# Slater determinant in second quantization using the creation operators help [closed]

$$\left\langle 0\left|\hat{\Psi}\left(x_{1}\right) c_{\alpha_{1}}^{\dagger}\right| 0\right\rangle=\left\langle 0\left|\varphi_{\alpha_{1}}\left(x_{1}\right)-c_{\alpha_{1}}^{\dagger} \hat{\Psi}\left(x_{1}\right)\right| 0\right\rangle=\varphi_{\alpha_{1}}\left(x_{1}\right)$$ $$\left\langle 0\left|\hat{\Psi}\left(x_{1}\right) \hat{\Psi}\left(x_{2}\right) c_{\alpha_{2}}^{\dagger} c_{\alpha_{1}}^{\dagger}\right| 0\right\rangle=\left\langle 0\left|\hat{\Psi}\left(x_{1}\right)\left(\varphi_{\alpha_{2}}\left(x_{2}\right)-c_{\alpha_{2}}^{\dagger} \hat{\Psi}\left(x_{2}\right)\right) c_{\alpha_{1}}^{\dagger}\right| 0\right\rangle$$

$$=\left\langle 0\left|\hat{\Psi}\left(x_{1}\right) c_{\alpha_{1}}^{\dagger}\right| 0\right\rangle \varphi_{\alpha_{2}}\left(x_{2}\right)-\left\langle 0\left|\hat{\Psi}\left(x_{1}\right) c_{\alpha_{2}}^{\dagger} \hat{\Psi}\left(x_{2}\right) c_{\alpha_{1}}^{\dagger}\right| 0\right\rangle$$

$$=\varphi_{\alpha_{1}}\left(x_{1}\right) \varphi_{\alpha_{2}}\left(x_{2}\right)-\varphi_{\alpha_{2}}\left(x_{1}\right) \varphi_{\alpha_{1}}\left(x_{2}\right)$$

I'm trying to do this but the reason why $$c_{\alpha_{1}}^{\dagger} \hat{\Psi}\left(x_{1}\right)$$

is zero eludes me, especially since in N=2 it's not

## closed as unclear what you're asking by Ben Crowell, ZeroTheHero, Buzz, Kyle Kanos, Emilio PisantyJan 21 at 14:38

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• I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. Please don't post images of text. It breaks search functionality and won't work for blind people. – Ben Crowell Jan 19 at 19:42
• Remade it in LaTex – user220348 Jan 19 at 22:40

It’s because the single $$\Psi$$ is expanded into a sum of creation and annihilation operators $$\sim a + a^\dagger$$, and so you just have $$\langle 0 | c^\dagger a_p | 0 \rangle$$ and $$\langle 0| c^\dagger a^\dagger_p |0 \rangle$$, both of which are zero because the creation operator $$c^\dagger$$ hits the vacuum on the left.