3
$\begingroup$

I'm studying Quantum Field Theory and encountered Wick's theorem: for the real Klein Gordon field $\phi(x)$, one has

$$ T(\phi(x_1)\cdots\phi(x_n)) = N(\phi(x_1)\cdots\phi(x_n) + \text{ all possible contractions}) $$

where $T$ denotes time ordering and $N$ denotes normal ordering. With suitable definitions of $T$ and $N$ this can be extended to Dirac fields $\psi(x)$ as well:

$$T(\psi_{a_1}(x_1)\cdots\psi_{a_n}(x_n)) = N(\psi_{a_1}(x_1)\cdots\psi_{a_n}(x_n) + \text{ all possible contractions}) $$

where the $a_i$ are spinor indices. But what a lot of books/notes on the topic do is the following. They prove the theorem for the KG field, and later for the Dirac field, and then they suddenly apply it to Yukawa theorey, e.g., to mixed expressions of the form

$$T(\psi_a(x_1)\bar{\psi}_b(x_2)\phi(x_3)\bar{\psi}(z)\psi(z)\phi(z)) = \dots$$

I guess the theorem must also hold for these kind of expressions, but I have not been able to find a reference that proves it. I understand the (canonical) proofs for the KG and Dirac case seperately, but I don't see how one would extend these to mixed expressions. If anyone knows how to prove the theorem for these cases, or knows a reference to such a proof, I would love to hear (well, read) it.

$\endgroup$
2
$\begingroup$

This is because when we write the Fourier expansion of the real field and spinor fields, we assume that the only commutators and anti-commutators which are nonzero are those involving the real field with itself and the spinor field with itself. This is from the canonical commutation relations (2.24), (3.88) in Peskin and Schroeder.

In other words, when we perform normal ordering, we don't need to commute/anti-commute spinor fields through real fields, because those things are identically zero.

$\endgroup$
1
$\begingroup$

Yes, Wick's theorem for free fields also works for mixed sectors, cf. my Phys.SE answer here. The crucial assumption is that the contractions are at the center of the (super) algebra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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