I am studying QFT and I found this integral on my lecture notes (for the context: we're trying to show that the covariant commutation relations are Lorentz invariant)

$$∫\frac{d^{3}p dp_{0}}{(2\pi)^{3}}\frac{1}{2ω_{p}}(δ(p_{0}-ω_{p})-δ(p_{0}+ω_{p}))e^{-ip\cdot x}$$

with $ω_{p}=\sqrt{m^{2}+\overrightarrow{p}^{2}}$. In the notes the latter is said to be equal to this other thing here

$$∫ \frac{d^{4}p}{(2\pi)^{3}}\frac{\epsilon(p_{0})}{2ω_{p}}(δ(p_{0}-ω_{p})+δ(p_{0}+ω_{p}))e^{-ip\cdot x}$$

where he introduced the sign function of $p_{0}$ defined as

$$\epsilon(p_{0})=\begin{cases}1\hspace{0.2cm}\text{if} p_{0}>0\\ -1 \text{if} p_{0}<0. \end{cases}$$

My question is probably trivial: how they can be equal? Can someone does the explicit calculation? Cause I tried and failed every time. To me we should split the first in one part with the + (over $\mathbb{R}^{+}$) subtracting the term in $\mathbb{R}^{-}$, but I keep failing.

Ok I think I solved my problem, you can just split the two integrals and use the Dirac delta to obtain something that goes like $\epsilon(ω_{p})e^{-iω_{p}x_{0}}+\epsilon(-ω_{p})e^{iω_{p}x_{0}}=e^{-iω_{p}x_{0}}-e^{iω_{p}x_{0}}$ and rewrite the last one in term of an integral with the Dirac delta function. Is that correct?


1 Answer 1


Yes, what you said is correct. You can also consider the following since it's much simpler ($a>0$):

$$ \int dx f(x) [\delta(x-a)-\delta(x+a)] = f(a) - f(-a) $$ $$ \int dx f(x)\epsilon(x) [\delta(x-a)+\delta(x+a)] = \epsilon(a)f(a) + \epsilon(-a)f(-a) = f(a) - f(-a) $$

So you can see why the equality holds: $\epsilon(x)$ is positive for the first term and negative for the second.


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