Solving Schrodinger's eqn using ladder operators for potential V

[Hamiltonian]

The Schrodinger equation:
$$-\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} + \hat V\psi = E\psi$$
$$\frac{1}{2m}[\hat p^2 + 2m\hat V ]\psi = E\psi$$
The ladder operators:
$$\hat a_\pm = \frac{1}{\sqrt{2m}}[\hat p \pm i\sqrt{2m\hat V}]$$

$$\hat a_\pm \hat a_\mp = \frac{1}{2m}[\hat p^2 + (2m\hat V) \mp i\sqrt{2m}[\hat p,\hat V ^{1/2}]]$$
The Hamiltonian:
$$\hat H = \frac{1}{2m}[\hat p^2 +2m\hat V]$$
The Hamiltonian in terms of the ladder operator:
$$\hat H = \hat a_\pm \hat a_\mp \pm {\frac{i}{\sqrt{2m}} [\hat p, \hat V^{1/2}]} $$
finding $\hat H(\hat a_- \psi)$:
$$\hat H(\hat a_- \psi) = [\hat a_-\hat a_+ \frac{i}{\sqrt{2m}}[\hat p, \hat V^{1/2}]](\hat a_- \psi)$$
$$\hat H(\hat a_- \psi) = \hat a_-[E - \frac{2i}{\sqrt{2m}}[\hat p, \hat V^{1/2}]](\psi)$$

If we apply the lowering operator repeatedly we will eventually reach a state with energy less than zero, which do not exist.
$$\hat a_- \psi_0 = 0$$
$$[\hat p - i(2m\hat V)^{1/2}]\psi_0 = 0$$
$$[-i\hbar\frac{d}{dx} -i \sqrt{2mV(x)}]\psi_0 = 0$$
$$\psi_0(x) = -\int {\frac{\sqrt{2mV(x)}}{\hbar}dx}$$
$$\psi_n(x) = A_n(\hat a_+)^n \psi_0(x)$$
where $\psi_0$ is the ground state wave function solution.
Is this a correct way to give a general solution of the Schrodinger equation for any potential?

 

[Nugatory]

Is this a correct way to give a general solution of the Schrodinger equation for any potential?

You can quickly check whether you have a general solution by trying it on the special case $V=0$. Try this and you’ll likely see that the harmonic oscillator potential has a property required for this approach to work that is not found in more general potentials.

 

[Hamiltonian]

$$\psi_0(x) = -\int {\frac{\sqrt{2mV(x)}}{\hbar}dx}$$
$$\psi_n(x) = A_n(\hat a_+)^n \psi_0(x)$$
where $\psi_0$ is the ground state wave function solution.
Is this a correct way to give a general solution of the Schrodinger equation for any potential?

I made a small error here, $\psi_0$ is supposed to be:
$$\psi_0(x) = e^{\int -\frac{\sqrt{2mV(x)}}{\hbar} dx}$$

You can quickly check whether you have a general solution by trying it on the special case $V=0$. Try this and you’ll likely see that the harmonic oscillator potential has a property required for this approach to work that is not found in more general potentials.

Is there a way to account for the case where $V(x) = 0$?
Leaving aside this case the final siltion should hold true for any arbitrary potential $V(x)$?

 

[Demystifier]

The "lowering operators" do not really lower energy. Instead, they multiply $\psi(x)$ by a function that depends on $x$ (except in the cases $V=0$ and $V\propto x^2$), so it cannot be interpreted as a number times $\psi(x)$ .

 

[vanhees71]

This technique is pretty much restricted to the harmonic oscillator and the angular-momentum eigenvalue problems, and the latter is even also reducible to the symmetric harmonic-oscillator problem in 2 dimensions. So one can expect that most of the problems, where this "factorization of the Hamiltonian" approach works, is due to symmetries reducing to the harmonic oscillators in $d$ dimensions and thus related to the Lie algebras of the SU(d) groups. Schrödinger has written (at least) two papers about this too:

https://www.jstor.org/stable/20490744
https://www.jstor.org/stable/20490756