Timeline for Creation and Annihilation operators in QFT - Part II

6 events

when toggle format	what		by	license	comment
Sep 13, 2020 at 5:08	comment	added	Igor Khavkine		@IamWill All correct. Don't see any reason for you to doubt yourself there.
Sep 12, 2020 at 20:01	comment	added	JustWannaKnow		@IgorKhavkine Maybe I could use the following: if $\mathcal{H} = L^{2}(\mathbb{R}^{3};\mathbb{C}^{2}) \cong L^{2}(\mathbb{R}^{3}\times \{+1,-1\})$ it seems that $\mathcal{H}_{n} \cong L^{2}(\mathbb{R}^{3}\times \{+1,1\})\otimes \cdots \otimes L^{2}(\mathbb{R}^{3}\times \{+1,-1\}) \cong L^{2}(\mathbb{R}^{3n}\times \{+1,-1\}^{n})$? (Don't know for sure if the last isomorphism holds tho).The latter seems more useful to define $\varphi(x,\sigma)$ and $\varphi^{\dagger}(x,\sigma)$ since $x \in \mathbb{R}^{3}$ and $\sigma$ is a spin variable, probably taking values $\{+1,-1\}$.
Sep 12, 2020 at 17:48	comment	added	Igor Khavkine		Your very last question is essentially this: what is $\mathcal{H}_n^-$? Not unexpectantly, it is always $\mathcal{H}_n^- = \bigwedge^n \mathcal{H}$, denoting the fully anti-symmetrized Hilbert tensor product. The further isomorphism $L^2(\mathbb{R}^3; \mathbb{C}^2) \cong L^2(\mathbb{R}^3\times\{+1,-1\})$ should make it clear what $\mathcal{H}_n^-$ is. Exercise: fill in the blank in $\mathcal{H}_n^- \subset L^2(\mathbb{R}^{3n}; ?)$.
Sep 12, 2020 at 16:19	history	edited	JustWannaKnow	CC BY-SA 4.0	added 193 characters in body
Sep 12, 2020 at 16:01	comment	added	gmvh		Your guess is correct.
Sep 12, 2020 at 15:50	history	asked	JustWannaKnow	CC BY-SA 4.0