How to find the least width of potential well so that it has only one bound state? I'm dealing with an asymmetric potential well which is given by,
$$
\begin{array}{ll}
V = 0 &\text{for} \, x < 0 \\
V = -V_0 & \text{for} \, 0 \leq x < d \\
V = 4V_0 & \text{for} \, x\geq d
\end{array}
$$
The task is to find the least value of $d$ for which a bound state exists.
I know that $E<0$ for a bound state. I've got the equation which quantifies the energy,
$$\dfrac{\cos(kd) +\frac{\alpha}{\kappa}\sin(kd)}{-\sin(kd) + \frac{\alpha}{\kappa}\cos(kd)} = -\dfrac{\kappa}{\beta}$$
where
$$\hbar\kappa = \sqrt{2m(E + V_0)}; \hbar\alpha = \sqrt{-2mE}; \hbar\beta = \sqrt{2m(4V_0-E)} \, .$$
How can I proceed to get the minimum value of $d$?
 A: disclaimer: There may be better ways to do this. This was just the first thing to pop into my head.
Solve your equation for d=, then differentiate and minimize.
First, because this gets messy fast, I'll use "t" to mean "$\tan(\kappa d)$." I'll do the same with "$s$" and "$c$."
$-\dfrac{\kappa}{\beta}=\dfrac{c+\dfrac{\alpha}{\kappa} s}{-s+\dfrac{\alpha}{\kappa}c}\\
.\\
=\dfrac{c+\dfrac{\alpha}{\kappa} s}{-s+\dfrac{\alpha}{\kappa}c}\cdot\dfrac{\dfrac{\kappa}{c}}{\dfrac{\kappa}{c}}\\
.\\
=\dfrac{\kappa+\alpha t}{-\kappa t+\alpha}\\
.\\
=\dfrac{-\kappa-\alpha t}{\kappa t-\alpha}\\
.\\
=\dfrac{-\kappa-\alpha t}{\kappa t-\alpha}\cdot\dfrac{\kappa}{\kappa}\\
.\\
=\dfrac{-\kappa^2-\alpha\kappa t}{\kappa(\kappa t-\alpha)}\\
.\\
=\dfrac{-\alpha(\kappa t-\alpha)-\alpha^2-\kappa^2}{\kappa(\kappa t-\alpha)}\\
.\\
\text{(This step takes a bit of trying to force a form, but it's a form like the denominator, so it shouldn't be too magical.)}\\
.\\
=\dfrac{-\alpha^2-\kappa^2}{\kappa(\kappa t-\alpha)}-\dfrac{\alpha}{\kappa}.$
From here, you have an expression with just one tangent function in it, so you can invert, differentiate, set to zero, and be on your way.
$\dfrac{-\alpha^2-\kappa^2}{\kappa(\kappa t-\alpha)}-\dfrac{\alpha}{\kappa}=-\dfrac{\kappa}{\beta}$
A: Assuming that the given equations are correct (with the correction $k=\kappa$). the minimum energy for a bound state is $E=0$ because this is the limit (highest energy for the wave function in the $V=0$ region for being a damped wave, i.e. where the real $\alpha$, corresponding to a damped wave, becomes zero and turns imaginary (propagating wave). Because of the higher potential $4V_0$, the damping constant $\beta$ is definitely real in the other region outside the well. With $E=\alpha=0$ the characteristic equation becomes $$\dfrac{\cos(\kappa d) }{\sin(\kappa d)} =\cot{(\kappa d)}= \dfrac{\kappa}{\beta}=\frac{1}{2}$$ From this follows $$\kappa d=arccot{(\frac{1}{2})}$$ Thus the lowest $d$ for a bound state is given by $$d=\hbar\frac{arccot{(\frac{1}{2})}}{ \sqrt{2mV_0}}$$ Note: For $arccot{(\frac{1}{2})}= arctan{(2)}$, the principal (i.e. lowest) value has to be taken.
