# How do I plot the energy eigenvalues and wave function solutions of a delta function double potential well with #V(chi) = -alpha(delta(chi+L)+delta(chi-L))#?

##### 1 Answer

Well, here it is... Here is the Excel sheet I made while doing this. Also, here are the even solution graphs.

This, by the way, results in a wave function solution that resembles the hydrogen molecule, except with internuclear distance

This is the case because

**DISCLAIMER:** *VERY LONG ANSWER!*

Here's the scenario (just change

We set up the wave function for a tunnelling free-particle. Call the regions

#psi_I = Ae^(-kchi) + Be^(kchi)#

#psi_(II) = Ce^(-kchi) + De^(kchi)#

#psi_(III) = Fe^(-kchi) + Ge^(kchi)# where

#k = sqrt(-(2mE)/ℏ^2)# , since#E < 0# for bound states.

Since the wave function must vanish at

#C = D# for even solutions and#C = -D# for odd solutions.#G = B# for even solutions and#G = -B# for odd solutions.

This simplifies down to **even solutions** being:

#psi_"even" = {(Be^(kchi),-oo < chi < -L),(C(e^(-kchi) + e^(kchi)),-L < chi < L),(Be^(-kchi),L < chi < oo):} #

and **odd solutions** being:

#psi_"odd" = {(-Be^(kchi),-oo < chi < -L),(C(e^(-kchi) - e^(kchi)),-L < chi < L),(Be^(-kchi),L < chi < oo):} #

**EVEN SOLUTION**

Continuity at

#Be^(-kL) = C(e^(kL) + e^(-kL))#

#=> B = C(e^(2kL) + 1)#

Next, consider the Schrodinger equation at just

#-ℏ^2/(2m) (d^2psi)/(dchi^2) - alphadelta(chi+L)psi = Epsi#

Integration near

#-ℏ^2/(2m) lim_(epsilon->0) int_(-L-epsilon)^(-L + epsilon) (d^2psi)/(dchi^2) dchi - alpha lim_(epsilon->0) int_(-L-epsilon)^(-L+epsilon) delta(chi+L)psidchi = lim_(epsilon -> 0) Eint_(-L-epsilon)^(-L+epsilon) psidchi#

Denote

#-ℏ^2/(2m) int_(-L^(-))^(-L^(+)) (d^2psi)/(dchi^2) dchi - alpha int_(-L^(-))^(-L^(+)) delta(chi+L)psidchi = 0#

Evaluating the integrals requires examination of the left and right sides of the discontinuous derivative.

#-ℏ^2/(2m) (Delta (dpsi(-L))/(dchi)) = alpha psi(-L^(pm))#

Now evaluate the

#-ℏ^2/(2m) (-kCe^(kL) + kCe^(-kL) - kBe^(-kL)) = alpha Be^(-kL)#

#C(-e^(kL) + e^(-kL)) - Be^(-kL) = -(2malpha)/(ℏ^2k) Be^(-kL)#

Multiply through by

#C(-e^(2kL) + 1) = B - (2malpha)/(ℏ^2k) B#

#C(e^(2kL) - 1) = B((2malpha)/(ℏ^2k) - 1)#

Seeing as we have an expression for

#cancelC(e^(2kL) - 1) = cancelC(e^(2kL) + 1)((2malpha)/(ℏ^2k) - 1)#

#e^(2kL) - 1 = (2malpha)/(ℏ^2k)e^(2kL) - e^(2kL) + (2malpha)/(ℏ^2k) - 1#

#e^(2kL) = ((2malpha)/(ℏ^2k) - 1)e^(2kL) + (2malpha)/(ℏ^2k)#

#1 = (2malpha)/(ℏ^2k) - 1 + (2malpha)/(ℏ^2k)e^(-2kL)#

#cancel2(1 - (malpha)/(ℏ^2k)) = cancel2[(malpha)/(ℏ^2k)e^(-2kL)]#

#color(blue)(e^(-2kL) = (ℏ^2k)/(malpha) - 1)#

Let

Clearly, there will ** always** be an intersection (always at least one bound state) as long as

**decreasing**function intersecting with an

**increasing**function of slope

*positive*

The x-intercept turns out to be

So the approximate **even solution energy** for

#1.27846 ~~ 2kL#

#=> k^2 ~~ 0.40861/L^2 = -(2mE)/(ℏ^2)#

#=> color(blue)(E ~~ -(0.20431ℏ^2)/(mL^2))#

If we take the limit as

**ODD SOLUTION**

Continuity at

#Be^(-kL) = C(e^(-kL) - e^(kL))#

#=> B = -C(e^(2kL) - 1)#

As before, consider the Schrodinger equation at just

#-ℏ^2/(2m) (d^2psi)/(dchi^2) - alphadelta(chi-L)psi = Epsi#

Integration near

#-ℏ^2/(2m) int_(L^(-))^(L^(+)) (d^2psi)/(dchi^2) dchi - alpha int_(L^(-))^(L^(+)) delta(chi-L)psi = 0#

