\rangle_0 = \frac{\alpha}{\sqrt\pi}\cdot\frac{2\cdot3\sqrt\pi}{8\alpha^5} = \frac{3}{4\alpha^4} \quad\Longrightarrow\quad \boxed{\Delta E_0^{(1)} = \frac{3b}{4\alpha^4}}. $$
2. First-excited-state shift. $$ \Delta E_1^{(1)} = b\langle x^4\rangle_1 = b\frac{2\alpha^3}{\sqrt\pi}\int_{-\infty}^{\infty}x^6e^{-\alpha^2x^2}dx = \frac{4b\alpha^3}{\sqrt\pi}\int_0^\infty x^6e^{-\alpha^2x^2}dx. $$ Applying the same formula with $2n=6$ ($n=3$): $$ \int_0^\infty x^6e^{-\alpha^2x^2}dx = \sqrt\pi\cdot\frac{6!}{3!}\left(\frac{1}{2\alpha}\right)^7 = \frac{15\sqrt\pi}{16\alpha^7}. $$ So $$ \langle x^4\rangle_1 = \frac{4\alpha^3}{\sqrt\pi}\cdot\frac{15\sqrt\pi}{16\alpha^7} = \frac{60}{16\alpha^4} = \frac{15}{4\alpha^4} \quad\Longrightarrow\quad \boxed{\Delta E_1^{(1)} = \frac{15b}{4\alpha^4}}. $$
3. Compare the two shifts. $$ \frac{\Delta E_1^{(1)}}{\Delta E_0^{(1)}} = \frac{15b/4\alpha^4}{3b/4\alpha^4} = 5. $$ $$ \boxed{\text{The first excited state receives a shift 5 times larger than the ground state.}} $$ This is physically sensible: $\psi_1$ has a larger spatial extent than $\psi_0$ (it has a node at the origin and larger probability weight further from $x=0$), so it samples the steeply-rising quartic potential $bx^4$ more strongly than the more tightly localized ground state.