Considering the exciton eigensystem $\mathcal{H} | \lambda \rangle = E_\lambda |\lambda\rangle$ with the Hermitian Hamiltonian $\mathcal{H}$ and wave function $|\lambda\rangle$. I'm thinking about the overlap $\langle \lambda|0 \rangle$, where $|0\rangle$ is the vacuum or ground state. If the overlap is zero, then for the oscillator strength which is proportional to $\langle \lambda|\mathbf{r}|0 \rangle$, where $\mathbf{r}$ is the position operator, we insert the completeness relation $\sum_{\mu}|\mu\rangle\langle \mu|=1$ between $\mathbf{r}$ and $|0\rangle$ $$ \langle \lambda|\mathbf{r}|0 \rangle = \sum_{\mu}\langle \lambda|\mathbf{r}|\mu\rangle\langle \mu|0 \rangle \tag{1} $$ which seems to be zero due to the overlap term. This is of course wrong. Then what should the overlap be or am I missing something?

Now going further for the finite momentum exciton $|\lambda \mathbf{q}\rangle$, for which we have $$ \mathcal{H} | \lambda \mathbf{q} \rangle = E_{\lambda \mathbf{q}} |\lambda \mathbf{q}\rangle \tag{2} $$ Then what the completeness relation should be? $\sum_{\lambda \mathbf{q}}|\lambda \mathbf{q}\rangle\langle \lambda \mathbf{q}|=1$?

Also, if the ground state or vacuum state has energy of $E_0$, I'd like to know the velocity of the ground state $\partial_\mathbf{q} E_0$ or $\langle 0|\mathbf{v}|0\rangle$ where $\mathbf{v}$ is the exciton velocity operator. Is that zero?


1 Answer 1


Your formula is correct. But this only gives $$\langle\lambda|r|0\rangle = \langle\lambda|r|0\rangle,$$ because a complete set of basis vectors includes the ground state term $|0\rangle\langle 0|$.

  • $\begingroup$ Can you please comment on the finite momentum exciton as well? $\endgroup$ Oct 27, 2021 at 20:44
  • $\begingroup$ The overlap has to be zero because the exciton and the ground state have dfferent energy eigenvalues. $\endgroup$
    – mike stone
    Oct 27, 2021 at 21:45
  • $\begingroup$ @mikestone THanks! For two general exciton states, does $\langle \lambda \mathbf{q}| \mu \mathbf{q}^\prime \rangle = \delta_{\lambda \mu} \delta_{\mathbf{q} \mathbf{q}^\prime}$ hold? $\endgroup$ Oct 27, 2021 at 23:47

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