# Homework Help: Dirac-Delta Potential (Scattering)

1. Oct 16, 2015

### Nicolaus

1. The problem statement, all variables and given/known data
Say there exists a delta potential V(x) = α(x-L). Solve the eigenvalue problem to yield the energy eigenstates.
Say a quanton is in an initial state given by Ψ(x)=Aeikox at t=0. What is Ψ(x,t)?

2. Relevant equations
Schrodinger Eqn.

3. The attempt at a solution
I solved the first part to find the energy eigenvectors (not-normalized), after applying the boundary condition (including the discontinuity at x-L), to get:
Ψk(x) = A(eikx + B/A(e-ikx) x<L
Ψk(x)=A(C/A)(eikx) x>L
B/A and C/A were obtained from the boundary conditions, and of course represent the reflection and transmission coefficients, respectively.
Since ψ(x,t) = ∫Ψk(x)⋅φ(k)dk
where φ(k) represents the coefficients of the wave function, obtained via Fourier Transform.
i.e. φ(k) = ∫Ψk*(x)⋅Ψ(x)dx
I'm having trouble, qualitatively, understanding how to compute the given integral given that the initial state has a well-defined momentum and is travelling from the left to the right; would the integral with B/A(-ikx) vanish since it represents a left moving quanton, which is orthogonal to the right moving quanton in the initial state?

2. Oct 18, 2015

### Nicolaus

Bump

3. Oct 18, 2015

### andrewkirk

The integral is an expansion in the X basis of $\langle\psi_k|\psi\rangle$. Note that $|\psi\rangle$ is simply $|p_0\rangle$ in the P (momentum) basis, where $p_0=\hbar k_0$ (or maybe $\frac{k_0}{\hbar}$. I'm in a rush and don't have time to check which it is). That suggests to me that the problem may be more tractable if we instead compute $\langle\psi_k|\psi\rangle$ by expanding in the P basis. I would take the case $x>L$ first, as it has only one component. What is the representation in the P basis of the ket that has representation $x\mapsto \psi_k(x)=\alpha e^{ikx}$ in the X basis?

4. Oct 20, 2015

### Nicolaus

I am getting Dirac-Delta Functions for φ(k). Does this sound right?

5. Oct 20, 2015

### andrewkirk

Yes.

6. Oct 20, 2015

### Nicolaus

Qualitatively, though, the initial state is a non-localized particle with well-defined energy, travelling to the right. Would this mean that the second integral (with the reflection coefficient) would be zero since that represents a particle moving in the left direction (reflected part), which would be orthogonal to right-going particle at t=0?

7. Oct 20, 2015

### andrewkirk

It wouldn't mean that, because the two integrals are not entire wave functions on their own. They are two parts of the same energy eigenket, which is (in the X basis):

$$\left[1-\Theta(x-L)\right](\alpha e^{ik_1 x}+\beta e^{-ik_1x})+\Theta(x-L) \gamma e^{ik_2x}$$
where $\Theta$ is the Heaviside function and $k_2$ is a function of $k_1$ and the potential step $V$.

We can't break off a spatial fragment of the ket and reason about that fragment being orthogonal to the initial state.