A Confusing eigensolutions of a wave function

[Yourong Zang]

Consider a potential cavity
$$V(r)=\begin{cases}\infty, &x\in(-\infty,0]\\\frac{\hslash^2}{m}\Omega\delta(x-a), &x\in(0,\infty)\end{cases}$$
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
$$k^2=\frac{2mE}{\hslash^2}$$
I can clearly understand the first part but in the second part, why is the amplitude of the function 1 and why is there a phase angle?

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

 

[BvU]

Hello Yourong Zang, $\qquad$ $\qquad$ !

Please post this kind of questions in the homework forum and use the template -- it's mandatory, see guidelines

Your problem is worked out here With a positive omega you have a delta barrier, not a cavity.

 

[Yourong Zang]

Hello Yourong Zang, $\qquad$ $\qquad$ !

Please post this kind of questions in the homework forum and use the template -- it's mandatory, see guidelines

Your problem is worked out here With a positive omega you have a delta barrier, not a cavity.

Thank you.

 

[DrClaude]

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