Find the wave function of a particle in a spherical cavity

[docnet]

Homework Statement

psb

Relevant Equations

psb

(a) Let the center of the concentric spheres be the origin at $r=0$, where r is the radius defined in spherical coordinates. The potential is given by the piece-wise function
$$V(r)=\infty, r<a$$
$$V(r)=0, a<r<R$$
$$V(r)=\infty, r<a$$
(b) we solve the Schrödinger equation and obtain
$$\psi(r)=C_1cos(\alpha r)+C_2sin(\alpha r)$$
we look for a wavefunction $\psi(r)$ that vanishes at $$r=a$$ and $$r=R$$
(c) we take $$sin(\alpha r)$$ and do a change of variables that translates $\psi$ by $a$ in the negative $r$ direction $$r=\hat{r}-a$$ We solve for an $α$ that solves the Schrödinger equation as well as the boundary condition at $r=R$. After making adjustments by trial and error we find
$$\alpha=\frac{n\pi}{R-a}$$
$$\psi=Nsin\Big(\frac{n\pi}{R-a}(r-a)\Big)$$
$$E=\frac{n^2\pi^2\hbar^2}{2m(R-a)^2}$$

 

[TSny]

(a) Let the center of the concentric spheres be the origin at $r=0$, where r is the radius defined in spherical coordinates. The potential is given by the piece-wise function
$$V(r)=\infty, r<a$$ $$V(r)=0, a<r<R$$ $$V(r)=\infty, r<a$$

The first and third equations above are identical.

(b) we solve the Schrödinger equation and obtain
$$\psi(r)=C_1cos(\alpha r)+C_2sin(\alpha r)$$

Should the left-hand side be $U(r)$ rather than $\psi(r)$?

 

[docnet]

Thank u! Re-try:

(a) Let the center of the concentric spheres be the origin at $r=0$, where r is the radius defined in spherical coordinates. The potential is given by the piece-wise function
$$V(r)=\infty, r<a$$
$$V(r)=0, a<r<R$$
$$V(r)=\infty, R<r$$
(b) we solve the Schrödinger equation and obtain
$$U(r)=C_1cos(\alpha r)+C_2sin(\alpha r)$$
we look for a wavefunction $U(r)$ that vanishes at $$r=a$$ and $$r=R$$
(c) we take $$sin(\alpha r)$$ and do a change of variables that translates $\psi$ by $a$ in the negative $r$ direction $$r=\hat{r}-a$$ We solve for an $α$ that solves the Schrödinger equation as well as the boundary condition at $r=R$. After making adjustments by trial and error we find
$$\alpha=\frac{n\pi}{R-a}$$
$$\psi=\frac{U(r)}{r}=\frac{N}{r}sin\Big(\frac{n\pi}{R-a}(r-a)\Big)$$
$$E=\frac{n^2\pi^2\hbar^2}{2m(R-a)^2}$$

 

[TSny]

Your work looks correct to me.

 

[docnet]