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

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^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.

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:

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#. 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#.

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.