Find the wave function of a particle in a spherical cavity

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Homework Statement
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Relevant Equations
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(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 Schrodinger 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 Schrodinger equation as well as the boundary condition at ##r=R##. After making adjustments by trial and error we find
$$\alpha=\frac{n\pi}{R-a}$$
$$\psi=Nsin\Big(\frac{n\pi}{R-a}(r-a)\Big)$$
$$E=\frac{n^2\pi^2\hbar^2}{2m(R-a)^2}$$
 
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docnet said:
(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 Schrodinger 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)##?
 
Thank u! Re-try:

Screen Shot 2021-02-28 at 4.38.21 PM.png


(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 Schrodinger 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 Schrodinger equation as well as the boundary condition at ##r=R##. After making adjustments by trial and error we find
$$\alpha=\frac{n\pi}{R-a}$$
$$\psi=\frac{U(r)}{r}=\frac{N}{r}sin\Big(\frac{n\pi}{R-a}(r-a)\Big)$$
$$E=\frac{n^2\pi^2\hbar^2}{2m(R-a)^2}$$
 
Your work looks correct to me.
 
:bow:
 
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