# How do I find the eigenvalues for the finite potential well of width 2L?

## Derive a transcendental equation for the allowed energies and solve it graphically in the finite well of the height ${V}_{0}$ $V \left(x\right) = \left\{\begin{matrix}- {V}_{0} & | x | \ge L \\ 0 & \text{elsewhere}\end{matrix}\right.$ $\alpha \tan \alpha = k$(even parity) $\alpha \cot \alpha = - k$(odd parity) i hope that u understand what $\alpha \mathmr{and} k$ are. but to mark the energy eigenvalues i had to make circle of radius $\alpha + k$ what is the physical reason behind this?

Mar 16, 2018

Here is the Excel sheet I made while doing this.

The radius of the circle just tells you what you set the height of your potential well to be. However, its radius is given by $\sqrt{{\alpha}^{2} {L}^{2} + {k}^{2} {L}^{2}}$ in your notation...

${\lim}_{{V}_{0} \to \infty} \text{Finite Well" = "Infinite Potential Well}$

If we choose V_0 = (20ℏ^2)/(2mL^2) then we get three bound states in the well. These energy levels are shown below.

$\underline{E \text{ "" "" "" "" "" "E//V_0" "" "color(white)(.)"Quantum Number"" "" }}$
(1.63948ℏ^2)/(2mL^2)" "" "0.081974" "" "1.28042

(6.44190ℏ^2)/(2mL^2)" "" "0.322095" "" "2.53809

(13.89150ℏ^2)/(2mL^2)" "color(white)(.)0.694575" "" "3.72713

The graph below utilizes a well width of $2 L = 4$ as an example, with the appropriate coefficients $B$, $C$, and $D$ to make the wave functions a practical scale. Well, let's start by properly defining the problem... instead of defining a tunnelling problem with a barrier of width $2 L$, which you have done, let's define a finite potential well of width $2 L$...

$V \left(x\right) = \left\{\begin{matrix}{V}_{0} & | x | \ge L \\ 0 & - L < x < L\end{matrix}\right.$

That way, we get: Then, let's define the eigenfunctions for each region $\left(I , I I , I I I\right)$:

${\psi}_{I} = A {e}^{- \alpha x} + B {e}^{\alpha x} , \text{ "" "" "" } - \infty < x < - L$
${\psi}_{I I} = C \sin \left(k x\right) + D \cos \left(k x\right) , \text{ } - L < x < L$
${\psi}_{I I I} = F {e}^{- \alpha x} + G {e}^{\alpha x} , \text{ "" "" "" } L < x < \infty$

where:

alpha = sqrt((2m(V_0 - E))/ℏ^2), $\text{ } E < {V}_{0}$ for bound states
k = sqrt((2mE)/ℏ^2)

Since the wave function must vanish at $\pm \infty$, $A = 0$ and $G = 0$. Furthermore, since the solution will have even and odd parity, we split the answer to the even and the odd solutions.

EVEN SOLUTIONS

For the even solutions, the $x = 0$ line is a reflection axis, giving $B = F$. $\cos$ is even with respect to $x = 0$, so $C = 0$ as well.

${\psi}_{I} = B {e}^{\alpha x} , \text{ "" "" } - \infty < x < - L$
${\psi}_{I I} = D \cos \left(k x\right) , \text{ } - L < x < L$
${\psi}_{I I I} = B {e}^{- \alpha x} , \text{ "" "" } L < x < \infty$

Now we apply the continuity conditions.

$\underline{{\psi}_{I} \left(- L\right) = {\psi}_{I I} \left(- L\right)}$:
$B {e}^{- \alpha L} = D \cos \left(- k L\right)$ $\text{ "" "" } \boldsymbol{\left(1\right)}$

$\underline{\frac{\mathrm{dp} s {i}_{I} \left(- L\right)}{\mathrm{dx}} = \frac{\mathrm{dp} s {i}_{I I} \left(- L\right)}{\mathrm{dx}}}$:
$\alpha B {e}^{- \alpha L} = - k D \sin \left(- k L\right)$ $\text{ } \boldsymbol{\left(2\right)}$

$\underline{{\psi}_{I I} \left(L\right) = {\psi}_{I I I} \left(L\right)}$:
$D \cos \left(k L\right) = B {e}^{- \alpha L}$ $\text{ "" "" "" } \boldsymbol{\left(3\right)}$

$\underline{\frac{\mathrm{dp} s {i}_{I I} \left(L\right)}{\mathrm{dx}} = \frac{\mathrm{dp} s {i}_{I I I} \left(L\right)}{\mathrm{dx}}}$
$k D \sin \left(k L\right) = \alpha B {e}^{- \alpha L}$ $\text{ "" "" } \boldsymbol{\left(4\right)}$

Taking $\frac{\left(2\right)}{\left(1\right)}$, we get the so-called transcendental equation:

$\textcolor{g r e e n}{\alpha = k \tan \left(k L\right)}$

Now, let $u = \alpha L$ and $v = k L$. Here we have specified the quantum number to be $v$.

