# Exciton state, ground state and completeness relation

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?

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 $$|0\rangle\langle 0|$$.
• @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? Commented Oct 27, 2021 at 23:47