# Twin delta potential

1. Aug 11, 2007

### ehrenfest

Hello, how would you derive the solutions of the twin delta function potential:
$$V(x) = -g \delta(x+a) + \delta(x -a )$$
?

Obviously the solutions outside the potential are Ae^Kx for x less than aand A e^-Kx for x greater than a. But how do you derive the hyperbolic cosine function for the interval between -a and a? Please use symmetry arguments because I need to understand those!

2. Aug 11, 2007

### Gokul43201

Staff Emeritus
Your potential is even about the origin. What does that tell you about solutions for the wavefunction?

3. Aug 11, 2007

### ehrenfest

That tells me that because the solutions are exponentials and the Schrodinger equation is a second-order DE, the wavefunctions need to be even or odd about the origin. Why does this mean the coefficients need to be the same for the even and odd parts of the function in the interval -a to a?

4. Aug 11, 2007

### Gokul43201

Staff Emeritus
That follows from continuity at -a and +a (writing psi in the middle as the sum of even and odd parts).

Last edited: Aug 11, 2007
5. Aug 11, 2007

### ehrenfest

So,

$$\psi(x) = C e^{Kx} + D e^{-Kx}$$

in the middle interval. So we apply

$$C e^{Ka} + D e^{-Ka} = Ae^{-Ka}$$

and

$$C e^{-Ka} + D e^{Ka} = Ae^{Ka}$$.

I do not see why that implies C = D. I guess a related issue I am confused about is how we know the coefficient for the the wavefunction for large positive and negative numbers is the same (A).

EDIT: those two last equation are not supposed to be the same (look at the tex pop up to see that). I think the tex compiler is just being stupid

Last edited: Aug 11, 2007
6. Aug 11, 2007

### Gokul43201

Staff Emeritus
You don't have to start by assuming they have the same coefficient, A. Let the coefficient in the region x>a, be B. From continuity of the derivatives, you'll find that A/B = C/D = h, say. Use this in the equation for the continuity of $\psi(x)$.

7. Aug 11, 2007

### ehrenfest

Do you mean the disconuity of the first derivatives at a and -a? The equation

$$\psi(a^+) - \psi(a^-) = \pm 2mg/\hbar^2 \psi(a)$$

?

The problem with that is that I do not have a single expression for psi(a), right?