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In the Quantum Field Theory(QFT), we work in the distributional sense, that the normalization of vacuum is \begin{equation}\langle0|0\rangle=2E(\vec{0})(2\pi)^3\delta(\vec{0})\end{equation} This fact is solved by renormalisation techniques in situations it is suitable. However, if we forget about the renormalisation and think about the problem not as if it was a problem with infinity, but as a distributional problem, with the delta distribution definition $$\delta(0)f(x)=\delta_0(f(x))=f(0)$$ how would we need to rewrite the first equation to satisfy this definition and what would be the meaning of the applied function?

Note: my "non-standard" writing $\delta(0)f(x)=\delta_0(f(x))$ has nothing to do with the convention used in the first equation and the energy function product with the Dirac delta. It is completely different convention and that is why I am asking how to rewrite the above. However, the energy function above should be a constant, an energy of vacuum, not a function, anyway.

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  • $\begingroup$ No, they are not. $\endgroup$
    – user74200
    Commented Feb 11, 2017 at 13:31
  • $\begingroup$ Hamiltonian is a three dimensional object and the normalisation is 4D. $\langle0|0\rangle=\langle p=0|p=0\rangle$ with $p=(p_0,p_1,p_2,p_3)$. Actually, even $\langle p|p\rangle=E(\vec{p})/E(\vec{0})\langle0|0\rangle=2E(\vec{p})(2\pi)^3\delta(\vec{0})$ $\endgroup$
    – user74200
    Commented Feb 11, 2017 at 13:38
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    $\begingroup$ I don't understand why questions about QFT are downvoted by so many people. $\endgroup$
    – Physics Guy
    Commented Feb 11, 2017 at 15:43

1 Answer 1

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The vacuum is always normalized as $\langle 0|0\rangle=1$. You seem to be confusing this with the normalization of the one-particle state $|{\bf 0}\rangle$ which is an example of
$$ \langle {\bf p}|{\bf p}'\rangle= (2\pi)^2 2E({\bf p}) \delta^3({\bf p}-{\bf p}'). $$ where ${\bf p}$ is the three-momentum. The one-particle state $|{\bf p}={\bf 0}\rangle $ describes a particle at rest and this is not the same as the vacuum.

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  • $\begingroup$ Nope, it is not three-momentum, it is four-momentum. $\langle x|x\rangle=1$, and since $p=\mathcal{F}x$, where $\mathcal{F}$ is a fourier transform, $\langle p|p\rangle=2E(p)(2\pi)^3\delta^3(0)$, if $p=0$, we get that for vacuum. The 2E(p) we get from the time component. Seriously. Don't try to correct the statement, it is not wrong, it is taken from a textbook. It is not wrong. $\endgroup$
    – user74200
    Commented Feb 11, 2017 at 19:24
  • $\begingroup$ Four-momentum zero is not a particle, it is nothing. It is a field with no mass and no momentum, an empty space, nothing. For a particle with no momentum the convention is $|k\rangle=|(m,\vec{0})\rangle$. This is single particle with no momentum. $\endgroup$
    – user74200
    Commented Feb 11, 2017 at 19:33

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