# Diagonalizing the free Hamiltonian $\hat{H}$ in terms of smeared wave packet operators

Consider the free real scalar field in (1+1)-dimensions. If the (normal-ordered) Hamiltonian is $$\hat{H}=\int_{-\infty}^{\infty} dk\ \sqrt{ k^2 + m^2 } \hat{a}_{k}^{\dagger} \hat{a}_{k}$$ where $\hat{a}_{k}$ and $\hat{a}_{k}^{\dagger}$ are the annihilation and creation operators for ordinary Minkowski particles.

Consider the wave-packet $f_{MN}:\mathbb{R} \to \mathbb{C}$ defined for all $N,M \in \mathbb{Z}$ as the following $$f_{MN}(k) = \begin{cases} \frac{1}{\sqrt{\mathcal{E}}} e^{- 2 \pi i N \frac{k}{\mathcal{E}}} \ \ \ \ \ \ \ , \ k\in\left[ (M-\frac{1}{2})\mathcal{E} , (M+\frac{1}{2})\mathcal{E} \right] \\ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ , \ \mathrm{otherwise} \end{cases}$$ This wave-packet focuses the momentum around $\sim M\mathcal{E}$ with a width of $\sim \mathcal{E}$.

These wave-packets are orthonormal in that $$\int_{-\infty}^{\infty} dk \ F_{NM}(k) F_{IJ}(k) = \delta_{NI}\delta_{MJ} \\ \sum_{N,M} F^{\ast}_{NM}(k)F_{NM}(p) = \delta(k-p)$$

Consider the smeared operators $$\hat{a}_{MN} \equiv \int_{-\infty}^{\infty} dk\ F^{\ast}_{MN}(k) \hat{a}_{k}$$ which can be inverted via the orthonormality conditions stated to give $$\hat{a}_{k} = \sum_{M,N} F_{MN}(k) \hat{a}_{MN}$$

MY QUESTION: Is it possible to put the Hamiltonian into a form $\sum_{NM} E_{MN} \hat{a}^{\dagger}_{MN}\hat{a}_{MN}$ for some energy $E_{MN}$?

I think that if this is possible at all, then it would have to rely on an approximation (maybe something like $\mathcal{E} \ll m$). So far I have been able to get the Hamiltonian in the form: $$\hat{H} = \sum_{ M N J} \hat{a}^{\dagger}_{MN} \hat{a}_{MJ} \int_{(M-\frac{1}{2})\mathcal{E}}^{(M+\frac{1}{2})\mathcal{E}} dk\ \sqrt{ k^2 + m^2 } \frac{1}{\mathcal{E}} e^{- 2 \pi i (N - J) k/\mathcal{E} }$$ But this integral is too difficult for me.

• Your Hamiltonian is already diagonal in the "k" basis. If you switch to represent the Hamiltonian in the "NM" basis it will not be diagonal anymore. The number operator will still be diagonal in your new basis, but not the energy operator. – hft Aug 6 '18 at 0:58