# Second quantization, creation and annihilation operators in space basis

I am working through some basic knowledge of second quantization. At the beginning, everything is neat and clean.

We have $$|n_{1}, n_{2}, ...\rangle = a_{1}^{\dagger}a_{2}^{\dagger}...|0\rangle$$, where for fermions $$n_{i} = 0$$ or $$1$$ and the creation and annihilation operators fulfill the anticommutation relation.

$$\lbrace a_{i}, a_{j}^{\dagger}\rbrace = \delta_{ij}$$

The book goes further, using free electrons in a box potential as an example, which is also fine:

$$|\psi_{0}\rangle = |P_{0\uparrow}, P_{0\downarrow}, ..., P_{F\uparrow}, P_{F\downarrow}\rangle = a_{P0\uparrow}^{\dagger}a_{P0\downarrow}^{\dagger}...|0\rangle$$,

since for free non-interacting electrons, their eigenstates are momentum states.

Next, for the non-interacting electrons, the author introduce the field operator, it starts confusing,

$$\Psi_{\sigma}^{\dagger}(r) = \sum_{P}\frac{e^{-iP\cdot r/\hbar}}{\sqrt{V}}a_{P\sigma}^{\dagger}$$

and the following three equations,

$$|r_{1\sigma_{1}}, ..., r_{n\sigma_{n}}\rangle = \Psi_{\sigma_{1}}^{\dagger}(r_{1}),...,\Psi_{\sigma_{n}}^{\dagger}(r_{n})|0\rangle$$

$$\Psi_{\sigma_{n+1}}^{\dagger}(r_{n+1})|r_{1\sigma_{1}}, ..., r_{n\sigma_{n}}\rangle = \sqrt{n+1}(-1)^{\delta_{n+1}}|r_{1\sigma_{1}} ... r_{n+1\sigma_{n+1}}\rangle$$

$$\Psi_{\sigma}(r_{i})|r_{1\sigma_{1}}, ..., r_{n\sigma_{n}}\rangle = \frac{1}{\sqrt{n}}\sum\delta(r-r_{i})(-1)^{\delta_{i}}|r_{1\sigma_{1}} ...,r_{i-1\sigma_{i-1}},r_{i+1\sigma_{i+1}},..., r_{n\sigma_{n}}\rangle$$

I understand these equations, however, I have several further questions:

1. What is the relation between $$\Psi_{\sigma_{i}}(r_{i})$$ and $$\Psi_{\sigma_{i}}^{\dagger}(r_{i})$$? commute? anticommute? or not well defined?

2. I guess what I am confused at the beginning is can two electron have the same $$|r\rangle$$? I know from Pauli exclusive principle, for electrons, $$|n_{1}, n_{2}, ... \rangle = |1,0, 1....\rangle$$. We can go further, for non-interacting free electron, the eigenstates are momentum states, so $$|P_{0\uparrow}, P_{0\downarrow}, ..., P_{F\uparrow}, P_{F\downarrow}\rangle = |1,1,...,1,1\rangle$$. But what about $$|r_{1\sigma_{1}}, ..., r_{n\sigma_{n}}\rangle$$? I know for most of systems, space basis are always not the eigenbasis, but are they superposition of eigenstates? So Pauli exclusive principle also means two fermions cannot have the same superposition of eigenstates?

3. I guess this the same question as (2). If we have a magnetic field along $$z$$ direction, we cannot have two electrons with the same spin state along $$z$$ and the same spatial wavefunction. But could we have two electrons with the same spin stat along $$x$$ and the same spatial wavefunction?

• "So Pauli exclusive principle also means two fermions cannot have the same superposition of eigenstates?" This is a strange question, because any state can be made as a superposition of eigenstates. And eigenstates of an operator is a red herring, anyway, because Pauli says that two fermions can't be in the same (single-particle) state, period. So, two electrons can both be in the same $\lvert r\rangle$ if they have different spin states, but they can't be in the same $\vert r\rangle$ if they're in the same spin state. For instance. Jan 22 at 2:55
• "...we cannot have two electrons with the same spin state along z and the same spatial wavefunction. But could we have two electrons with the same spin stat along x and the same spatial wavefunction?" I'm confused by this question. Why do you think $z$ is privileged? I guess you're thinking about putting the electrons in the $\hat{S}_z$ eigenstates versus the $\hat{S}_x$ eigenstates in these two different scenarios, but why should that be different with respect to the statistics and the anti-symmetry of the wave function? Jan 22 at 2:58
• I'm asking these questions because I'm trying to drill down to where your misconceptions might be. It seems like your possible misconceptions or misunderstandings have nothing to do with QFT, really, and more to do with just what the Pauli exclusion principle is really saying? Jan 22 at 3:00
• I also agree with march. It is worrisome that you seem to be shaky in your understanding. However, to your questions, (1) is something a good textbook would derive; you should also take it up as a homework problem and derive it for yourself. The answer to (2) and (3) will be a corollary of (1). Jan 22 at 3:58
• @march thanks, I am affirmed now.
– QFT
Jan 22 at 4:20

1. What is the relation between $$\Psi_{\sigma_{i}}(r_{i})$$ and $$\Psi_{\sigma_{i}}^{\dagger}(r_{i})$$? commute? anticommute? or not well defined?

You can work the relationship out based on the previously-described relationship between the $$a$$ and the $$a^\dagger$$ operators.

You will arrive at something along the lines of $$\{\psi_{\sigma}(\vec x), \psi_\rho^\dagger(\vec y)\}=\delta_{\sigma\rho}\delta^3(\vec x - \vec y)\;.$$

1. I guess what I am confused at the beginning is can two electron have the same $$|r\rangle$$?

"Have the same $$|r\rangle$$" is somewhat ambiguous. But if what you really want to know is the value of $$\psi_\sigma(\vec r)^\dagger\psi_{\sigma}^\dagger(\vec r)|0\rangle$$ then you can figure it out using your previously-described anti-commutation relations for the $$a^\dagger$$ operators. (Spoiler alert: The value when the spin is the same and the $$\vec r$$ is the same is zero.)

1. I guess this the same question as (2).

Then it is the same answer as 2. You can work it out using your previously-described anti-commutation relations for the $$a$$ and $$a^\dagger$$ operators.