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

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

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

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

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

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    $\begingroup$ 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. $\endgroup$ – Ben Crowell Jan 19 at 19:42
  • $\begingroup$ Remade it in LaTex $\endgroup$ – user220348 Jan 19 at 22:40
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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.

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  • $\begingroup$ That doesn't make sense to me. Ψ here is the field operator, even if I expand it I don't see how I get the second and third row from (25). $\endgroup$ – user220348 Jan 19 at 18:58

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