- #1
Yourong Zang
- 5
- 0
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.
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: