# Field operators for states with spin

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.

• 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 $a_{i\sigma}^\dagger$. Jan 26, 2017 at 0:22
• 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.
– Soap
Jan 26, 2017 at 0:30
• @ACuriousMind (I forgot to identify you in the previous comment).
– Soap
Jan 26, 2017 at 1:07

$$\hat{\psi}_\sigma (\vec{r} ) = \langle \vec{r},\sigma |\hat{a}\rangle = \sum_{i \sigma'}{\langle \vec{r},\sigma} | i,\sigma' \rangle \langle{i, \sigma'}| \hat{a} \rangle$$
If we can separate $$|\vec{r},\sigma\rangle$$ and $$|i,\sigma \rangle$$ as $$| \vec{r}\rangle |\sigma\rangle$$ and $$| i\rangle |\sigma\rangle$$ respectively, the field operator becomes
$$\hat{\psi}_\sigma (\vec{r} ) = \sum_{i\sigma'} \langle \vec{r} | i \rangle \delta_{\sigma \sigma'} \langle i,\sigma' | \hat{a} \rangle = \sum_{i} \phi_i (\vec{r}) \langle i,\sigma|\hat{a} \rangle = \sum_{i} \phi_i(\vec{r} ) \hat{a}_{i\sigma }$$.