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.

  • 1
    $\begingroup$ 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, 2017 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, 2017 at 0:30
  • $\begingroup$ @ACuriousMind (I forgot to identify you in the previous comment). $\endgroup$
    – Soap
    Jan 26, 2017 at 1:07

1 Answer 1


Thank you for your detailed explanation. Even though you ask a question, I resolved my persistent questions thanks to your explanation. I think we can approach this problem as following method, although it is not that different from yours;

$\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

$\hat{\psi}_\sigma (\vec{r} ) = \sum_{i\sigma'} \langle \vec{r} | i \rangle \delta_{\sigma \sigma'} \langle i,\sigma' | \hat{a} \rangle = \sum_{i} \phi_i (\vec{r}) \langle i,\sigma|\hat{a} \rangle = \sum_{i} \phi_i(\vec{r} ) \hat{a}_{i\sigma }$.

I'm not sure my answer is right, and I'm also a student struggling to learn about 2nd quantization. But I've searched many materials online, and organized the answer like this.


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