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Given the Dirichlet boundary condition, I am to show that the functions that satisfy $$(\nabla ^2 + k_{lmn}^2) \psi_{lmn} (x,y,z) = 0$$ are given by $$\psi_{lmn} = (\frac{\pi}{2x})^{1/2} J_{l+1/2}(x) Y_{lm} (\theta, \phi)$$ for a hollow sphere or radius $a$, where the green function $G$ can be expanded as $$G(\vec{x},\vec{x'}) = \sum_n a_n(\vec{x'}) \psi_n (\vec{x}).$$

Note that I can use the solution of the Helmhotz equation, which is given to me as

$$\Psi = \left\{ \begin{array}{c} J_m(\rho \sqrt{k^2-\alpha^2}) \\ Y_m(\rho \sqrt{k^2 - \alpha^2}) \end{array} \right\} \left\{ \begin{array}{c} e^{i\alpha z} \\ e^{-i \alpha z} \end{array} \right\} \left\{ \begin{array}{c} e^{i m \phi} \\ e^{-i m \phi} \end{array} \right\} $$

where the brackets express a linear combination of their arguments.

I am confused as to how to proceed from there and how to use the BC's to make it look like the desired result.

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    $\begingroup$ Hi user40119, welcome to Physics.SE. You've demonstrated that you're able to use TeX markup. Please edit your question to turn the image into TeX markup too. $\endgroup$ – Brandon Enright Feb 7 '14 at 2:38
  • $\begingroup$ I think there is a typo in your $\psi_{lmn}$. Should it really be independent of $n$? $\endgroup$ – Chris Mueller Apr 9 '14 at 16:54
  • $\begingroup$ @BrandonEnright: OP might not have known how to do the arrays. I did it for him, but OP should at least look at the TeX so he/she knows for next time. $\endgroup$ – DanielSank Oct 25 '14 at 5:07
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    $\begingroup$ Please improve the title, it's too general given what you're actually asking. Titles should be as specific as possible. $\endgroup$ – DanielSank Dec 30 '14 at 6:05
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I cannot comment, that's why I post an answer.

I argee with answers above: you don't need Green's function to solve this equation, you just need to separate variables by entering spherical coordinate system like this $\psi(x,\theta,\phi)=R(r)\psi_{ang}(\theta,\phi)$. After doing this you'll get two separate equations on functions $R(r)$ and $\psi_{ang}(\theta,\phi)$: Bessel equation for $R(r)$, see Bessel functions and equation for angles. Solution of second equation is represented by spherical harmonics.

P.S. It is also easier to search for $R(r) = \frac{\chi(x)}{x}$. Than you'll exactly get Bessel equation.

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I do not think that you need the Green's function to obtain the required result. In view of the symmetry of the problem it is natural to switch to spherical coordinates. Then you end up with a radial part that is a linear combination of a spherical Bessel and Neumann function for each $l$. The boundary conditions and behaviour in $0$ then fix the coefficients. See, for instance, A. Messiah, Quantum Mechanics I, in particular the part about central potentials and the Appendix.

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  • $\begingroup$ I agree with this - I cannot see what Green's functions would have to do with the initial problem you submit, and concur that it all has to do with central potentials, which will be treated in a number of references. $\endgroup$ – Dominique Geffroy Feb 8 '14 at 22:42
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The proposed solution that is given in the last expression is in cylindrical coordinates, which would be an inconvenient choice for this problem. When the Helmholtz equation is solved in spherical coordinates, which would be more convenient for the problem at hand, one obtains solutions given by the product of spherical Bessel functions (Bessel functions with half-integer indices), Legendre polynomials (having another index) and harmonic functions.

To solve the problem for the Dirichlet boundary condition [let's say $\Psi(r=r_0)=0$] one would impose this on the radial part of the solution, which is the spherical Bessel function. There would be different solutions, because the spherical Bessel function become zero at different points. These different solutions represent different modes, distinguished by an addition modal index. Added to the two modal indices that are already present for the solutions, one ends up with the three indices, as given in the first expression.

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