I'm just starting to learn about field operators. The construction I saw was the following:

We want an operator $\Psi^\dagger(x)$ which, acting on the vacuum state $\vert 0 \rangle$ gives the state with well defined position $\vert x \rangle$ (in one dimension). Well, we have $$ \vert x\rangle=\underset{i}{\sum}\vert\phi_{i}\rangle\langle\phi_{i}\vert x\rangle=\underset{i}{\sum}\phi_{i}^{*}(x)a_i^{\dagger}\vert0\rangle $$ (with $\{\vert\phi_i\rangle\}$ an eigenbasis of some observable $A$). So $$\Psi^{\dagger}(x)=\underset{i}{\sum}\phi_{i}^{*}(x)a_i^{\dagger}$$

Now I want to do the same for the case when one has to consider the spin. Thus, I want an operator $\Psi_\sigma^\dagger(x)$ which, acting on the vacuum state $\vert 0 \rangle$ gives the state $\vert x \,\sigma \rangle$. In this case: $$ \vert x\,\sigma\rangle=\underset{i}{\sum}\vert\Phi_{i}\rangle\langle\Phi_{i}\vert x\,\sigma\rangle=\underset{i}{\sum}\Phi_{i\sigma}^{*}(x)\,a_{i}^{\dagger}\vert0\rangle $$ (with $\{\vert\Phi_i\rangle\}$ an eigenbasis of some observable $A$ acting on the state space $\mathcal{E}=\mathcal{E_x}\otimes\mathcal{E_\sigma}$). So $$\Psi_{\sigma}^{\dagger}(x)=\underset{i}{\sum}\Phi_{i\sigma}^{*} (x)\,a_{i}^{\dagger}$$ I now try something a little bit different: instead of using the closure relation of the basis $\{\vert\Phi_i\rangle\}$, I use the one of $\{\vert\phi_i\rangle\}$: $$ \vert x\,\sigma\rangle=(\underset{i}{\sum}\vert\phi_{i}\rangle\langle\phi_{i}\vert x\rangle)\otimes\vert\sigma\rangle=\underset{i}{\sum}\langle\phi_{i}\vert x\rangle(\vert\phi_{i}\rangle\otimes\vert\sigma\rangle)=\underset{i}{\sum}\phi_{i}^{*}(x)a_{i\sigma}^{\dagger}\vert0\rangle $$ (Note: I am not completely sure of my second equal sign). So $$\Psi_{\sigma}^{\dagger}(x)=\underset{i}{\sum}\phi_{i}^{*}(x)a_{i\sigma}^{\dagger}$$

I would like you to please confirm this (the last one in particular, which I haven't seen anywhere), since I'm not completely sure of it.

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    $\begingroup$ In your first formula, where does the $a^\dagger\lvert 0\rangle$ come from? Should it be $a_i^\dagger\lvert 0\rangle$, and if yes, how do you know such an operator exists? Same question for the $a_{i\sigma}^\dagger$. $\endgroup$ – ACuriousMind Jan 26 '17 at 0:22
  • $\begingroup$ Yes, you're right, I'll correct it. Well honestly I don't know the answer to that. In my classes, we motivated their existence considering an analogy with a system of N oscillators, but it was hardly rigorous and I did not understand it completely. $\endgroup$ – Soap Jan 26 '17 at 0:30
  • $\begingroup$ @ACuriousMind (I forgot to identify you in the previous comment). $\endgroup$ – Soap Jan 26 '17 at 1:07

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