# Delta function potential; Schrodinger Equation

1. Feb 18, 2014

### unscientific

1. The problem statement, all variables and given/known data

Consider the TISE for a particle of mass m moving along the x-axis and interacting an attractive delta function potential at origin:

Part(a): What is the difference between a bound state particle and a free particle?

Part(b): Show $\psi _{(x)} = exp (-|k|x)$ is a solution of TISE, given conditions:

Part(c): Explain why symmetric solution is of form:

Part(d): Use boundary conditions to show the following:

Part(e): Investigate this solution graphically

Part(f): Find the significance of this result

2. Relevant equations

3. The attempt at a solution

Part(a)

A bound particle as energy E<0 while a continuum eigenfunction gives energy E>0.

Part (b)

TISE is given by:

$$\hat {H}\psi = E\psi$$

Using first boundary condition provided we get

$$2k = \frac{\lambda}{a}$$

Substituting $\psi = e^{-k|x|}$ into TISE and using what we have just found:

$$-\frac{\hbar ^2}{2m}(\frac{\lambda}{2a})^2 - \frac{\hbar ^2 \lambda}{2ma}\delta_{(x)} = E$$

I'm not sure why the delta function goes to zero here?

$$E = -\frac{\hbar ^2}{2m}(\frac{\lambda}{2a})^2$$

Part (c)

For |x|>a, V = 0:
$$-\frac{\hbar^2}{2m}\frac{\partial ^2 \psi}{\partial x^2} = E\psi$$

Solving this equation gives an exponential $\psi = exp (-k|x|)$.

For -a<x<a, the particle is bounded and symmetric about origin so $\psi \propto exp(kx) + exp(-kx)$ which is $\propto cosh(kx) = A cosh(kx)$.

Part (d)

Boundary Conditions:

At x = a, $\psi_{|x|<a} = \psi_{|x|>a}$.
Thus we get:

$$e^{-ka} = A cosh (ka)$$

Second boundary condition:
At x = a, $\frac{\partial \psi}{\partial x}_{left} = - \frac{\partial \psi}{\partial x}_{right}$

$$e^{-ka} = A sinh (ka)$$

Using boundary condition earlier,

$$tanh (ka) = 1$$
$$tanh(ka) = 2 - 1$$

Can I simply substitute in $2k = \frac{\lambda}{a}$ here?

$$tanh(ka) = \frac{2ka}{ka} - 1$$
$$tanh (ka) = \frac{\lambda}{y} - 1$$

Part (e)

We can see that $y_0 < \lambda$.

Since $tanh(y_0) < 1$, thus $y_0 > \frac{\lambda}{2}$.

Since for x≠a both delta functions are positive, it implies that the second term corresponding to the two delta functions $\delta_{(x-a)}$ and $\delta_{(x+a)}$ in the schrodinger equation are >0. Thus the total energy subtract that is less than $-\frac{\hbar ^2\lambda^2}{8ma^2}$.

Part(f)

Two delta function corresponds to electron being attracted to the two protons in nucleus (which is more massive so can treat as stationary). This is a more stable (favourable) configuration, so

$$H + H^+ \rightarrow H_2^+$$

2. Feb 19, 2014

### vela

Staff Emeritus
You're verifying the solution for x>0 and x<0. At x=0, there's a cusp, so $\psi''$ isn't defined, so the Schrodinger equation doesn't really apply.

This isn't correct. Integrate the Schrodinger equation between $x=a-\varepsilon$ and $x=a+\varepsilon$ and take the limit as $\varepsilon \to 0^+$. This will yield the proper boundary condition on the derivatives.

3. Feb 19, 2014

### unscientific

So the equation$exp(-k|x|)$ here only applies when x≠0. We can't find out what the solution is when x = 0? Is it because due to the nature of the delta function it is a sum of infinite series at x = 0?

But graphically you can see that it is symmetric about 0. You can also see that at x-ε the gradient is opposite that of at x+ε.

4. Feb 19, 2014

### vela

Staff Emeritus
At x=0, the lefthand side of the Schrodinger equation isn't defined, so it doesn't really tell you anything about $\psi$. The continuity requirement, however, allows you to fill in the hole at x=0.

It may look that way in the picture when you have a single delta function, but this time you have two. You no longer have the symmetry you had before. You need to justify the boundary condition mathematically.