# Commutation relations (QFT curved spacetime Wald)

So I'm currently reading Wald's QFT on curved spacetime. In section 2.3, he constructs the quantum theory of $$n$$ decoupled harmonic oscillators by choosing $$\mathcal{H}$$ as the set of positive frequency solutions and then choosing the Hilbert space to be $$\mathcal{F}_s(\mathcal{H})$$ - the symmetric Fock space of $$\mathcal{H}$$. He then defines the Heisenberg position operator to be $$q^\prime_{iH}=\xi_i(t)a_i+\bar{\xi}_i(t)a^\dagger_i$$ where $$\xi_i(t)$$ is the solution with the $$i$$th harmonic oscillator excited and $$a_i$$ is the annihilation associated with said solution. Also, the momentum operator is defined as $$p^\prime_{iH}=dq^\prime_{iH}/dt$$. So the question is: how do I go about showing that, with these definitions, $$[q^\prime_{iH},p^\prime_{jH}]=i\delta_{ij}?$$ And I ask this because, although Wald defined $$[a_i,a^\dagger_j]$$, I get terms like $$[\xi_i(t)a_i,\bar{\xi}_j(t)a^\dagger_j]$$ when I expand the equation above. And in all my ignorance, I don't know for certain what to do in this case.

• If I understand the notation correctly, $\xi_i(t)$ and $\bar \xi_j(t)$ are numbers, not operators. In that case $$[\xi_i(t)a_i,\bar{\xi}_j(t)a^\dagger_j]=\xi_i(t)\bar{\xi}_j(t)[a_i,a^\dagger_j]$$ and the only remaining commutator is known. Does this solves the issue?
– Gold
Commented Feb 16, 2022 at 13:43
• Well, that is something I'm a little bit confused. It's stated that $\xi_i(t) \in \mathcal{H}$, so it's a curve in phase space. If I consider it as a "number", as you suggested, this actually solves the issue. And thank you. Commented Feb 16, 2022 at 17:01
• Actually, now I see that it's supposed to be a number indeed. I got confused with the notation. Commented Feb 16, 2022 at 17:23