Inside the well
$\psi_2'' = -k^2\psi$
$k = \frac{\sqrt{2mE}}{\hbar}$
$\psi = A\sin(kx) + B\cos(kx)$
Outside the well on the left
For bound states $E>V_0$
$\psi_1'' = \alpha^2\psi$
$\alpha = \frac{\sqrt{2m(V_0-E)}}{\hbar}$
$\psi_1 = Fe^{\alpha x} + Ge^{-\alpha x}$
This is the wavefunction of the left side so it must go to 0 as x approaches negative of infinite
so $G = 0$
and similarly
Outside the well on the right
$\psi_3 = He^{\alpha x} + Ie^{-\alpha x}$
with $H =0$
Because the wavefunction must be continuous
$\psi_2'(-a/2) = \psi_1'(-a/2)$
and
$\psi_2'(a/2) = \psi_3'(a/2)$
Now to have a antisym. $\psi$ the antisym. 402
operators must be removed by imposing certain equalities on the constants.For example sine is a antisym. function so it can be removed by imposing that A = 0
And if you are through your course you may encounter or encountered this problem.
For symmetric case we get
$\alpha L/2 = kL/2\tan(kL/2)$
where we can define a parameter describing the size of the potential $g_0^2 = \frac{2ma^2V_0}{\hbar}$ and another one $\phi = k^2L/2$
$\alpha L/2 = \sqrt{g_0^2 - \phi^2}= \phi\tan\phi$
Anti-symmetric we get
$\alpha = -k\cot(kL/2)= \sqrt{g_0^2 - \phi^2}= -\phi\cot\phi $
Since $g_0^2$ is a constant we have to assign a value to it. e.g 0.02
https://www.wolframalpha.com/input/?i=sqrt(0.002-x%5E2)+%3D+xtan(x) for even
https://www.wolframalpha.com/input/?i=sqrt(0.002-x%5E2)+%3D+-xcot(x) for odd
For even parity there is atleast one solution but not for odd so it is not necessary that you will have one bound state when not given the parity of the wavefunction
For very small and shallow wells the particle will tend to remain outside because the electron achieves a high energy and thus does not remain localized as the energy eigen-values are given by $E_n = \frac{4\hbar^2\phi^2}{2mL^2}$ We can also show that the wavefunction is decaying as the width becomes less.The simple reason is that the uncertainty in position increases because there is very less space for the electron to exist Taking odd functions in the consideration
$\psi = Asin(kx)$
$|\psi_n|^2 = A^2\int_0^{1 \times 10^{-10}}\sin^2(kx)$
$= A^2[\frac{x}{2} -\frac{1}{4k}\sin{2kx}]_0^{1 \times 10^{-10}}$
$= A^2[\frac{1 \times 10^{-10}}{2} -\frac{1}{4 \times 1\times 10^{-24}}\sin{2 \times 1\times 10^{-24} \times 1 \times 10^{-10}}]$
To show that less energy particles also have a very less probability of staying inside the well which they do not generally you can consider $k = 1\times 10^(-24)$
$=A^2\times 3. 402