# 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 $d$?

• I would suggest the Dirac potential, but that is symmetric, if you Google "More on the Asymmetric Infinite Square Well: Energy Eigenstates with Zero Curvature", you should get an associated pdf which deals with modifying the standard well to something that resembles what your question is. – user108787 Oct 18 '16 at 17:17

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.

$\dfrac{-\alpha^2-\kappa^2}{\kappa(\kappa t-\alpha)}-\dfrac{\alpha}{\kappa}=-\dfrac{\kappa}{\beta}$

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.

• I don't get why minimum energy for bound state is $E=0$. Even if $E<0$, $\alpha$ is still real and there will be still exponential damping. – crimson Oct 18 '16 at 20:06
• When $E<0$ then you have a real $\alpha$ corresponding to a damped wave. When $E=0$, $\alpha=0$, and when $E>0$, $\alpha$ becomes imaginary corresponding to a propagating wave. – freecharly Oct 18 '16 at 20:32