1
$\begingroup$

I am having trouble tracking down the source of the claim that only singlet excitons (para-excitons) may be excited optically (ignoring the effect of singlet-triplet state mixing due to spin-orbit interaction). See, for example, section II of Overhauser or the last few paragraphs of Gross.

The claim seems reasonable when compared with atomic theory in the LS coupling limit, where we have the selection rule $\Delta S=0$. However, if we consider the spin component of the exciton wavefunction: \begin{equation} |S=0,M=0\rangle =\left.\frac{1}{\sqrt{2}}\left(|\uparrow\rangle_e|\downarrow\rangle_h-|\downarrow\rangle_e|\uparrow\rangle_h\right)\right\}\,\,\text{Singlet} \\ \left.\begin{array}{lr} &|S=1,M=1\rangle=|\uparrow\rangle_e|\uparrow\rangle_h \\ &|S=1,M=0\rangle =\frac{1}{\sqrt{2}}\left(|\uparrow\rangle_e|\downarrow\rangle_h+|\downarrow\rangle_e|\uparrow\rangle_h\right) \\ &|S=1,M=-1\rangle=|\downarrow\rangle_e|\downarrow\rangle_h \end{array}\right\} \,\,\text{Triplet} \end{equation} the $M=0$ states for both the singlet and the triplet exciton feature an electron and hole paired up with oppositely aligned spins (one is spin-up while the other is spin-down). If we consider the fact that a hole has a spin opposite to that of the state from which the electron was removed, either of these two exciton states corresponds to an electron promoted to the conduction band while preserving its spin orientation. On this basis I would expect that both states should give rise to optically allowed transitions. What am I missing here?

$\endgroup$
1
$\begingroup$

You can understand the selection rule by calculating the transition dipole matrix. For example, for the singlet excition $|0,0\rangle$: $$\begin{align} \langle0|\mu|0,0\rangle & = \frac{1}{\sqrt2}[\langle \uparrow|_e \ \mu \ (\hat K|\downarrow \rangle_h)- \langle \downarrow|_e \ \mu \ (\hat K|\uparrow \rangle_h)]\\ & = \frac{1}{\sqrt2}(-\mu\langle \uparrow|\uparrow \rangle- \mu\langle \downarrow| \downarrow \rangle)=-\sqrt2\mu \end{align}$$ With $\hat K$ the time reversal operator, $\mu$ the transition dipole moment and $|0\rangle$ the vacuum state. Similarly, you can solve that the transition dipole matrices of all the triplet states are zero.

$\endgroup$
  • $\begingroup$ This is what I was looking for. That extra minus sign from K operating on the first term is crucial! Thanks! $\endgroup$ – ashwmk Jul 22 '18 at 3:16
0
$\begingroup$

The $S=1, m=0$ state is a triplet state just like the $S=1, m=\pm 1$ states. All three triplet substates are equally affected by the $\Delta S=0$ selection rule.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.