I have a guilty suspicion this should be obvious. What is the difference between these two expectations taken over the same measure ($\int \mathrm{d}\mu(\bar\psi,\psi)\exp{\sum \bar\psi A\psi}$ for example)?

Let $$f_{1}=\langle \prod_{i}\bar\psi_{i}^{a_{i}}\psi_{i}^{b_{i}}\rangle$$ and $$f_{2}=\langle \prod_{i}|\psi_{i}|^{2a_{i}}\rangle$$

I know that each of them produces a determinant with elements from the inverse of $A$, but I have a suspicion that determinant in $f_{2}$ will be twice as big as in $f_{1}$, and the elements would be different. Obviously, in $f_{1}$ $(a_{i},b_{i})\in \{0,1\}$, and in $f_{2}$ the only contributions come from $a_{i}\in\{0,1\}$, so the expectations certainly aren't equivalent. I have a feeling that one may be written as a pfaffian and the other as a determinant, but I'm unsure.

  • $\begingroup$ What do you mean by $| \psi_i |^2$ here? Is it $\psi_i^* \psi_i$? $\endgroup$ – Darkseid Oct 3 '17 at 21:10
  • $\begingroup$ @FedorIndutny Yes, sorry. $\endgroup$ – TeeJay Oct 3 '17 at 21:11
  • $\begingroup$ So essentially it is: $\langle \prod_{i} \overline{\psi}^{a_i} \gamma^0 \psi^{a_i} \rangle$ $\endgroup$ – Darkseid Oct 3 '17 at 21:17
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    $\begingroup$ Btw, why the title mentions Wick's theorem? $\endgroup$ – Darkseid Oct 3 '17 at 21:43
  • $\begingroup$ There are no Dirac matrices in this problem, $\bar\psi$ is simply the set of the addition $n$ variables of the Grassmann algebra. $\endgroup$ – TeeJay Oct 4 '17 at 0:27

There is no determinant in the expectation values of product of fermionic operators. It appears promptly in numerator of correlation function only to be immediately canceled out by denominator:

$$ \langle A \rangle = \frac{\int d\mu \, e^{iS} A}{ \int d\mu \, e^{iS} } $$

The expectation value of product of fermionic fields will be given by the products of inverse matrices in all possible pairing configurations.

Given that your action has only kinetic term for the fields (apparently) in the first case when $a_i \neq b_i$, the expectation value is going to be zero due to charge conservation. The latter one is going to be more complicated as you'd have to split the measure into left-handed and right-handed fermions:

$$ \int {\cal{D}}\psi_L {\cal{D}}\psi^\dagger_L {\cal{D}}\psi_R {\cal{D}}\psi^\dagger_R $$

$|\psi|^2$ then turns into $\psi_L^\dagger \psi_L + \psi_R^\dagger \psi_R$, and generally you'll get perturbative answer, because $\overline{\psi} A \psi$ is going to separate into interaction terms between left-/right-handed fields.

  • $\begingroup$ The first statement is wrong. Call the inverse of $(A^{-1})_{ij}=D_{ij}$, then an expansion of the sort I'm talking about is something like $\sum_{\pi \in S_{n}} (-1)^{\mathrm{sgn} (\pi)}\prod_{i=1}^{n}D_{i,\pi (i)}$, which is obviously a determinant. $\endgroup$ – TeeJay Oct 4 '17 at 0:26
  • $\begingroup$ Oh, sorry. This wasn't clear for me from the question. $\endgroup$ – Darkseid Oct 4 '17 at 0:35

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