Non-trivial solution for a linear set of coefficients involved in the phonon modes of a semiconductor quantum dot I am trying to use the method outlined in this linked paper (T. Takagahara, Journal of Luminescence, 70 (1996), pp. 129-143) to find the phonon-exciton coupling in a spherical PbS quantum dot. In Eq 2.5 (the first equation below), there is a need to find the coefficients $p$ and $q$, which can be found by using Eq 2.6 (the second equation below). I have tried solving that linear system by using Mathematica's NullSpace and LinearSolve commands and can only come up with $p=q=0$, which implies that no phonon in the system displaces any particle. If anyone is familiar with this method and could give some insight, it would be very helpful. The system is
$u=p_{lm}L_{lm}(hR)+q_{lm}N_{lm}(kR)$
Where
\begin{equation}
 \left( \begin{array}{ccc}
\alpha & \beta\\
\gamma & \delta\end{array} \right)\left( \begin{array}{ccc}
p \\
q\end{array} \right)= \left( \begin{array}{ccc}
0 \\
0 \end{array} \right)
\end{equation}
and
$\alpha=-\sigma^2hRj_l(hR)+2(l+2)j_{l+1}(hR)$
$\beta=lkRj_l(kR)-2l(l+2)j_{l+1}(kR)$
$\gamma=-\sigma^2hRj_l(hR)+2(l-2)j_{l-1}(hR)$
$\delta=(l+1)[2(l-1)j_{l-1}(kR)-kRj_l(kR)]$
$\sigma=\sqrt((\lambda+2\mu)/\mu)$
Here, $u$ is the displacement vector field in the crystal caused by a given phonon mode, and $L_{lm}$ and $N_{lm}$ represent the longitudinal and transverse portions of that displacement, respectively. $j_l$ is a spherical bessel function of order $l$. For $l=0$, $\beta\to0$ and so $p$ and $q$ both are pushed to 0, but I know that the $l=0$ mode must be non-zero. For $l=1,2,3...$ I still get $p=q=0$, and no atomic displacement due to the phonons. Am I missing something obvious here?
 A: (I will write $\ell$ instead of $l$, because it confuses me less frequently.)
Assuming that $\delta$ definition is a typo, and it is supposed to be
$$
\delta=(\ell+1)[2(\ell-1)j_{\ell-1}(kR)-kRj_\ell(kR)]
$$
Your matrix's determinant is nonzero...
\begin{align}
\alpha\delta-\beta\gamma &= \ell \left[R j_{\ell}(kR) k - 2 j_{\ell+1}(kR) \left(\ell + 2\right)\right] \left[R h j_{\ell}(hR) \sigma^{2} - 2 j_{\ell-1}(hR) \left(\ell - 2\right)\right]\\
&\quad + \left(\ell + 1\right) \left[R j_{\ell}(kR) k - 2 j_{\ell-1}(kR) \left(\ell - 1\right)\right]\left[R h j_{\ell}(hR) \sigma^{2} - 2 j_{\ell+1}(hR) \left(\ell + 2\right)\right]\\
&=2 R^{2} h j_{\ell}(hR) j_{\ell}(kR) k \ell \sigma^{2} 
+ R^{2} h j_{\ell}(hR) j_{\ell}(kR) k \sigma^{2} 
- 2 R h j_{\ell}(hR) j_{\ell-1}(kR) \ell^{2} \sigma^{2} \\
&\quad+ 2 R h j_{\ell}(hR) j_{\ell-1}(kR) \sigma^{2} 
- 2 R h j_{\ell}(hR) j_{\ell+1}(kR) \ell^{2} \sigma^{2} 
- 4 R h j_{\ell}(hR) j_{\ell+1}(kR) \ell \sigma^{2} \\
&\quad
- 2 R j_{\ell}(kR) j_{\ell-1}(hR) k \ell^{2} 
+ 4 R j_{\ell}(kR) j_{\ell-1}(hR) k \ell 
- 2 R j_{\ell}(kR) j_{\ell+1}(hR) k \ell^{2} \\
&\quad
- 6 R j_{\ell}(kR) j_{\ell+1}(hR) k \ell 
- 4 R j_{\ell}(kR) j_{\ell+1}(hR) k 
+ 4 j_{\ell-1}(hR) j_{\ell+1}(kR) \ell^{3} \\
&\quad
- 16 j_{\ell-1}(hR) j_{\ell+1}(kR) \ell 
+ 4 j_{\ell-1}(kR) j_{\ell+1}(hR) \ell^{3} 
+ 8 j_{\ell-1}(kR) j_{\ell+1}(hR) \ell^{2} \\
&\quad
- 4 j_{\ell-1}(kR) j_{\ell+1}(hR) \ell 
- 8 j_{\ell-1}(kR) j_{\ell+1}(hR)\\
&\neq0
\end{align}
Its kernel will therefore be trivial.
