The original system is a free particle, and there are no restrictions on its space. Assume that the perturbation potential is $\epsilon \frac{\mu\omega^2}{2} x^2$ so that $$ \hat{H}=\hat{H}_0+\epsilon\hat{H}'=-\frac{\hbar^2}{2\mu}\frac{d^2}{dx^2}+\epsilon\frac{\mu\omega^2}{2}x^2\,. $$ Assume that the wave function of a free particle with momentum $p'$ is $$\psi_{p'}^0=\frac{1}{\sqrt{2\pi\hbar}}e^{ip'x/\hbar}\,.$$
The superscript number indicates the number of perturbation levels. The first level perturbation fixed state Schrödinger equation is $$\hat{H}'\psi_{p'}^0+\hat{H}_0\psi_{p'}^1=E^1_{p'}\psi_{p'}^0+E_{p'}^0\psi_{p'}^1$$
If we use the unperturbed perturbation wave function to represent the level 1 perturbation wave function: $$\psi_{p'}^1=\frac{1}{\sqrt{2\pi\hbar}}\int a_{p'}^1(p)e^{i\frac{p}{\hbar}x} dp$$
After multiplying both sides by the complex conjugate of the unperturbed wave function to integrate over the full space, we can obtain $$\psi_{p''}^{0*}\hat{H}'\psi_{p'}^0+\psi_{p''}^{0*}\hat{H}_0\psi_{p'}^1=E^1_{p'}\psi_{p''}^{0*}\psi_{p'}^0+E_{p'}^0\psi_{p''}^{0*}\psi_{p'}^1\,,$$ in which case $$-\frac{\mu\omega^2\hbar^2}{2}\delta''(p'-p'')+E_{p''}^0a_{p'}^1(p'')=E^1_{p'}\delta(p'-p'')+E_{p'}^0a_{p'}^1(p'')$$ and $$E^1_{p'}\delta(p'-p'')+\frac{\mu\omega^2\hbar^2}{2}\delta''(p'-p'')=(E_{p''}^0-E_{p'}^0)a_{p'}^1(p'')\,.$$
Considering $E^0_{p'}=\frac{p'^2}{2\mu}$, we can obtain $$E^1_{p'}\delta'(p'-p'')+\frac{\mu\omega^2\hbar^2}{6}\delta^{(3)}(p'-p'')=\frac{p'+p''}{2\mu}a_{p'}^1(p'')$$ so that $$a_{p'}^1(p'')=E^1_{p'}\frac{2\mu}{p'+p''}\delta'(p'-p'')+\frac{\mu\omega^2\hbar^2}{6}\frac{2\mu}{p'+p''}\delta^{(3)}(p'-p'')\,,$$ in which case $$\psi_{p'}^1=\frac{1}{\sqrt{2\pi\hbar}}\int [E^1_{p'}\frac{2\mu}{p'+p''}\delta'(p'-p'')+\frac{\mu\omega^2\hbar^2}{6}\frac{2\mu}{p'+p''}\delta^{(3)}(p'-p'')]e^{i\frac{p''}{\hbar}x}dp''\,,$$ where \begin{align} k&=\frac{p'}{\hbar}\,, \\ \psi_{p'}^1(x)&=\frac{1}{\sqrt{2\pi\hbar}}[\frac{\mu E^1_{p'}}{2\hbar^2}\frac{i2kx-1}{k^2}+\frac{\mu^2\omega^2}{24\hbar^2}\frac{i(-4k^3x^3+6kx)+(6k^2x^2-3)}{k^4}]e^{ikx} \,, \\ C&=\frac{12\mu E^1_{p'}}{\mu^2\omega^2}\,, \\ A&=C\frac{2kx}{k^2}+\frac{-4k^3x^3+6kx}{k^4}=-\frac{4}{k}x^3+(C\frac{2}{k}+\frac{6}{k^3})x\,, \\ B&=C\frac{-1}{k^2}+\frac{6k^2x^2-3}{k^4}=\frac{6}{k^2}x^2-(C\frac{1}{k^2}+\frac{3}{k^4})\,, \\ D&=\frac{\mu^2\omega^2}{24\hbar^2\sqrt{2\pi\hbar}}\,, \\ \psi^1_k&=D\sqrt{A^2+B^2}\exp\left[i\left(kx+\arctan\frac{A}{B}\right)\right]\,. \end{align}
At this point, we can see that $$\psi^1_k\varpropto z(x)\psi^0_{p'}\,,$$ where $z=\sqrt{A^2+B^2}e^{i\arctan\frac{A}{B}}$ and as $x\rightarrow+\infty$, $z(x)\rightarrow-\infty$.
This does not make sense, because the 1D harmonic oscillator or other 1D bound state wave functions should tend to finite values at infinity and in the process of solving also can not let the energy level discrete.
Is there something I'm missing? Or is there no solution to this problem.
