1
$\begingroup$

In this answer https://physics.stackexchange.com/a/376193/274751 two different conventions for the normalization of momentum eigenstates are mentioned. This convention amounts to the choice of $N(p)$ in the following definition of a momentum eigenstate (Weinberg, p. 64, eq. 2.5.5):

$$ |p \rangle = N(p) U(L(p))| k \rangle $$

The inner product between momentum eigenstates, in terms of this $N(p)$, becomes (Weinberg, p. 67, eq. 2.5.17):

$$ \langle p | p' \rangle = |N(p)|^2 \left(\frac{p_0}{k_0}\right) \delta^3(\vec{p}-\vec{p}') $$

Weinberg chooses $N(p) = \sqrt{\frac{k_0}{p_0}}$, so that

$$ \langle p'| p \rangle = \delta^3(\vec{p}-\vec{p}') $$

The other convention is the one mentioned in the linked answer, and in it, we have

$$ \langle p'| p \rangle = 2p_0 \delta^3(\vec{p}-\vec{p}') $$

My question is what would be the $N(p)$, as defined by Weinberg, that is used for this second convention? From a simple comparison with eq. 2.5.17 we have that $N(p) = \sqrt{2k_0} = \sqrt{2mc}$, but this doesn't look right. Why does it not depend on $p$? Are there any further differences between the treatments that should be taken into account here?

$\endgroup$
1
  • 3
    $\begingroup$ It seems that you have your answer already. Why doesn't it look right? Who said $N(p)$ can't be constant with respect to $p$? $\endgroup$
    – Javier
    Aug 25, 2021 at 23:14

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.