Dirac-Delta Potential (Scattering)

In summary: We can only reason about the entire ket being orthogonal to the initial state.In summary, the problem involves solving for the energy eigenvectors for a delta potential and then using the Schrodinger equation to find the wave function at a later time. The integral for the wave function involves expansion in both the position and momentum bases, and results in Dirac-Delta functions. The two parts of the integral cannot be considered separately as they are components of the same energy eigenket.
  • #1
Nicolaus
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0

Homework Statement


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)?

Homework Equations


Schrodinger Eqn.

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 traveling 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?
 
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  • #2
Bump
 
  • #3
Nicolaus said:
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 traveling 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?
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
I am getting Dirac-Delta Functions for φ(k). Does this sound right?
 
  • #5
Nicolaus said:
I am getting Dirac-Delta Functions for φ(k). Does this sound right?
Yes.
 
  • #6
Qualitatively, though, the initial state is a non-localized particle with well-defined energy, traveling 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
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.
 

1. What is a Dirac-Delta Potential?

The Dirac-Delta Potential, also known as the Dirac Delta function, is a mathematical concept used in quantum mechanics to describe a potential energy function that is highly concentrated at a single point in space. It is often represented by the symbol δ and has the properties of being infinite at its center and zero everywhere else.

2. How is the Dirac-Delta Potential used in scattering problems?

In scattering problems, the Dirac-Delta Potential is used to model the interaction between particles at a specific point in space. This allows us to study the behavior of particles as they approach and interact with this potential, providing valuable insights into the scattering process.

3. What are the properties of the Dirac-Delta Potential?

Some of the key properties of the Dirac-Delta Potential include its infinite height at the center, its narrow width, and its ability to represent a point-like particle or interaction. It is also a continuous and differentiable function, making it useful in mathematical calculations.

4. How does the strength of the Dirac-Delta Potential affect the scattering process?

The strength of the Dirac-Delta Potential at the point of interaction can greatly influence the scattering process. A stronger potential will result in a larger deflection of the particle, while a weaker potential will have a smaller effect. This makes the Dirac-Delta Potential a useful tool for studying the effects of different interaction strengths on scattering behavior.

5. What are the limitations of using the Dirac-Delta Potential in scattering problems?

While the Dirac-Delta Potential is a useful concept in scattering problems, it also has its limitations. For example, it is a simplification of real-world interactions and does not take into account the finite size of particles or the effects of other forces. Additionally, its infinite nature can lead to mathematical inconsistencies, requiring careful use and interpretation in calculations.

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