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I am trying to properly understand relation (2.5.5) of Steven Weinberg's QFT.

In (2.5.2), It is shown that a state like $U(\Lambda)\Psi_{p,\sigma}$ has a momentum eigenvalue of ${\Lambda}p$. Consequently, it should generally be expanded in terms of all possible $\Lambda p$ state-vectors without any constraints on their $\sigma$ label, i.e. $$U(\Lambda)\Psi_{p,\sigma} = \sum_{\sigma^\prime} C_{\sigma \sigma^\prime} (\Lambda,p) \Psi_{\Lambda p,\sigma^\prime}\tag{2.5.3}.$$

Now the question: assume we have two momenta $k$ and $p$ related to each other through a Lorentz transformation $L$ in the form $p=Lk$. How could the following relation hold? In other words, why do we have the constraint not to include all possible $\sigma$ labels in the following expansion?

$$ U(L) \Psi_{k,\sigma} = \frac{1}{N(p)} \Psi_{Lk,\sigma}$$

Is this some sort of a definition (or constraint) that we are introducing to study the consequences? In other words, is this in fact the definition of "Standard Lorentz Transformation"?

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  • $\begingroup$ yes, you are right the momentum on the right hand side should be $p=Lk$. $\endgroup$
    – moha
    Commented Mar 13, 2022 at 17:35

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$\Psi_{p,\sigma} = U(L(p)) \Psi_{k,\sigma}$ is a definition for the state $\Psi_{p,\sigma}$. This formula does not apply to any two momenta or for any $L(p)$. You choose a reference momentum $k$ and you choose an $L(p)$ such that $L(p)k=p$ and then you define $\Psi_{p,\sigma}$ as shown above.

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  • $\begingroup$ The question is: why have we assumed the discrete quantum label does not change for such a transformation while, generally, we should not assume such a thing, should we? $\endgroup$
    – moha
    Commented Mar 13, 2022 at 17:40
  • $\begingroup$ It's not an assumption. It's a definition! $\Psi_{k,\sigma} = U(L(p)) \Psi_{k,\sigma}$ defines the discrete label $\sigma$. $\endgroup$
    – Prahar
    Commented Mar 13, 2022 at 17:41
  • $\begingroup$ This cannot be a definition of the discrete quantum label. This can, at best, be a definition of the standard Lorentz transformation as a Lorentz Trans whose operator representation does not affect the quantum discrete label. Still, you have not given your opinion on this argument: there is no implicit or explicit constraint on a Lorentz tran not to change discrete quantum labels, is there? If your agree, then the next question is what is special about the standard Lorentz tranformation that keeps these labels invariant? Pay attention i am not talking about the little group. $\endgroup$
    – moha
    Commented Mar 15, 2022 at 21:47
  • $\begingroup$ @moha - There is nothing special about the standard Lorentz transformation. We have chosen to define the $\sigma$ label on the state $\Psi_{p,\sigma}$ by identifying it with the $\sigma$ label on the reference state $\Psi_{k,\sigma}$. Note that the $\sigma$ label on $\Psi_{k,\sigma}$ hasn't been defined yet (and that's where the little group will come in), but what we have done here is defined the $\sigma$ label on all other states in terms of the reference state. $\endgroup$
    – Prahar
    Commented Mar 15, 2022 at 22:25
  • $\begingroup$ @moha - You are thinking of the equation $\Psi_{p,\sigma} = U(L(p)) \Psi_{k,\sigma}$ as a constraint on $L(p)$. Instead you should think of it as an equation which constrains the $\sigma$ label on $\Psi_{p,\sigma}$ and matches/identifies it with the $\sigma$ label on $\Psi_{k,\sigma}$. $\endgroup$
    – Prahar
    Commented Mar 15, 2022 at 22:27

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