Consider Schrodinger equation for a quantum harmonic oscillator with small quartic perturbation \[ \left(-\frac{1}{2}\frac{d^2}{dx^2} + \frac{x^2}{2} + \frac{g x^4}{4}\right)\psi(x) = E_0(g)\psi(x) \tag{30} \] and make the ansatz \[ \psi(x) = e^{-x^2/2} \sum_{n=0}^\infty \left(\frac{g}{4}\right)^n B_n(x) \qquad \text{with} \quad B_0(x) = 1 \tag{31} \] \[ E_0(g) = \sum_{k=0}^\infty a_k \left(\frac{g}{4}\right)^k. \tag{32} \] We already know that \(a_0 = \frac{1}{2}\) from the unperturbed oscillator. We want to find the first two corrections \(a_1\) and \(a_2\).
Find a recurrence relation for \(B_k(x)\) and \(a_k\)
Solve the relation by assuming \(B_i(x) = \sum_{j=1}^{2i} x^{2j} (-1)^i B_{i,j}\).
Considering different powers of \(x\), find the following relations \[ a_n = (-1)^{n+1} B_{n,1} \tag{33} \] \[ 2j B_{n,j} = (j+1)(2j+1) B_{n,j+1} + B_{n-1,j-2} - \sum_{k=1}^{n-1} B_{n-k,1} B_{k,j} \tag{34} \]
Find \(a_1\) and \(a_2\). You can check that your result agrees with the usual perturbation theory.