As before, we now get:

#-ℏ^2/(2m) (Delta (dpsi(L))/(dchi)) = alpha psi(L^(pm))#

Now evaluate the

#-ℏ^2/(2m) (-kBe^(-kL) + kCe^(-kL) + kCe^(kL)) = alpha Be^(-kL)#

#(ℏ^2k)/(2malpha) (Be^(-kL) - Ce^(-kL) - Ce^(kL)) = Be^(-kL)#

Again multiply through by

#(ℏ^2k)/(2malpha) (B - C - Ce^(2kL)) = B#

#(ℏ^2k)/(2malpha) (-C - Ce^(2kL)) = B(1 - (ℏ^2k)/(2malpha))#

#(ℏ^2k)/(2malpha) C(1 + e^(2kL)) = B((ℏ^2k)/(2malpha) - 1)#

#C(1 + e^(2kL)) = B(1 - (2malpha)/(ℏ^2k))#

Now plug in

#cancelC(1 + e^(2kL)) = cancelC(1 - e^(2kL))(1 - (2malpha)/(ℏ^2k))#

#1 + e^(2kL) = 1 - (2malpha)/(ℏ^2k) - e^(2kL) + e^(2kL)(2malpha)/(ℏ^2k)#

#cancel2e^(2kL) - e^(2kL)(cancel2malpha)/(ℏ^2k) = - (cancel2malpha)/(ℏ^2k)#

#1 - (malpha)/(ℏ^2k) = - (malpha)/(ℏ^2k)e^(-2kL)#

#color(blue)(e^(-2kL) = 1 - (ℏ^2k)/(malpha))#

Now we again let

The y-intercept turns out to be

**HOW DO THE WAVE FUNCTIONS LOOK?**

We'll get **two** bound states if **one** if **For the rest of this answer, I choose** **.**

The **even solution energy** for

#2.21772 ~~ 2kL#

#=> k^2 ~~ 1.22957/L^2 = -(2mE)/(ℏ^2)#

#=> color(blue)(E ~~ -(0.61479ℏ^2)/(mL^2))#

In this case we can then get **even solution**:

#k = sqrt(-(2m)/ℏ^2 cdot -(0.61479ℏ^2)/(mL^2))#

#= sqrt(2 cdot (0.61479)/(L^2)) = 1.10886/L# where I simply chose

#B = 1# and#C = 0.0981705# to match#psi# at the boundaries.

#=> psi_(I) = Be^(1.10886x//L)#

#=> psi_(II) = C(e^(-1.10886x//L) + e^(1.10886x//L))#

#=> psi_(III) = Be^(-1.10886x//L)#

This results in (the

**This looks remarkably like the hydrogen bonding molecular orbital wave function plot!**

Then, we get **odd solution energy** for

#1.59362 ~~ 2kL#

#=> k^2 ~~ 0.63491/L^2 = -(2mE)/(ℏ^2)#

#=> color(blue)(E ~~ -(0.31745ℏ^2)/(mL^2))#

The other bound state (**odd solution**) has:

#k = sqrt(-(2m)/ℏ^2 cdot -(0.31745ℏ^2)/(mL^2))#

#= sqrt(2 cdot (0.31745)/(L^2)) = 0.79681/L#

#=> psi_(I) = -Be^(0.79681x//L)#

#=> psi_(II) = C(e^(-0.79681x//L) - e^(0.79681x//L))#

#=> psi_(III) = Be^(-0.79681x//L)#

Here I just chose

which looks like the

hydrogen antibonding molecular orbital wave function plot!

(Again, the

What if we choose

If

Hence, we discard the odd solution and **only have an even solution**:

#0.73884 ~~ 2kL#

#=> k^2 ~~ 0.13647/L^2 = -(2mE)/(ℏ^2)#

#=> color(blue)(E ~~ -(0.06824ℏ^2)/(mL^2))#

With this scenario,

#k = sqrt(-(2m)/ℏ^2 cdot -(0.06824ℏ^2)/(mL^2))#

#= sqrt(2 cdot (0.06824)/(L^2)) = 0.36943/L#

You can plug it back in if you wish, but I'm not going to plot this one, because it's unphysical; since this entire problem can be used to model something not unlike