= Solution
The <parity operator> acts as $P\psi(x)=\psi(-x)$, with $P^2=I$. For an even <potential energy> $[H,P]=0$, so <energy eigenstates> can be chosen even or odd, with <parity> <eigenvalues> $+1$ and $-1$. One-dimensional <bound states> are nondegenerate; the <ground state> is <nodeless> and even.
For the <finite square well>, square integrability excludes the growing exterior exponential, while even <parity> excludes the interior sine. Thus the interior solution is proportional to $\cos kx$, with matching decaying tails and
$$
\boxed{k=\frac{\sqrt{2m(V_0+E_0)}}{\hbar},\qquad
q=\frac{\sqrt{-2mE_0}}{\hbar}}.
$$
Continuity at $a$ gives $B\cos ka=Ae^{-qa}$ and $-Bk\sin ka=-qAe^{-qa}$. Their ratio yields
$$
q=k\tan ka,\qquad
\boxed{\sqrt{-E_0}=\sqrt{V_0+E_0}\tan\left(\frac{a\sqrt{2m(V_0+E_0)}}{\hbar}\right)}.
$$
The conditions at $-a$ follow from the same <parity>.
For a narrow weak well, $ka\ll1$, hence $q\simeq k^2a\simeq2mV_0a/\hbar^2$ and
$$
\boxed{E_0\simeq-\frac{2mV_0^2a^2}{\hbar^2}}.
$$
This is self-consistent because $|E_0|/V_0\simeq2mV_0a^2/\hbar^2\ll1$. It is the <shallow finite-square-well energy>, equivalently the attractive <delta potential> limit with integrated strength $2aV_0$. An arbitrarily weak attractive well of this form still has its even bound <ground state>.
The first excited state must be odd. Replace the interior cosine by $\sin kx$ and choose exterior tails with opposite signs. The matching condition becomes $q=-k\cot ka$. Its first root requires $\pi/2<ka<\pi$. At the binding threshold $q\to0$, so
$$
\boxed{\frac{a\sqrt{2mV_0}}{\hbar}>\frac{\pi}{2}}
$$
is necessary and sufficient for the first odd <bound state>. Equality gives a zero-energy nonnormalizable threshold solution, not an additional <bound state>.
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