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It is claimed that

$$\{c_\alpha,c_\beta \} = c_\alpha c_\beta + c_\beta c_\alpha = 0$$

where $c_\alpha$ and $c_\beta$ are the fermionic annihilation operators in second quantization. Why is that true? How does it relate to the antisymmetric properties in first quantization framework?

This is what I understood, taking a 2 state system for instance:

$$|1 1\rangle = \frac{1}{\sqrt{2}}[\psi_1(P_1)\psi_2(P_2) - \psi_1(P_2)\psi_2(P1)]$$

Then $(c_1 c_2 + c_2 c_1)|1 1\rangle$ would result in $2 |00\rangle$ which is not $0$.

What am I missing?

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  • $\begingroup$ Your top relation projects out the symmetric part of the product of two operators, leaving the antisymmetric part unscathed. $\endgroup$ – Cosmas Zachos Jan 17 at 16:21
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The answer lies in the definition of $| 11 \rangle$ in terms of $c^\dagger$s. OK, first of all let's use different indexes to represent orbital functions ($\psi$) and particle indexes. I'm gonna use numbers to represent orbitals and Greek alphabets to represent particles. Furthermore, the positions in ket also matter, i.e. it represents the arrangement of particles in real-space representation of orbitals. So using this convention $| 1 1\rangle$ becomes

$$ | \alpha \beta \rangle \equiv \frac{1}{\sqrt{2}}[\psi_1(\alpha)\psi_2(\beta) - \psi_1(\alpha)\psi_2(\beta)] \equiv c_{\alpha}^\dagger c_{\beta}^{\dagger} |0\rangle$$

The problem with the representation using real-space orbitals is that these orbitals are scalar functions and the commutation between two same functions are zero. Therefore, to form an anti-symmetric function we have to play with the orbital and particles indexes in such a way that when the particle indexes are swapped the whole function becomes the negative of the original one.

It's not the case with the fermionic operators because the antisymmetric properties are in its definition through the commutation relation.

So back again to your problem! So if we apply again the annihilation operators to $|\alpha\beta\rangle$, the equation becomes

$$ (c_\alpha c_\beta + c_\beta c_\alpha)\ |\alpha \beta\rangle = (c_\alpha c_\beta + c_\beta c_\alpha)\ c_{\alpha}^\dagger c_{\beta}^{\dagger} |0\rangle$$

and using the anti-commutation relations of the operators, i.e $\{c_i,c_j^\dagger\} = \delta_{ij}$ and $\{c_i^\dagger,c_j^\dagger\} = \{c_i,c_j\} = 0$ you will get

$$\left( (-1) \left[ 1 - n_\alpha\right] \left[ 1 - n_\beta\right] + \left[ 1 - n_\alpha\right] \left[ 1 - n_\beta\right] \right) |0\rangle = 0 |0\rangle$$

which was already expected from the anti-commutation relation! Here, $n_i \equiv c^{\dagger}_i c_i$, i.e. the number operator!

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