A Fundamental Set of Solutions for the Quantum Harmonic Oscillator The quantum harmonic oscillator has a Hamiltonian given by
$\displaystyle-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{1}{2}m\omega^2x^2\psi=E\psi$.
This is a spectral problem, but we know that the ground state energy (i.e. the smallest eigenvalue $E$) is given by $E=\frac{1}{2}\hbar\omega$. This turns the problem into a 2nd order linear ODE - which begs for two linearly independent solutions (cf. the [Encyclopedia of Mathematics entry][1]). However, any first course in quantum mechanics should reveal that "the" solution is given by a Gaussian function (cf. the [Wikipedia article][2] and [this][3]). What is the other solution?
My thoughts:
(a) The other solution is not normalizable (i.e. not $L^2(\mathbb{R})$).
(b) The existence of two linearly independent solutions only holds for finite intervals $(\alpha, \beta)$ (notation as in the Encyclopedia of Mathematics article). But even so, we can artificially restrict the domain to a finite interval, so what is the other solution?
I look forward to some clarification.
[1]: https://encyclopediaofmath.org/wiki/Fundamental_system_of_solutions#:~:text=A%20set%20of%20real%20(complex,(complex)%20numbers%20C1%E2%80%A6
[2]: https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator
[3]: http://math-wiki.com/images/8/8a/Quantum_harmonic_oscillator_lecture.pdf
 A: If you make an ansatz of the form $\psi(x) = e^{\alpha x^2}$, you get
$$\frac{\partial \psi}{\partial x} = -2 \alpha x e^{-\alpha x^2}$$
and
$$\frac{\partial^2 \psi}{\partial x^2} = -2\alpha e^{-\alpha x^2}+4\alpha^2x^2 e^{-\alpha x^2}.$$
That means that we get the equation
$$ -\frac{\hbar^2}{2m}(-2\alpha +4\alpha^2x^2)+\frac{1}{2}m\omega^2 x^2 = E$$
To cancel the quadratic term, we can have $\alpha = \pm \frac{m \omega}{2 \hbar}$. The solution with negative $\alpha$ is not normalizable, so your first reason is correct. This would also yield a negative energy, which is something that physicists are allergic to.
A: The correct answer is (a) : the other ones are not is $L^2$.

*

*Simplifications :

Making the change of variable   $x=\sqrt{\frac{h}{m\omega}}y$ the equation becomes :
$$-\frac{d^2\phi}{dy^2} + y^2\phi=e\phi.$$
where $e=\frac{2E}{h\omega}$.

*

*Lowest energy state.

For the lowest energy state, $e=1$ :
$$-\frac{d^2\phi}{dy^2} + y^2\phi=\phi.$$
The usual trick to solve this is to "factor" the differential equation thusly :
$$\left(-\frac{d}{dy}+y\right)\left[\left(\frac{d}{dy}+y\right)\phi(y)\right]=0. \quad (E)$$

*

*Resolution :

This means, $f:y\mapsto \left(\frac{d}{dy}+y\right)\phi(y)=\phi'(y)+y\phi(y)$ is a solution to :
$$-g'(y)+yg(y)=0 \quad(F1)$$
Solutions of $F1$ can be written as $c_0\exp(\frac{y^2}{2})$
This means that $\phi $ is a solution to (E) iff there exists $c_0\in\mathbb{R}$ such that $\phi$ is a solution to
$$\phi'(y)+y\phi(y)=c_0\exp(\frac{y^2}{2}) \quad (F2)$$
Let's fix $c_0\in\mathbb{R}$.

*

*solutions of the homogeneous part can be written as $y\mapsto \mu\exp\left(-\frac{y^2}{2} \right)$


*to find a particular solution to $(F2)$, use the "Variation of parameters" method. All computations done, you'll find $y\mapsto c_0 e^{-y^2/2}\int_{0}^{y}e^{t^2}dt$ is such a solution.
Therefore, solutions of (E) can be written as :
$$\phi : y\mapsto c_0\exp\left(\frac{-y^2}{2}\right)\int_{0}^{y}\exp(t^2)dt+\mu\exp\left(\frac{-y^2}{2}\right)$$
But this is never in $L^2$ unless $c_0= 0$. Indeed for $y>0$:
$$\exp\left(\frac{-y^2}{2}\right)\int_{0}^{y}\exp(t^2)dt \ge \exp\left(\frac{-y^2}{2}\right)\int_{y/\sqrt{2}}^{y}\exp(t^2)dt\ \\\ge\exp\left(-\frac{y^2}{2}\right) (y-\frac{y}{\sqrt{2}})\exp\left(\frac{y^2}{2}\right)\\
\ge y-y/\sqrt{2} \to_{y\to+\infty} +\infty$$
Therefore the $c_0\exp\left(\frac{-y^2}{2}\right)\int_{0}^{y}\exp(t^2)dt$ part is never in $L²$ unless of course $c_0=0$
