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?

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

1 Answer 1


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.