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Let be $$ P^\mu |p> = p^\mu |p> $$ i.e. $|p>$ is the eigen-vector of the 4-momentum operator.

Where does the following Lorentz-covariant completeness relation come from?

$$ \int d^4p \theta(p^0) \delta(p^2 + m^2)|p><p| = 1 . $$

Qmechanic
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apt45
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    hint: act with both sides on $|p'\rangle$, and use $\langle p|p'\rangle\propto \delta(\boldsymbol p-\boldsymbol p')$, together with $\mathrm dp\ \Theta(p^0)\delta(p^2+m^2)\propto \mathrm d\boldsymbol p$. – AccidentalFourierTransform Jan 15 '16 at 18:21
  • @AccidentalFourierTransform thank you. I would like to know how to derive it from relativistic quantum mechanics. Can you give me some references? – apt45 Jan 15 '16 at 18:26
  • Because I would like to understand better the theory than calculus... This completeness relation has to come from some Lorentz invariant principle or Lorentz transformation of states $|p>$... – apt45 Jan 15 '16 at 18:32
  • I would like to understand also the minus sign to my question... – apt45 Jan 15 '16 at 18:35
  • Thanks for your answer. I searched here a question like this one, but I didn't find it. I tried to derive the relation, but the problem is that I understood that all of my efforts were wrong, so I don't see the importance of write them here... – apt45 Jan 15 '16 at 18:56
  • Related: http://physics.stackexchange.com/q/83260/2451 – Qmechanic Jan 15 '16 at 23:46

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Two things are required to prove this. First, you haven't really defined your state $|p\rangle$ in its entirety since you haven't defined what normalization you are using for this state. In the formula above, it seems to me that you are using the normalization $$ \langle {\bf p} | {\bf p}' \rangle = \frac{2 E_p }{ 2\pi } \delta^3 ( {\bf p} - {\bf p'} ) $$ Then, the completeness relation is $$ 1 = \int d^3 p \frac{2\pi}{2 E_p } | {\bf p} \rangle \langle {\bf p} | $$ Finally, you have to show that (do this yourself) $$ \int d^3 p \frac{1}{2 E_p } = \int \frac{d^4 p}{ 2\pi } \theta ( p^0 ) \delta (p^2 + m^2 ) $$ which then implies $$ 1 = \int d^4 p \theta ( p^0 ) \delta ( p^2 + m^2 ) | {\bf p} \rangle \langle {\bf p} | $$

Prahar
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