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

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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 $V_0$
$V(x) = {(-V_0 ,|x|>=L),(0, "elsewhere"):}$

$alphatanalpha = k$ (even parity)
$alphacotalpha = -k$ (odd parity)

i hope that u understand what $alpha and k$ are.

but to mark the energy eigenvalues

i had to make circle of radius $alpha + k$
what is the physical reason behind this?

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Answer 1 · 2018-03-16T05:03:47

[Here is the Excel sheet I made while doing this.](https://www.dropbox.com/s/hrwleaifu9oos9c/GRAPH_-_Finite_Potential_Well_Solutions.xlsx?dl=0)

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 $sqrt(alpha^2L^2 + k^2L^2)$ in your notation...

$lim_(V_0 -> oo) "Finite Well" = "Infinite Potential Well"$

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

$ul(E" "" "" "" "" "" "E//V_0" "" "color(white)(.)"Quantum Number"" "" ")$
$(1.63948ℏ^2)/(2mL^2)" "" "0.081974" "" "1.28042$

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

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

The graph below utilizes a well width of $2L = 4$ as an example, with the appropriate coefficients $B$ , $C$ , and $D$ to make the wave functions a practical scale.

Eigenfunctions of the Finite Well

Well, let's start by properly defining the problem... instead of defining a tunnelling problem with a barrier of width $2L$ , which you have done, let's define a finite potential well of width $2L$ ...

$V(x) = {(V_0, |x| >= L), (0, -L < x < L):}$

That way, we get:

![https://upload.wikimedia.org/]( useruploads.socratic.org )

Then, let's define the eigenfunctions for each region $(I,II,III)$ :

$psi_(I) = Ae^(-alphax) + Be^(alphax), " "" "" "" "-oo < x < -L$
$psi_(II) = Csin(kx) + Dcos(kx), " "-L < x < L$
$psi_(III) = Fe^(-alphax) + Ge^(alphax), " "" "" "" "L < x < oo$

where:

$alpha = sqrt((2m(V_0 - E))/ℏ^2)$ , $" "E < V_0$ for bound states
$k = sqrt((2mE)/ℏ^2)$

Since the wave function must vanish at $pmoo$ , $A = 0$ 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.

$psi_(I) = Be^(alphax), " "" "" "-oo < x < -L$
$psi_(II) = Dcos(kx), " "-L < x < L$
$psi_(III) = Be^(-alphax), " "" "" "L < x < oo$

Now we apply the continuity conditions.

$ul(psi_(I)(-L) = psi_(II)(-L))$ :
$Be^(-alphaL) = Dcos(-kL)$ $" "" "" "bb((1))$

$ul((dpsi_(I)(-L))/(dx) = (dpsi_(II)(-L))/(dx))$ :
$alphaBe^(-alphaL) = -kDsin(-kL)$ $" "bb((2))$

$ul(psi_(II)(L) = psi_(III)(L))$ :
$Dcos(kL) = Be^(-alphaL)$ $" "" "" "" "bb((3))$

$ul((dpsi_(II)(L))/(dx) = (dpsi_(III)(L))/(dx))$
$kDsin(kL) = alphaBe^(-alphaL)$ $" "" "" "bb((4))$

Taking $((2))/((1))$ , we get the so-called transcendental equation:

$color(green)(alpha = ktan(kL))$

Now, let $u = alphaL$ and $v = kL$ . Here we have specified the quantum number to be $v$ .

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

$u_0^2 -= u^2 + v^2 = alpha^2L^2 + k^2L^2$

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

So now for the even solution, we can graph

$alphaL = kLtan(kL)$

or

$u = sqrt(u_0^2 - v^2) = vtanv$

Now we can simply use Excel to graph $sqrt(u_0^2 - v^2)$ vs. $v$ and $vtanv$ 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 $v_1 ~~ 1.28042$ and $v_3 ~~ 3.72713$ +%3D+vtanv). 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:

$psi_(I) = Be^(alphax), " "" "" "-oo < x < -L$
$psi_(II) = Csin(kx), " "-L < x < L$
$psi_(III) = -Be^(-alphax), " "" "L < x < oo$

Now we apply the continuity conditions as before.

$ul(psi_(I)(-L) = psi_(II)(-L))$ :
$Be^(-alphaL) = Csin(-kL)$ $" "" "" "bb((1))$

$ul((dpsi_(I)(-L))/(dx) = (dpsi_(II)(-L))/(dx))$ :
$alphaBe^(-alphaL) = kCcos(-kL)$ $" "" "bb((2))$

$ul(psi_(II)(L) = psi_(III)(L))$ :
$Csin(kL) = -Be^(-alphaL)$ $" "" "" "bb((3))$

$ul((dpsi_(II)(L))/(dx) = (dpsi_(III)(L))/(dx))$
$kCcos(kL) = alphaBe^(-alphaL)$ $" "" "" "bb((4))$

Taking $((2))/((1))$ as before, we obtain the other transcendental equation:

$color(green)(alpha = -kcot(kL))$

Using the same substitutions as before, we again graph $sqrt(u_0^2 - v^2)$ vs. $v$ and $-vcotv$ 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 $v_2 ~~ 2.53809$ +%3D+-vcotv).

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 $u_0 -> oo$ , we recover the solutions to the infinite potential well. In that case the circular graph we have has a radius of $oo$ .

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.

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

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