# Eigensolution of the wave function in a potential field.

• Yourong Zang
In summary, the conversation discusses the solution to a scattering problem with a delta potential. It is explained that the amplitude of the wave function is set to 1 to represent an incoming wave from the right, and the reflected wave is also taken into account. The ratio of these two amplitudes is determined by the boundary conditions, and the phase angle is on the right due to the limitation of only being able to have a sine factor at 0.
Yourong Zang
1. Homework Statement
Consider a potential field
$$V(r)=\begin{cases}\infty, &x\in(-\infty,0]\\\frac{\hslash^2}{m}\Omega\delta(x-a), &x\in(0,\infty)\end{cases}$$
The eigenfunction of the wave function in this field suffices
$$-\frac{\hslash^2}{2m}\frac{d^2\psi}{dx^2}+\frac{\hslash^2}{m}\Omega\delta(x-a)\psi=E\psi$$
A textbook gives the following solution:
$$\psi(x)=\begin{cases}Asin(kx), &x\in(0,a)\\ sin(kx+\phi), &x\in(a,\infty)\end{cases}$$
where
$$k^2=\frac{2mE}{\hslash^2}$$

2. Homework Equations
I can clearly understand the first part but in the second part, why does the amplitude of the function equal to 1 and why is there a phase angle?

And is this wave
$$\psi(x)=\sin(kx+\phi)$$
called something like the "excitation mode"?

3. The Attempt at a Solution
A solution about delta potential is not what I want. There is an infinite potential on the left+a delta potential.

Last edited:
Yourong Zang said:
A solution about delta potential is not what I want
It can still help you understand why the amplitudes are different for this scattering problem

The amplitude is set to one to describe an incoming wave from the right (*). Can't be normalized, so it's a mathematical simplification to study solutions of the SE

(*) plus the reflected wave: because of the step everything has to go back again

BvU said:
The amplitude is set to one to descibe an incoming wave from the right. Can't be normalized, so it's a mathematical simplification to study solutions of the SE
So I guess what I need is only the ratio of these two amplitudes (A in this case)?

Right ! Follows from boundary conditions (as in the link in the other thread)

I suppose the ##\phi## is on the right because at 0 you can only have a sine factor

Yourong Zang
BvU said:
Right ! Follows from boundary conditions (as in the link in the other thread)

I suppose the ##\phi## is on the right because at 0 you can only have a sine factor
Great, thank u.

## 1. What is an eigensolution?

An eigensolution is a special type of solution to a mathematical equation that satisfies certain conditions. In the context of the wave function in a potential field, an eigensolution is a specific solution that describes the behavior of a particle in the given potential field.

## 2. How is the wave function related to the eigensolution in a potential field?

The wave function is a mathematical representation of the state of a particle in a potential field. The eigensolution of the wave function describes the specific energy levels and corresponding probabilities of the particle in the given potential field.

## 3. What is the significance of finding eigensolutions in a potential field?

Finding eigensolutions in a potential field allows us to understand the behavior of a particle in that field and predict its energy levels and probabilities. This is important in various fields such as quantum mechanics and chemistry, where the behavior of particles is governed by potential fields.

## 4. How are eigensolutions calculated for the wave function in a potential field?

Eigensolutions are calculated by solving the Schrödinger equation, which is a differential equation that describes the behavior of quantum systems. The solutions to this equation are the eigenfunctions, and the corresponding energies are the eigenvalues.

## 5. Can there be multiple eigensolutions for a given potential field?

Yes, there can be multiple eigensolutions for a given potential field. Each eigensolution represents a different energy level and probability for the particle in that field. The number of eigensolutions depends on the complexity of the potential field and the constraints imposed on the system.

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