I have to solve the following problem:
Consider the potential well: $$ V(x)=-V_0, \hspace{10px} |x|<a/2 $$ and $0$ everywhere else. $a$ is also a positive constant and so is $V_0$. Find the wave function everywhere for an energy $0>E>-V_0$. Also determine the equation that gives the bound states. Verify that there is always at least 1 bound state in the well.
And I have done that, it is simply solving the time independent Schrodinger equation:
$$ \frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi $$
and afterwards apply continuity of the wave function as well as its derivative. After all that work I get to the following equations:
$$ k\tan\left(\frac{ka}{2}\right)=\tilde{k}, \tag{1} $$ $$ \frac{1}{k}\tan\left(\frac{ka}{2}\right)=-\frac{1}{\tilde{k}}, \tag{2} $$
where $k^2=\frac{2m|E|}{\hbar^2}$ and $\tilde{k}^2=\frac{2m}{\hbar^2}(V_0-|E|)$. But these equations can't be both simultaneously true. In fact if the first is true, then without any loss of generality we will have only even solutions, whereas if the second is true then it only admits odd solutions. However, for proving that there is always at least one bound state, I defined:
$$ \chi=\frac{ka}{2} \\ \eta=\frac{\tilde{k}a}{2} $$
and so if $(1)$ is true then we get:
$$ \chi^2+\eta^2=\frac{mV_0a^2}{2\hbar^2} \\ \chi \tan(\chi)=\eta $$
which is a system of equations that is always satisfied for any $V_0>0$. On the other hand if we do the same admitting $(2)$ true:
$$ \chi \cot (\chi)=-\eta \\ \chi^2+\eta^2=\frac{mV_0a^2}{2\hbar^2} $$
which is only satisfied for a certain $V_{0_{min}}$.
Now here's my question:
I have seen that when the potential is symmetric there can be even and odd solutions to the same equation. But I thought that these solutions were independent and each would give me a different set of solutions, $\psi_n$ such that the general solution would consist of linear combinations of those particular functions but could describe the same wave function. In other words, I thought that in a way the even and odd solutions would simply be different basis for the same quantum system. Again in another way, in fourier analysis, any periodic signal can be decomposed as a sum of complex exponentials, so that was, in a way, my pilar to my understanding, even though we had two different sets of coefficients the end result could still be the same. But based on this result, how could one basis (even) allow any $V_0$ and secure bound states, but on another basis (odd) that does not hold? Shouldn't they describe the same reality?
Also, the potential is just a shifted version of a assymmetric potential which will not give even and odd solutions, but is essentially the same problem only with a shifted reference frame. How can we obtain these two different results?
PS: I have seen other questions on this stackexchange but none answered my doubts.