# A Quantized Dirac field calculations

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1. Apr 13, 2017

### Nod

Hi everyone!

I'm having a problem with calculating the fermionic propagator for the quantized Dirac field as in the attached pdf. The step that puzzles me is the one performed at 5.27 to get 5.28. Why can I take outside (iγ⋅∂+m) if the second term in 5.27 has (iγ⋅∂-m)? And why there's a difference of the D(x-y) and D(y-x)?

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• ###### qft.dvi - five.pdf
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2. Apr 13, 2017

### Demystifier

Since $p$ is an integration variable, in the second term in (5.27) you can replace $p$ with $-p$. (More precisely, introduce a new variable $p'=-p$ and then remove the prime since it is a dummy variable.) With a little extra work, that should resolve your first question. The answer to the second question should be obvious from (5.29).

3. Apr 13, 2017

### Demystifier

A more interesting comment on that page of Tong lectures is that the anticommutation relation does not violate causality because the fermion field is not an observable, while bilinear observables commute (not anticommute). That's an important lesson to those who like to interpret fields as "fundamental" objects.

4. Apr 13, 2017

### Nod

I also thought like this, but still I have doubts:
If I replace $p$ with $-p$, then instead of
$\displaystyle{\not} p = γ_μ p^μ = γ_0 p^0 +γ_i p^i$
I'll have
$\displaystyle{\not}p = γ_μ p^μ = γ_0 p^0 -γ_i p^i$.
But the last equation is not equal to $-\displaystyle{\not}p$ , because for that also the energy part $γ_0 p^0$ must change the sign!

5. Apr 13, 2017

### stevendaryl

Staff Emeritus
I think it's pretty straightforward:

$D(x-y) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_p} e^{-i p \cdot(x-y)}$

So: $(i \displaystyle{\not} \partial + m) D(x-y) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_p} (+\displaystyle{\not} p + m) e^{-i p \cdot(x-y)}$

$D(y-x) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_p} e^{-i p \cdot(y-x)} = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_p} e^{+i p \cdot(x-y)}$

So: $(i \displaystyle{\not} \partial + m) D(y-x) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_p} (- \displaystyle{\not} p + m) e^{-i p \cdot(x-y)}$

Subtract the two and you get:

$(i \displaystyle{\not} \partial + m) D(x-y) - (i \displaystyle{\not} \partial + m) D(y-x)$
$= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2E_p} ((+ \displaystyle{\not} p + m) e^{-i p \cdot(x-y)} - (- \displaystyle{\not} p + m) e^{-i p \cdot(x-y)})$

You have to remember that $\partial$ acts on $x$, not $y$, and that $i \displaystyle{\not}\partial e^{\mp i p \cdot (x-y)} = \pm \displaystyle{\not}p e^{-i p \cdot (x-y)}$

6. Apr 13, 2017

### Nod

Thank you for explanation! Now I get where the signs come from :)