Double delta function well

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.27.

We can extend the case of the particle in a delta function well to the case of a particle in a double delta function well. That is, the potential is

$\displaystyle V(x)=-\alpha\left[\delta(x+a)+\delta(x-a)\right] \ \ \ \ \ (1)$

where ${\alpha}$ gives the strength of the well.

Since the potential is an even function, any solution can be expressed as a linear combination of even and odd solutions. Consider even solutions first. In regions away from the delta functions, the Schrödinger equation is, since ${V=0}$:

$\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}\psi(x)=E\psi(x) \ \ \ \ \ (2)$

The most general even solution of this equation is (with ${\kappa\equiv\sqrt{-2mE}/\hbar}$; remember ${E}$ is negative so ${\kappa}$ is real)

$\displaystyle \psi_{e}(x)=\begin{cases} Ae^{-\kappa x} & x>a\\ Be^{-\kappa x}+Ce^{\kappa x} & 0

We can narrow down the number of constants by applying Born’s conditions. At all points the wave function must be continuous, so applying this condition at ${x=a}$ gives

 $\displaystyle Ae^{-\kappa a}$ $\displaystyle =$ $\displaystyle Be^{-\kappa a}+Ce^{\kappa a}\ \ \ \ \ (4)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle B+Ce^{2\kappa a} \ \ \ \ \ (5)$

At ${x=0}$, the continuity condition gives us no information, and at ${x=-a}$ we just repeat the condition at ${x=a}$.

Another condition is that the derivative of the wave function must be continuous at all points where the potential is finite. This means we can apply this condition at ${x=0}$ to get

 $\displaystyle -B\kappa+C\kappa$ $\displaystyle =$ $\displaystyle B\kappa-C\kappa\ \ \ \ \ (6)$ $\displaystyle C-B$ $\displaystyle =$ $\displaystyle B-C\ \ \ \ \ (7)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle C \ \ \ \ \ (8)$

Thus we get

$\displaystyle \psi_{e}(x)=\begin{cases} B\left(1+e^{2\kappa a}\right)e^{-\kappa x} & x>a\\ B\left(e^{-\kappa x}+e^{\kappa x}\right) & -a

The constant ${B}$ could be found by normalizing the wave function but we won’t need to do that to find the energies.

Following the same logic as in the single delta function well, we calculate the difference in the derivative across the delta function at ${x=a}$. We get

 $\displaystyle \psi_{e\left(a+\right)}'$ $\displaystyle =$ $\displaystyle -B\kappa\left(1+e^{2\kappa a}\right)e^{-\kappa a}\ \ \ \ \ (10)$ $\displaystyle \psi_{e\left(a-\right)}'$ $\displaystyle =$ $\displaystyle B\left(-\kappa e^{-\kappa a}+\kappa e^{\kappa a}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -B\kappa e^{-\kappa a}\left(1-e^{2\kappa a}\right) \ \ \ \ \ (12)$

The change in derivative across ${x=a}$ is then

 $\displaystyle \Delta\psi_{e}'$ $\displaystyle =$ $\displaystyle \psi_{e\left(a+\right)}'-\psi_{e\left(a-\right)}'\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2B\kappa e^{\kappa a} \ \ \ \ \ (14)$

As in the single delta function well, we must also have

 $\displaystyle \Delta\psi_{e}'$ $\displaystyle =$ $\displaystyle -\frac{2m\alpha}{\hbar^{2}}\psi(a)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2m\alpha}{\hbar^{2}}B\left(e^{-\kappa a}+e^{\kappa a}\right)\ \ \ \ \ (16)$ $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle \frac{m\alpha}{\hbar^{2}}\left(1+e^{-2\kappa a}\right) \ \ \ \ \ (17)$

To get a condition on the energy, we need ${\kappa}$, but this is a transcendental equation in ${\kappa}$, so the only way we can solve it is numerically. We can see the solutions graphically, however, if we plot the left and right sides of the equation and look for intersections. To make this easier, we’ll introduce the auxiliary variable ${\xi\equiv a\kappa}$ to get

$\displaystyle \xi=\frac{ma\alpha}{\hbar^{2}}\left(1+e^{-2\xi}\right) \ \ \ \ \ (18)$

Now if we specify ${\alpha}$ we can get a numerical solution. If ${\alpha=\hbar^{2}/ma}$, we get

$\displaystyle \xi=\left(1+e^{-2\xi}\right) \ \ \ \ \ (19)$

In the plot below, we draw ${y=\xi}$ in red and ${y=\left(1+e^{-2\xi}\right)}$ in blue.

