1
$\begingroup$

This is a screen shot of one page in David Tong's QFT notes.

This is a screen shot of one page in David Tong's QFT notes.

To clarify some of the notations, $p$ means the 4-vector momentum, while $\vec{p}$ denotes the 3-momentum, and similarly for $x$ and $\vec{x}$.

I have two questions:

  • In my understanding, causality forbids the propagation of a particle from one point to another that is spacelike separated, that is to say, the two point function has to vanish for spacelike intervals. Is this requirement equivalent to the equation (2.86) in the image that operators commute?

  • How is the integral in equation (2.89) zero? I couldn't get to understand David's arguments.

$\endgroup$
  • 2
    $\begingroup$ Tyler answered your second question, and I think this will answer your first one. $\endgroup$ – gravitaunt Sep 21 '16 at 2:34
  • $\begingroup$ I just found that the following chapter in David's notes expands a little on the mathematical equivalence of transition amplitude being zero and the operators commute for spacelike intervals. But the link you attached provids many physical insights into this! $\endgroup$ – JamieBondi Sep 21 '16 at 15:31
2
$\begingroup$

Let's just look at the second term in the integral (ignoring some pesky constants, and seeing that $E_p=E_{-p}$) \begin{align} \int_{-\infty}^{\infty} d^3p \frac{1}{2E_p}e^{-i\mathbf{p}\cdot\mathbf{x}}=-\int_{\infty}^{-\infty} d^3p\frac{1}{2E_p}e^{i\mathbf{p}\cdot\mathbf{x}}=\int_{-\infty}^{\infty} d^3p \frac{1}{2E_p}e^{i\mathbf{p}\cdot\mathbf{x}} \end{align} In the second equality I sent $\mathbf{p}\to-\mathbf{p}$. So, the whole integral will vanish...

$\endgroup$
  • 1
    $\begingroup$ This does not seem right: We have $\exp(\pm i\mathbf{p}\cdot\mathbf{x})= \cos(\mathbf{p}\cdot\mathbf{x})\pm i\sin(\mathbf{p}\cdot\mathbf{x})$. So the contribution of the cosine should remain nonzero. Or do I miss something? $\endgroup$ – flippiefanus Sep 21 '16 at 4:53
  • $\begingroup$ I don't see why that would affect anything....it will still cancel the first integral. $\endgroup$ – ClassicStyle Sep 21 '16 at 5:47

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.