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The equation (2.5.12) on page 65 says that: $$ \left(\boldsymbol{\Psi}_{k',\sigma'},\boldsymbol{\Psi}_{k,\sigma}\right)=\delta^3\left(\boldsymbol{k}'-\boldsymbol{k}\right)\delta_{\sigma '\sigma}. $$ I am wondering why it uses $\delta^3\left(\boldsymbol{k}'-\boldsymbol{k}\right)$ of the three-vector momentum rather than $\delta^4\left(k'^{\mu}-k^{\mu}\right)$ of the four-momentum one.

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    $\begingroup$ Could it be this: since $k^\mu k_\mu = m^2$, equality of the spatial components implies equality of the temporal components up to a sign. But $k^0 \ge 0$, so it is in fact equality. $\endgroup$ – Robin Ekman Jul 20 '14 at 22:02
  • $\begingroup$ @RobinEkman Are you saying that we are only considering the states of one particle here? $\endgroup$ – richardfatman Jul 20 '14 at 22:56
  • $\begingroup$ Yes, I think in that section Weinberg is considering only one-particle states. If it were multiple-particle states there would have to be several $\delta$:s. $\endgroup$ – Robin Ekman Jul 20 '14 at 23:45
  • $\begingroup$ @RobinEkman Thank you for your reply. However, I checked the page again and I think Weinberg is talking about the normalization of these states with standard momentum $k^\mu$. If we are talking about the states of only one type of particle, $k^\mu$ is just fixed (equals to (0,0,0,M) according to Table 2.1 on page 66). So I think here those states are not states of a particle of a a particular mass m. Am I correct? $\endgroup$ – richardfatman Jul 21 '14 at 15:55
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    $\begingroup$ the k' here is not a standard momentum while k is. Please refer to \url{physics.stackexchange.com/questions/24766/…} $\endgroup$ – Zheng Liu Dec 3 '14 at 11:11
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(An answer from a student also studying Weinberg's book)

Basically I'll be repeating what @Robin Ekman has said in his comment, but with some clarifications.

First of all, here Weinberg is talking about the states of a single massive particle. That is, $\Psi_{k,~\sigma}$ and $\Psi_{k',~\sigma'}$ are two possible momentum eigenstates of this particle, and the former happens to be chosen as the "standard" state. Now as Ekman points out, for such a particle, the 4-momentum $k^\mu$ is constrained on the hyperboloid $k^\mu k_\mu = m$ with $k^0>0$ in the momentum space. Therefore, only $3$ independent parameters are needed to "enumerate" all the momentum eignvalues (that is, to completely parametrize the hyperboloid in mathematical jargon). These $3$ parameters can be conveniently chosen as the 3-momentum $\mathbf{k}$, which are just the spatial components of $k^\mu$. In this sense, we can write (without any ambiguity) $$\left|{\mathbf{k}',~\sigma'}\right.\rangle := \Psi_{k',~\sigma'},$$ where $\mathbf{k}'$ is the spatial part of $k'$.

Next, considering the labels $k'$ and $\sigma'$ together, we can see that they are meant to form a complete set of quantum numbers. To be more specific, the states $$\left\{\left.\left|{\mathbf{k}',~\sigma'}\right.\rangle\right| \mathbf{k}'\in \mathbb{R}^3,\sigma'=\text{all the allowed discrete eigenvalues}\right\}$$ for an orthonormal basis of the single particle's Hilbert space. Here the "allowed discrete eigenvalues" are determined from the representation of the little group. The relation $$(\Psi_{k',~\sigma'},\Psi_{k,~\sigma})=\delta(\mathbf{k}-\mathbf{k}')\delta_{\sigma~\sigma'}$$ is just (part of) the orthonormality condition in that Hilbert space.

It might be useful here to point out that the "induced volume element" of the hyperboloid, under parametrization $\mathbf{k}$, is $$d^3\mathbf{k}/\sqrt{\mathbf{k}^2+m^2},$$ as is given by the second equation on Page 67 of Weinberg's book.

Also when Weinberg moves on to massless particles, the hyperboloid in the present discussion changes to the null surface (the "future light cone" $k^\mu k_\mu = 0,~ k^0 > 0$), and the logic is much the same!

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  • $\begingroup$ Does this mean that the normalization of 1-particle states is not Lorentz-invariant? (And is this a problem?) $\endgroup$ – Physics Llama Jun 18 '20 at 22:45

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