We see there is a solution around ${\xi=1.1}$, but to get this more accurately, we can use software like Maple’s fsolve command to solve the equation numerically. We get ${\xi=1.108857553}$. From this we can get the energy as

 $\displaystyle E$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}\kappa^{2}}{2m}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}\xi^{2}}{2ma^{2}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -0.614782\frac{\hbar^{2}}{ma^{2}} \ \ \ \ \ (22)$

For ${\alpha=\hbar^{2}/4ma}$, the equation becomes

$\displaystyle 4\xi=\left(1+e^{-2\xi}\right) \ \ \ \ \ (23)$

In the plot below, ${y=4\xi}$ is in red and ${y=\left(1+e^{-2\xi}\right)}$ is in blue.

A solution exists around ${\xi=0.4}$ and using Maple again we discover that ${\xi=0.3694175156}$, giving an energy of ${E=-0.068235\hbar^{2}/ma^{2}}$.

For the odd solution, we start off with the most general odd wave function:

$\displaystyle \psi_{o}(x)=\begin{cases} Ae^{-\kappa x} & x>a\\ Be^{-\kappa x}-Ce^{\kappa x} & 0

As before, the continuity condition at ${x=a}$ gives us

 $\displaystyle Ae^{-\kappa a}$ $\displaystyle =$ $\displaystyle Be^{-\kappa a}-Ce^{\kappa a}\ \ \ \ \ (25)$ $\displaystyle A$ $\displaystyle =$ $\displaystyle B-Ce^{2\kappa a} \ \ \ \ \ (26)$

This time the continuity of the derivative at ${x=0}$ gives us nothing new, but the continuity of the wave function itself gives us

 $\displaystyle B-C$ $\displaystyle =$ $\displaystyle -B+C\ \ \ \ \ (27)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle C \ \ \ \ \ (28)$

Thus the function is

$\displaystyle \psi_{o}(x)=\begin{cases} B\left(1-e^{2\kappa a}\right)e^{-\kappa x} & x>a\\ B\left(e^{-\kappa x}-e^{\kappa x}\right) & -a

From here we follow the same argument as above.

 $\displaystyle \psi_{o\left(a+\right)}'$ $\displaystyle =$ $\displaystyle -B\kappa\left(1-e^{2\kappa a}\right)e^{-\kappa a}\ \ \ \ \ (30)$ $\displaystyle \psi_{o\left(a-\right)}'$ $\displaystyle =$ $\displaystyle B\left(-\kappa e^{-\kappa a}-\kappa e^{\kappa a}\right)\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -B\kappa e^{-\kappa a}\left(1+e^{2\kappa a}\right) \ \ \ \ \ (32)$

The change in derivative across ${x=a}$ is then

 $\displaystyle \Delta\psi_{o}'$ $\displaystyle =$ $\displaystyle \psi_{o\left(a+\right)}'-\psi_{o\left(a-\right)}'$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2B\kappa e^{\kappa a}$

As in the single delta function well, we must also have

 $\displaystyle \Delta\psi_{o}'$ $\displaystyle =$ $\displaystyle -\frac{2m\alpha}{\hbar^{2}}\psi(a)\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2m\alpha}{\hbar^{2}}B\left(e^{-\kappa a}-e^{\kappa a}\right)\ \ \ \ \ (34)$ $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle \frac{m\alpha}{\hbar^{2}}\left(1-e^{-2\kappa a}\right) \ \ \ \ \ (35)$

In terms of ${\xi}$ we get

$\displaystyle \xi=\frac{ma\alpha}{\hbar^{2}}\left(1-e^{-2\xi}\right) \ \ \ \ \ (36)$

If ${\alpha=\hbar^{2}/ma}$, we get

$\displaystyle \xi=\left(1-e^{-2\xi}\right) \ \ \ \ \ (37)$

In the plot below, we draw ${y=\xi}$ in red and ${y=\left(1-e^{-2\xi}\right)}$ in blue.

We see there is a solution near ${\xi=0.8}$ and using Maple, we find ${\xi=0.79681213}$, with corresponding energy ${E=-0.317455\hbar^{2}/ma^{2}}$.

There is also an intersection at ${\xi=0}$ for any value of ${\alpha}$. However, this would correspond to ${E=0}$, which would mean that ${d^{2}\psi/dx^{2}=0}$ everywhere. This would imply ${\psi(x)=Ax+B}$ which isn’t normalizable so it isn’t a physically acceptable solution.

