6
$\begingroup$

When deriving LSZ reduction theorem Weinberg in his QFT book have assumed n-point generalized Green functions, $$ G(q_{1},...,q_{n}) = \int d^{4}x_{1}...d^{4}x_{n}e^{-i\prod_{i =1}^{n}q_{j}x_{j}} \langle |\hat {T}\left( \hat {O}_{l}(x_{1})\hat {A}_{2}(x_{2})...\hat A_{n}(x_{n})\right) |\rangle , \quad (1) $$ where $\hat {O}_{l}(x)$ transforms under the irreducible representation of the Lorentz group as some free field $\hat {\Psi}_{l}(x)$. By insertion between $\hat {O}_{l}(x_{1})$ and $\hat {A}_{2}(x_{2})$ functional unit $$ \sum_{i, \sigma}\int d^{3}\mathbf p | (\mathbf p , \sigma )_{i}\rangle \langle (\mathbf p , \sigma )_{i}| $$ and by allocation of one-particle states from it he have "reduced" (with some hints) $(1)$ to the form $$ G(q_{1},...,q_{n}) \to f(q)\sum_{\sigma}\langle | \hat {O}_{l}(0)| (\mathbf q_{1}, \sigma )\rangle \times $$

$$ \times \int d^{4}x_{2}...e^{-iq_{2}x_{2}-...}\langle (\mathbf q_{1}, \sigma ) |\hat {T}\left( \hat {A}(x_{2})...\right) | \rangle \delta (q_{1} + ... + q_{n}). \qquad (2) $$ Here $f(q)$ contains the pole of the first order $\frac{1}{q^{2} - m^{2} - i\varepsilon}$ and $q = q_{1} + ... + q_{r}$.

After that he says that in $(2)$ there is equality $\hat {O}_{l}(0)| (\mathbf q_{1}, \sigma )\rangle = \frac{1}{\sqrt{(2 \pi )^{3}}}Nu^{\sigma}_{l}(\mathbf q_{1})| \rangle $.

So I have the question: why was factor $N$ (in comparison with free field-like expression $\hat {O}_{l}(0)| (\mathbf q_{1}, \sigma )\rangle = \frac{1}{\sqrt{(2 \pi )^{3}}}u^{\sigma}_{l}(\mathbf q_{1})| \rangle $) appeared? What is its physical sense? Is its appearance connected with the fact that $| \rangle$ doesn't refer to the "usual" vacuum? Can you also comment this statement, if you please?

$\endgroup$
  • 1
    $\begingroup$ (a) You should have read the entire section 10.3 before you asked. It is the subject of this section. (b) That said, what Weinberg states here is IMO not a proof, but the common belief that QFT should be interpreted in this way. (c) You can safely think |> is a vacuum, or no-particle state. (d) My impression is Weinberg is your first QFT textbook. If so, I recommend you to read another before Weinberg. (Say Srednicki?) $\endgroup$ – teika kazura Jul 5 '14 at 0:19
  • $\begingroup$ @teikakazura : it is not my first QFT textbook. For example, I have read Srednicki book yesterday for understanding the meaning of $N$. But there is also not written why do we must introduce this factor. Also - why do for interacting fields all is simple (we also add factor $N$ to the definition of the free field). Also I have read Peskin and Schroeder book, but there also aren't explanations. $\endgroup$ – Andrew McAddams Jul 5 '14 at 9:22
  • $\begingroup$ The section 10.3 in my editing of the book refers to the LSZ reduction formula proof. Also, Weinberg only concludes there that he should be assume that one of operators of his formula from paragraph 10.2 transforms as free field under the irreducible rep of the Lorentz group. The remaining results of 10.3. The main remaining results were obtained only from this statement. $\endgroup$ – Andrew McAddams Jul 5 '14 at 9:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.