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



\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}







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?

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    $\begingroup$ Hi Geoff. First of all, I think this is probably a good question, and I hope it finds an answer on this site. That being said, I have some remarks: 1) Try to use [mathjax] to make sure your equations etc. are properly rendered, to improve readability. 2) It is not really useful to include some numerical values in your question, as they are really not of interest to anyone except you, and in particular the exact values will have no impact on the answer that you seem to be looking for, so they are superfluous. I will now edit your question to reflect these tips. [continued] $\endgroup$ – Danu Sep 24 '14 at 18:53
  • $\begingroup$ ...if you disagree with my edit, you can always rollback my edit and comment, telling me what you think :) $\endgroup$ – Danu Sep 24 '14 at 18:54
  • $\begingroup$ @GeoffDiederich that's okay :) I would also suggest you expand your question a little to make it completely self-contained. I am having trouble trying to improve it right now, because everything is taken out of context and I can't make a whole lot of sense out of it. Could you please show all the relevant equations (and like I said, the numbers really aren't what's important here; we're interested in the conceptual question). $\endgroup$ – Danu Sep 24 '14 at 18:58
  • $\begingroup$ @GeoffDiederich Is your definition of $\delta$ correct? Should the second factor's second term be $-kRj_{1}(kR)$ or $-kRj_{\ell}(kR)$? $\endgroup$ – Alex Nelson Sep 25 '14 at 20:08

(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.

  • $\begingroup$ I was just told this and knew it was something simple that I was thinking was more complicated. Thanks for answering. $\endgroup$ – Geoff Diederich Sep 26 '14 at 15:49

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