For ${\alpha=\hbar^{2}/4ma}$, the equation becomes

$\displaystyle 4\xi=\left(1-e^{-2\xi}\right) \ \ \ \ \ (38)$

In the plot below, ${y=4\xi}$ is in red and ${y=\left(1-e^{-2\xi}\right)}$ is in blue.

In this case, there is no intersection except the non-physical one at ${\xi=0}$, so for this value of ${\alpha}$, there is no bound state with an odd wave function.

16 thoughts on “Double delta function well”

1. Spence

Thanks! I have been looking all over for this (not for homework–I’m trying to prepare for an exam!) and this site contains just the info I was looking for: an easy-to-follow and step-by-step solution to the “double delta” potential problem.

1. growescience

If you’re wondering why there is
only
a single exponential term for ${psi_{e}}$ in equation 3 for the regions ${x>a}$ and ${x<-a}$ it is because we require the wave function to be finite everywhere and if we had a ${e^{+kappa x}}$ term for ${x>a}$, then ${psi_{e}rightarrowinfty}$ as ${xrightarrowinfty}$. Similarly, if we had a ${e^{-kappa x}}$ term for ${x<-a}$, then ${psi_{e}rightarrowinfty}$ as ${xrightarrow-infty}$.

2. Antara dey

Why the constants are taken same in two interval 0<x<a and -a<xa and x<-a. ? Is that we consider the even & odd solutions as the potensial is eve?

1. gwrowe Post author

In equation 3, we’re looking for even solutions so ${\psi_{e}\left(-x\right)=\psi_{e}\left(x\right)}$. In equation 24, the solution is odd so ${\psi_{e}\left(-x\right)=-\psi_{e}\left(x\right)}$.

3. RHB

I’m not sure why in eq. (3) there are 4 equations modeling it… I had thought there would only be 3, with the region between -a and a being taken all together. What is the intuition for treating that section as two different regions, -a -> 0 and 0 -> a? Thanks!

1. gwrowe Post author

In equation 3, we’re looking for even solutions so ${\psi_{e}\left(-x\right)=\psi_{e}\left(x\right)}$. To do this we can write down the general solution for ${x>0}$ (which is the first two lines in equation 3), and then take the mirror image to get the general solution for ${x<0}$. You’re right that nothing special happens with the potential at ${x=0}$; but here we’re looking explicitly for even solutions so we need to consider the regions ${x<0}$ and ${x>0}$ separately to ensure that ${\psi_{e}\left(-x\right)=\psi_{e}\left(x\right)}$.

4. Corey Loescher

For the graphical solution of (38), why can’t xi be negative?
I know that xi=kappa*a, and that a cannot be negative, since from alpha=hbar^2/(4ma), this would make alpha negative, creating a positive potential peak instead of a potential well.
This leads to an evaluation of whether or not kappa can be negative. From Eqn 2.128 in Griffiths, it is obvious that kappa for a single delta-function well must be positive, since attempting to normalize the solution gives B=sqrt(kappa). However, this doesn’t mean that normalizing the double well potential for psi_o will require kappa to be positive. Is there some other reason that kappa must be negative for this particular problem that I’m missing?

I made the bold assumption that kappa might be allowed to be negative, and attempted to normalize the solution. My solution was |B|^2=kappa*(e^(2*kappa*a)-kappa*a-2) utilizing the solution for kappa*a that my graphing calculator spat out, kappa*a=-.6282156, which is |B|^2=kappa*(-1.087116), which means that kappa must be negative to make the solution normalizable in this particular case.

Could you double-check my calculation of B with respect to this problem? Otherwise, the last answer should not be that there is no solution, but instead, E=-.1973274*hbar^2/(ma^2) for alpha=hbar^2/(4ma).

1. gwrowe Post author

You could choose ${\kappa}$ to be negative, but then you’d need to alter the solutions in eqns 3 and 24 so that ${\psi=Ae^{\kappa x}}$ for ${x>a}$ (and also flip the sign of the exponent for ${x). The parameter ${\xi=\kappa a}$ would then always be negative, and you’d end up with the same equations as in the post so you’d get the same solutions. The point is that you can make ${\kappa}$ either positive or negative, but not both.

1. Corey Loescher

I think I see what you mean. In other words, if kappa were negative, the previous solution for psi_o would blow up at positive and negative infinity.

Thanks for the quick reply. This site is a fantastic resource for anyone needing help with understanding quantum mechanical concepts.