# Two-particle operators in QFT and the factor 1/2

I am learning about QFT through the book Quantum Field Theory for the Gifted Amateur and I am having trouble understanding the factor 1/2 in the definition of two particle field operators. In the book they first introduce an arbitrary second-quantized two-particle operator $$\hat{A}$$, $$\hat{A} = \sum_{\alpha \beta \gamma \delta} \mathcal{A}_{\alpha \beta \gamma \delta} \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} \hat{a}_{\gamma} \hat{a}_{\delta},$$ where $$\mathcal{A}_{\alpha \beta \gamma \delta} = \langle \alpha,\beta | \gamma,\delta \rangle.$$ Then they state that a two-particle operator written as a function of spatial coordinates is given by $$\hat{V} = \frac{1}{2} \int d^3x\, d^3y \hat{\psi}^{\dagger}(\textbf{x}) \hat{\psi}^{\dagger}(\textbf{y}) V(\textbf{x},\textbf{y}) \hat{\psi}(\textbf{y}) \hat{\psi}(\textbf{x}),$$ where the factor $$1/2$$ is needed so that we don't double count the interactions. Now my question is, why don't we need to include this factor in the definition of $$\hat{A}$$ where we defined the operator in terms of the momentum-basis?

• Page / equation number? Commented Nov 28, 2021 at 11:38
• Page 44, Eqns. 4.51 and 4.52. Commented Nov 28, 2021 at 11:39

It basically comes down to the fact that in the first expression, we sum over two-particle states, but in the second expression, we sum over one-particle states.

In my opinion it should be written as

$$\hat{A} = \sum_{(\alpha \beta), (\gamma \delta)} \mathcal{A}_{(\alpha \beta), (\gamma \delta)} \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} \hat{a}_{\gamma} \hat{a}_{\delta}.$$

where $$(\alpha \beta)$$ and $$(\gamma \delta)$$ label two-particle states. In terms of one-particle states $$|\alpha\rangle$$, these can be written as

$$|\alpha \beta\rangle=\frac{1}{\sqrt{2}}\left(|\alpha\rangle|\beta\rangle\pm|\beta\rangle|\alpha\rangle\right)$$

with the $$+$$ for bosons and $$-$$ for fermions. We also have that $$\mathcal{A}_{(\alpha \beta), (\gamma \delta)}=\langle\alpha \beta|\hat{A} |\delta \gamma\rangle$$ (note that $$\gamma$$ and $$\delta$$ are backwards in the state, different to what is written in the book).

Now consider e.g.

$$\sum_{(\gamma \delta)} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}.$$

The sum over $$(\gamma \delta)$$ is half the sum over $$\gamma$$ and $$\delta$$ separately because the two-particle state with $$\gamma \leftrightarrow \delta$$ is the same, i.e. $$|\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}= |\gamma\delta\rangle\hat{a}_{\delta} \hat{a}_{\gamma}$$. Thus

$$\sum_{(\gamma \delta)} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}=\frac{1}{2}\sum_{\gamma \delta} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}.$$

We can then write $$|\delta\gamma\rangle$$ in terms of $$|\gamma\rangle$$ and $$|\delta\rangle$$ to give

$$\sum_{\gamma \delta} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}=\frac{1}{\sqrt{2}}\sum_{\gamma \delta} \left(|\delta\rangle|\gamma\rangle\pm|\gamma\rangle|\delta\rangle\right) \hat{a}_{\gamma} \hat{a}_{\delta}=\sqrt{2}\sum_{\gamma \delta} |\delta\rangle|\gamma\rangle \hat{a}_{\gamma} \hat{a}_{\delta}$$

using that $$\hat{a}_{\gamma} \hat{a}_{\delta}=\pm\hat{a}_{\delta} \hat{a}_{\gamma}$$. Hence overall we find that

$$\sum_{(\gamma \delta)} |\delta\gamma\rangle\hat{a}_{\gamma} \hat{a}_{\delta}=\frac{1}{\sqrt{2}}\sum_{\gamma \delta} |\delta\rangle|\gamma\rangle \hat{a}_{\gamma} \hat{a}_{\delta}.$$

The same argument shows that

$$\sum_{(\alpha \beta)} \langle\alpha\beta| \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} =\frac{1}{\sqrt{2}}\sum_{\alpha \beta} \langle\alpha|\langle\beta| \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger}.$$

Putting these together with $$\mathcal{A}_{(\alpha \beta), (\gamma \delta)}=\langle\alpha \beta|\hat{A} |\delta \gamma\rangle$$, we finally get

$$\hat{A} = \sum_{(\alpha \beta), (\gamma \delta)} \mathcal{A}_{(\alpha \beta), (\gamma \delta)} \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} \hat{a}_{\gamma} \hat{a}_{\delta}= \frac{1}{2}\sum_{\alpha \beta \gamma \delta} \left(\langle\alpha|\langle \beta|\right)\hat{A} \left(|\delta\rangle |\gamma\rangle\right) \hat{a}_{\alpha}^{\dagger} \hat{a}_{\beta}^{\dagger} \hat{a}_{\gamma} \hat{a}_{\delta}.$$

This second expression is the form that $$\hat{V}$$ is in: we have

$$\left(\langle\boldsymbol{x}|\langle \boldsymbol{y}|\right)\hat{V} \left(|\boldsymbol{z}\rangle |\boldsymbol{w}\rangle\right)=V(\boldsymbol{x},\boldsymbol{y})\delta(\boldsymbol{x}-\boldsymbol{z})\delta(\boldsymbol{y}-\boldsymbol{w})$$

and replacing the sum with an integral gives

$$\hat{V} = \frac{1}{2} \int \mathrm{d}^3x \int \mathrm{d}^3y\, \hat{\psi}^{\dagger}(\boldsymbol{x}) \hat{\psi}^{\dagger}(\boldsymbol{y}) V(\boldsymbol{x},\boldsymbol{y}) \hat{\psi}(\boldsymbol{y}) \hat{\psi}(\boldsymbol{x}).$$