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I have this general formula for the propagator.

$$ D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\omega_k}[e^{-i[\omega_kt-\vec k\cdot\vec x]}\Theta(t)+e^{i[\omega_kt-\vec k\cdot\vec x]}\Theta(-t)] $$

I am supposed to verify that it decays exponentially for spacelike separation. To impose this condition I use $t=0$ and $x>0$. The symbols $\vec k$ and $\vec x$ are 3-vectors and $\omega_k=+\sqrt{\vec k^2+m^2}$. Also, for the Heaviside step function $\Theta(0)=0.5$.

This is problem 1.3.1 in Zee's QFT in a Nutshell book, and the solution to this problem is in the back of the book. He writes that the first step should look like

$$ D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}e^{-i\vec k\cdot\vec x} $$

but when I plug it in the values for $x$ and $\Theta$, I get

$$ D(x)=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}\dfrac{1}{2}[e^{i\vec k\cdot\vec x}+e^{-i\vec k\cdot\vec x}]=-i\iiint\dfrac{ d^3k}{\big( 2\pi \big)^32\sqrt{\vec k^2+m^2}}\cos(\vec k\cdot\vec x) $$

What is the reasoning behind throwing out one of the exponentials instead of using both of them? I haven't done my integral yet, so I don't know if it still produces exponential decay, but it seems like won't, so I want to know what the discrepancy is before I start.

Qmechanic
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Jonathan Tooker
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    Is you expression not the same as his? The contribution of $k$ and $-k$ are the same by symmetry. – mike stone Jun 03 '18 at 21:20
  • My starting expression with $\Theta$ is the same as his, but I do not get his expression that he starts with in the solution, which he supposedly gets with $t=0$ and $\Theta(0)=1/2$. You're right this is symmetric under $k\to-k$ but I don't think I can swap the sign of only one $k$, and swapping the sign of both leaves the expression invariant (as expected of symmetry under $k\to-k$). Note that since cosine is even, my expression preserves the $k\to-k$ symmetry. – Jonathan Tooker Jun 03 '18 at 21:27
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    You can change the sign of the first exponential : split the integral into two, change the sign, put everything back together and you're done. – Adam Jun 03 '18 at 21:40
  • Hi Adam, Thank you, but I am not satisfied. Sign flip symmetry on $\vec k$ comes from the initial expression $k(x'-y')$ where it doesn't matter if $x'$ or $y'$ comes first, and then we define $x=x'-y'$ to get the "$x$" that appears in the expression for $D(x)\equiv D(x'-y')$ that I used above. Therefore, if I split and swap $k\to-k$ as you suggest, and then seek to recombine, I will have one term where $x=x'-y'$ and another where $x=y'-x'$, and that is not allowed. Can you think of some other reason? Or am I wrong perhaps? – Jonathan Tooker Jun 03 '18 at 21:50

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