From the definitions of $\alpha$ and $k$, we define

${u}_{0}^{2} \equiv {u}^{2} + {v}^{2} = {\alpha}^{2} {L}^{2} + {k}^{2} {L}^{2}$

= (2mL^2(V_0 - E))/ℏ^2 + (2mL^2E)/ℏ^2 = (2mL^2V_0)/ℏ^2

So now for the even solution, we can graph

$\alpha L = k L \tan \left(k L\right)$

or

$u = \sqrt{{u}_{0}^{2} - {v}^{2}} = v \tan v$

Now we can simply use Excel to graph $\sqrt{{u}_{0}^{2} - {v}^{2}}$ vs. $v$ and $v \tan v$ vs. $v$ on the same graph, and see where they intersect.

The intersections give the quantum numbers for each eigenvalue.

Suppose V_0 = (20ℏ^2)/(2mL^2), so that ${u}_{0}^{2} = 20$. Then we would get: For the even solutions we got ${v}_{1} \approx 1.28042$ and ${v}_{3} \approx 3.72713$. We'll see another one in the odd solutions later.

For now, each eigenvalue is given in terms of the quantum number ${v}_{n}$:

v_n^2 = k_n^2L^2 = (2mL^2E_n)/ℏ^2

=> color(green)(E_n = (ℏ^2v_n^2)/(2mL^2))

Now let's move on to the odd solutions.

ODD SOLUTIONS

Now the origin is a rotation axis, so that $B = - F$ and $D = 0$ instead. This gives new eigenfunctions:

${\psi}_{I} = B {e}^{\alpha x} , \text{ "" "" } - \infty < x < - L$
${\psi}_{I I} = C \sin \left(k x\right) , \text{ } - L < x < L$
${\psi}_{I I I} = - B {e}^{- \alpha x} , \text{ "" } L < x < \infty$

Now we apply the continuity conditions as before.

$\underline{{\psi}_{I} \left(- L\right) = {\psi}_{I I} \left(- L\right)}$:
$B {e}^{- \alpha L} = C \sin \left(- k L\right)$ $\text{ "" "" } \boldsymbol{\left(1\right)}$

$\underline{\frac{\mathrm{dp} s {i}_{I} \left(- L\right)}{\mathrm{dx}} = \frac{\mathrm{dp} s {i}_{I I} \left(- L\right)}{\mathrm{dx}}}$:
$\alpha B {e}^{- \alpha L} = k C \cos \left(- k L\right)$ $\text{ "" } \boldsymbol{\left(2\right)}$

$\underline{{\psi}_{I I} \left(L\right) = {\psi}_{I I I} \left(L\right)}$:
$C \sin \left(k L\right) = - B {e}^{- \alpha L}$ $\text{ "" "" } \boldsymbol{\left(3\right)}$

$\underline{\frac{\mathrm{dp} s {i}_{I I} \left(L\right)}{\mathrm{dx}} = \frac{\mathrm{dp} s {i}_{I I I} \left(L\right)}{\mathrm{dx}}}$
$k C \cos \left(k L\right) = \alpha B {e}^{- \alpha L}$ $\text{ "" "" } \boldsymbol{\left(4\right)}$

Taking $\frac{\left(2\right)}{\left(1\right)}$ as before, we obtain the other transcendental equation:

$\textcolor{g r e e n}{\alpha = - k \cot \left(k L\right)}$

Using the same substitutions as before, we again graph $\sqrt{{u}_{0}^{2} - {v}^{2}}$ vs. $v$ and $- v \cot v$ vs. $v$ on the same graph and find the intersection(s) to get the quantum number(s). In this case our well only had three energy levels. Here we again had ${u}_{0}^{2} = 20$ and found one quantum number ${v}_{2} \approx 2.53809$.

COMBINED SOLUTIONS

And combining them onto one graph, we get the three energy levels in this finite well with V_0 = (20ℏ^2)/(2mL^2): color(blue)(E_1 = (1.63948ℏ^2)/(2mL^2))

color(blue)(E_2 = (6.44190ℏ^2)/(2mL^2))

color(blue)(E_3 = (13.89150ℏ^2)/(2mL^2))

And of course, if we let ${u}_{0} \to \infty$, we recover the solutions to the infinite potential well. In that case the circular graph we have has a radius of $\infty$.

The choice of ${u}_{0}^{2}$ (and thus ${V}_{0}$ in units of ℏ^2/(2mL^2)) just sets what the height of the potential well is.