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

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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?

Derive a transcendental equation for the allowed energies and solve it graphically in the finite well of the height

i hope that u understand what

but to mark the energy eigenvalues

i had to make circle of radius

what is the physical reason behind this?

##### 1 Answer

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

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

If we choose **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

Well, let's start by **properly defining the problem**... instead of defining a tunnelling problem with a *barrier* of width ** finite potential well** of width

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

That way, we get:

Then, let's define the **eigenfunctions** for each region

#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 **even** and the **odd** solutions.

**EVEN SOLUTIONS**

For the even solutions, the

#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 **transcendental equation**:

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

Now, let **quantum number** to be

From the definitions of

#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

*The intersections give the quantum numbers for each eigenvalue.*

Suppose

For the even solutions we got

For now, each **eigenvalue** is given in terms of the quantum number

#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 **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 **transcendental equation**:

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

Using the same substitutions as before, we again graph

In this case our well only had three energy levels. Here we again had

**COMBINED SOLUTIONS**

And combining them onto one graph, we get the three energy levels in this finite well with

#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

The choice of