# Does the time-independent Schrodinger equation in 1D have an exact and general solution?

The (time-independent) Schrödinger equation is for sure the most important equation in quantum mechanics: $$-\frac{\hbar^2}{2m}\nabla^{2}\psi(\vec{r}\,)+V(\vec{r}\,)\psi(\vec{r}\,)=E\,\psi(\vec{r}\,).$$ Let’s consider the one-dimensional equation, $$\frac{d^2\psi(x)}{dx^{2}}-\frac{2m}{\hbar^2}\left[V(x)-E\,\right]\psi(x)=0.$$ We can also rewrite the equation as $$\frac{d^{2}psi(x)}{dx^2}+ S(x)\psi(x)=0,$$ where $$S(x)=-\frac{2m}{\hbar^{2}}\left[V(x)-E\,\right].$$ Is there is an exact general eigenvalue-eigenfunction solution for such equation? And if it’s not possible to get the exact eigenvalue-eigenfunction solution, can the equation be solved in a pure mathematical sense? I’m not talking here about the WKB approximation method; I’m talking about an exact and general solution. So does this equation have analytical solutions?

• @Lubavius The problem with 1D hydrogen atom is that one needs to cut off the divergence in energies. But with a cutoff the solution is a textbook case - if I am not mistaken, it was first published in the American Journal of Teachers, as having only pedagogical value. This paper later became very cited in the context of excitons in carbon nanotubes. – Vadim Oct 16 '20 at 10:28

No.
If we view the Schrödinger equation simply as a second order differential equation, then no, it doesn't have a general solution that we can find. Let me add that, from the mathematical standpoint, the equation alone is not even sufficient to have a well posed problem; it needs to be supplemented by the boundary conditions. The solutions for the same type of potential may be easy to get for one set of boundary conditions and hard or even impossible for the other. For example, the harmonic potential case, $$V(x) = \frac{m\omega^2x^2}{2}$$ is exactly solvable with open boundary conditions, but solving it with hard wall conditions, e.g., at $$x= \pm a$$, already requires solving transcendental equations; i.e. the solution is no longer exact. (However, some particular cases, such as the case of one hard wall at $$x=0$$ are still doable.)

But yes.
One may question the very notion of an exact solution: usually it means a solution in terms of simple functions that can be handled with pan and paper. Some people would generalize this to using special functions — Bessel functions, hypergeometric functions, etc. However, this notion of "exact" ultimately traces itself to our ability to calculate the numbers. When we need to calculate a sine, an exponent, or a Bessel function, most of us turn to a computer... but once we accept that computers can be used, almost any one particle Schrödinger equation is solvable. Note that this is not true for many-particle problems, where the problems easily get NP-hard, i.e. impossible using modern computational power (but possibly solvable with a quantum computer).

And formally, yes.
Using the evolution operator one can construct a formally exact solution, evolving from an exactly solvable case, which is quite a common procedure when developing all kinds of perturbation expansions.

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Oct 18 '20 at 0:36

For a few potential functions $$V(x)$$, there are exact analytical solutions. For a general potential, there aren’t, so one must use either numerical techniques or analytic approximations.

There is no general solution. But this is not only due to different possible potential functions, but also due to the boundary conditions. No differential equation can be solved without boundary conditions, and those can vary according to the problem.

Moreover the case distinction $$E<0$$ and $$E>0$$ leads to very different solutions. The first leads to descriptions of bound states, whereas the second one leads to scattering solutions, both very different.

• "No differential equation can be solved without boundary conditions" —actually, every differential equation can. E.g. Bessel equation has two linearly independent solutions: $J_n(r)$ and $Y_n(r)$. – Ruslan Oct 16 '20 at 8:58
• @Ruslan: It'd be more accurate to say that "no differential eigenvalue problem" can be solved without boundary conditions — the Schrödinger problem is an eigenvalue problem, after all. – Michael Seifert Oct 16 '20 at 13:05

It should be noted that the most famous analytic solution can be acquired for $$V(x) \propto x^2$$, which is the harmonic oscillator. For $$V(x) \propto x^3$$ you already need perturbation theory.

• Thanks for pointing out my mistake, Stark effect is of course linear perturbation of the Coulomb potential. I've edited the answer. – Cream Oct 16 '20 at 6:50