# Hybrid infinite-finite square well

Required math: calculus

Required physics: Schrödinger equation

References: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.40.

As a sort of hybrid of the finite and infinite square wells, suppose we have the potential

$\displaystyle V(x)=\begin{cases} \infty & x<0\\ -\frac{32\hbar^{2}}{ma^{2}} & 0\le x\le a\\ 0 & x>a \end{cases} \ \ \ \ \ (1)$

We can look for bound states (${E<0}$) for this potential. From our analysis of the infinite and finite square wells, we can write down the wave function’s general form:

$\displaystyle \psi(x)=\begin{cases} 0 & x<0\\ A\cos(lx)+B\sin(lx) & 0\leq x\leq a\\ Ce^{-\kappa a} & x>a \end{cases} \ \ \ \ \ (2)$

Here, ${l\equiv\sqrt{2m\left(E+32\hbar^{2}/ma^{2}\right)}/\hbar}$ and ${\kappa\equiv\sqrt{-2mE}/\hbar}$. Note that for a bound state, ${E<0}$.

Using the continuity of ${\psi}$ at ${x=0}$:

 $\displaystyle \psi(0)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (4)$

At ${x=a}$: continuity of ${\psi}$ gives

$\displaystyle B\sin(la)=Ce^{-\kappa a} \ \ \ \ \ (5)$

Continuity of ${\psi'}$ gives

$\displaystyle lB\cos(la)=-\kappa Ce^{-\kappa a} \ \ \ \ \ (6)$

Combining the last two gives

$\displaystyle \tan(la)=-\frac{l}{\kappa} \ \ \ \ \ (7)$

We can now define the auxiliary variables

 $\displaystyle z$ $\displaystyle \equiv$ $\displaystyle la\ \ \ \ \ (8)$ $\displaystyle z_{0}$ $\displaystyle \equiv$ $\displaystyle \frac{a}{\hbar}\sqrt{2m\frac{32\hbar^{2}}{ma^{2}}}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 8 \ \ \ \ \ (10)$

Also,

 $\displaystyle \left(l^{2}+\kappa^{2}\right)\hbar^{2}$ $\displaystyle =$ $\displaystyle 2m\frac{32\hbar^{2}}{ma^{2}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{\hbar}{a}z_{0}\right)^{2}\ \ \ \ \ (12)$ $\displaystyle z_{0}^{2}$ $\displaystyle =$ $\displaystyle \left(l^{2}+\kappa^{2}\right)a^{2}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle z^{2}+\kappa^{2}a^{2}\ \ \ \ \ (14)$ $\displaystyle \kappa a$ $\displaystyle =$ $\displaystyle \sqrt{z_{0}^{2}-z^{2}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{64-z^{2}} \ \ \ \ \ (16)$

The equation to be solved for ${z}$ is therefore:

 $\displaystyle \tan(z)$ $\displaystyle =$ $\displaystyle -\frac{la}{\kappa a}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{-z}{\sqrt{64-z^{2}}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{\sqrt{64/z^{2}-1}} \ \ \ \ \ (19)$

This is a transcendental equation which can be solved graphically or numerically. The graph looks like this, with ${\tan(z)}$ in green and ${-\frac{1}{\sqrt{64/z^{2}-1}}}$ in red:

[The vertical green lines are the asymptotes of ${\tan z}$; I couldn’t find a way to get Maple not to draw them.]

As can be seen from the graphical solution, there are 3 bound states. We can solve the equation numerically (using Maple’s ‘fsolve’, for example) to find the three solutions and corresponding energies.

From small to large ${z}$, we get

 $\displaystyle z_{1}$ $\displaystyle =$ $\displaystyle 2.785902114\ \ \ \ \ (20)$ $\displaystyle E_{1}$ $\displaystyle =$ $\displaystyle -28.119\frac{\hbar^{2}}{ma^{2}}\ \ \ \ \ (21)$ $\displaystyle z_{2}$ $\displaystyle =$ $\displaystyle 5.521446430\ \ \ \ \ (22)$ $\displaystyle E_{2}$ $\displaystyle =$ $\displaystyle -16.757\frac{\hbar^{2}}{ma^{2}}\ \ \ \ \ (23)$ $\displaystyle z_{3}$ $\displaystyle =$ $\displaystyle 7.957321494\ \ \ \ \ (24)$ $\displaystyle E_{3}$ $\displaystyle =$ $\displaystyle -0.3405\frac{\hbar^{2}}{ma^{2}} \ \ \ \ \ (25)$

For the highest energy, we can find the probability that the particle is outside the box (that is, its position is outside the range ${[0,a]}$).

From the boundary conditions, we find

$\displaystyle C=\frac{\sin(la)}{e^{-\kappa a}}B \ \ \ \ \ (26)$

so the integral of the square modulus of the wave function over its entire range is

$\displaystyle \left[\int_{0}^{a}\sin^{2}(lx)dx+\frac{\sin^{2}(la)}{e^{-2\kappa a}}\int_{a}^{\infty}e^{-2\kappa x}dx\right]B^{2} \ \ \ \ \ (27)$

The probability of the particle being outside the box is therefore the second term in the above expression divided by the entire expression:

$\displaystyle p_{3}=\frac{B^{2}\frac{\sin^{2}(la)}{e^{-2\kappa a}}\int_{a}^{\infty}e^{-2\kappa x}dx}{\left[\int_{0}^{a}\sin^{2}(lx)dx+\frac{\sin^{2}(la)}{e^{-2\kappa a}}\int_{a}^{\infty}e^{-2\kappa x}dx\right]B^{2}} \ \ \ \ \ (28)$

Using the value of ${z=7.957321494}$ given above, we can get values for ${l}$ and ${\kappa}$ in terms of ${a}$. Plugging these into the above expression and evaluating (using Maple) gives

$\displaystyle p_{3}=0.5420 \ \ \ \ \ (29)$

Note that this answer is very sensitive to the value of ${z}$. Even a minor rounding error will cause a significant difference in the value of ${p}$. For example, rounding ${z}$ to 7.96 gives ${p=0.550}$.

It’s interesting that for this barely bound state, the probability of the particle being outside the box is actually greater than that for being inside.

We can do similar calculations for the other two energies, and find:

 $\displaystyle p_{2}$ $\displaystyle =$ $\displaystyle 0.0702\ \ \ \ \ (30)$ $\displaystyle p_{1}$ $\displaystyle =$ $\displaystyle 0.0143 \ \ \ \ \ (31)$

Since these energies are lower, the probability of being found outside the well is much smaller, as we would expect.

## 8 thoughts on “Hybrid infinite-finite square well”

1. alex

when you found the probability p3, did you have to do that fraction? shouldnt the numerator just be equal to 1 because of normalization?

1. growescience

Since we didn’t normalize the wave function (that is, we never determined the constant ${B}$ in equation 2) then, yes, we do have to work out both numerator and denominator in equation 28. Actually, if you wanted to normalize the wave function then the denominator determines ${B}$ since

$\displaystyle B=\left[\int_{0}^{a}\sin^{2}(lx)dx+\frac{\sin^{2}(la)}{e^{-2\kappa a}}\int_{a}^{\infty}e^{-2\kappa x}dx\right]^{-1/2}$

1. gwrowe Post author

There is no transmission for a bound state. The wave function does extend into the region where ${V\left(x\right)=0}$, but it decays exponentially (see equation 2) so its amplitude tends to zero as ${x\rightarrow\infty}$.

2. apashanka das

sir here the transmission probability is there, although the particle is in the highest energy bound state which is greater than 50 percent chance ,is it consistent with the above?please explain