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I was reviewing the first few chapters of Weinberg Vol I and found a hole in my understanding in page 112, where he tried to show in the asymptotic past $t=−∞$, the in states coincide with a free state. In particular, he argued the integral

$$\tag{1} \int d\alpha\frac{e^{-iE_{\alpha}t}g(\alpha)T_{\beta\alpha}^+\Phi_\beta} {E_\alpha-E_\beta+i\epsilon}$$

would vanish, where $d\alpha=d^3\mathbf{p}$ (also involves discrete indices like spin, but of no relevance here). In his argument, he used a contour integration in the complex $E_{\alpha}$ plane, in which the integral of central interest is the integration along real line

$$\tag{2} \int_{-\infty}^\infty dE_\alpha\frac{e^{-iE_{\alpha}t}g(\alpha)T_{\beta\alpha}^+\Phi_\beta} {E_\alpha-E_\beta+i\epsilon}$$

I don't see how to obtain (2) from (1), since the lower bound of energy is the rest mass, in the best case I could get something like $\int_{m}^\infty dE_\alpha\cdots$, but how could one extend this onto the whole real line.

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cross-posted: physicsforums.com/showthread.php?t=649171 –  Jia Yiyang Nov 5 '12 at 11:27
    
Sorry, where do you exactly see the integral from -infinity to +infinity? Otherwise your integral from m to infinity is completely unnatural. What's more natural would be a double-line integral going around the m...infinity integral twice, back and forth. Quite generally, integrals over "open contours" that start at a random place of the complex plane such as m are completely silly and they don't allow one to use the power of complex calculus. We always deal with closed contours or contours going to infinity somewhere. –  Luboš Motl Nov 5 '12 at 14:07
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I think this may destroy the analyticity of the integrand. –  Jia Yiyang Nov 7 '12 at 3:15
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The point is an analytic continuation for $g(\alpha)$ from real line to complex plane should be unique, and the resulting function in most cases cannot have $g(\alpha)=0$ for $E_\alpha<0$ –  Jia Yiyang Nov 7 '12 at 16:54
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@JingyuanChen: If you force $g(\alpha)$ to be 0 for $E_\alpha<m$, $g(\alpha)$ is not analytic in its original domain(i.e. whole real line), you cannot do a analytic continuation for it. For example, think about the analytic continuation of $f(x)=\sum\limits_{k=0}^\infty x^k$, which is $\frac{1}{1-x}$ when $|x|<1$, undefined when $|x|>1$. Its analytic continuation should be $f(z)=\frac{1}{1-z}$, which is not 0 for $|x|>1$. In this example had you forced $f(x)=0$ when $|x|>1$, continuation would've been impossible. –  Jia Yiyang Nov 7 '12 at 17:29

1 Answer 1

up vote 2 down vote accepted

1) OP is basically wondering how Weinberg on the middle of p. 112 can extend the integration region from$^1$

$${\cal J}^{\pm}_{\beta}~=~ \int_{m_{\alpha}}^{\infty} \!dE_{\alpha}\frac{e^{-iE_{\alpha}t}g(E_{\alpha})T_{\beta\alpha}^{\pm}} {E_{\alpha}-E_{\beta}\pm i0^{+}}$$

to include the negative real axis

$${\cal J}^{\pm}_{\beta}~=~ \int_{-\infty}^{\infty} \!dE_{\alpha}\frac{e^{-iE_{\alpha}t}g(E_{\alpha})T_{\beta\alpha}^{\pm}} {E_{\alpha}-E_{\beta}\pm i0^{+}},$$

where $g:E_{\alpha}\mapsto g(E_{\alpha})$ is a meromorphic function?

2) That Weinberg (implicitly) assumes meromorphicity of the $g:D\subseteq \mathbb{C}\to \mathbb{C}$ function can be deduced further down on p. 112, where he writes that

[...] we can close the contour of integration for the integration variable $E_{\alpha}$ [...],

which is a clear reference to the residue theorem, which in turn assumes meromorphicity. Also Weinberg writes on the same page$^1$

[...] The functions $g(E_{\alpha})$ and $T_{\beta\alpha}^{\pm}$ may, in general, be expected to have some singularities at values of $E_{\alpha}$ with finite [...] imaginary parts [...]

So there is little doubt that Weinberg assumes meromorphicity of $g$.

3) On the other hand, on the bottom of p. 109, Weinberg writes$^1$

[...]Therefore, we must consider wave-packets, superpositions $\int\! dE_{\alpha}~g(E_{\alpha})\Psi_{\alpha}$ of states, with an amplitude $g(E_{\alpha})$ that is non-zero and smoothly varying over some finite range $\Delta E$ of energies.[...]

Now according to the identity theorem for holomorphic functions, if a function $g:D\subseteq \mathbb{C}\to \mathbb{C}$ is zero on a subset $S\subseteq D$ that has an accumulation point $c$ in the domain $D$, then $g\equiv 0$ is identically zero. However, any interval $I\subseteq \mathbb{R}$ on the real line of non-zero length has accumulation points. So if Weinberg in above quote literally means that $g$ is mathematically zero outside some finite interval $I\subseteq \mathbb{R}$, then $g\equiv 0$ would be identically zero in the whole complex plane.

Of course Weinberg doesn't mean that. He just means that $g$ outside some finite range takes so small values, that to the precision $\epsilon$ that we are working, it doesn't matter whether we include the integration region $\mathbb{R}\backslash I$, or not.

In particular, mathematically speaking, Weinberg has only proven the condition

$$\tag{3.1.12} \int_{m_{\alpha}}^{\infty} \!dE_{\alpha}~ e^{-iE_{\alpha}t} g(E_{\alpha}) \Psi^{\pm}_{\alpha}~\longrightarrow~ \int_{m_{\alpha}}^{\infty} \!dE_{\alpha}~ e^{-iE_{\alpha}t} g(E_{\alpha}) \Phi_{\alpha} ~\text{for}~ t\to\mp\infty \qquad $$

within some precision $\epsilon$. However, the precision $\epsilon$ can be made arbitrarily fine by preparing more and more sharply defined wavepackets $g$.

4) If one would like to have a concrete example of a $g$ function, one may think of a Lorentzian function (aka. Breit–Wigner or Cauchy distribution),

$$g(E_{\alpha}) ~=~ \frac{1}{\pi}\frac{\delta}{(E_{\alpha}-E_0)^2+\delta^2}, \qquad \int_{-\infty}^{\infty}\! dE_{\alpha}~g(E_{\alpha})~=~1,$$

for appropriate choices of constants $E_0$ and $\delta$.

5) Finally, one should not loose sight of Weinberg's main goal in Section 3.1, namely to argue the $\pm i0^{+}$ prescription in the Lippmann-Schwinger equations

$$\tag{3.1.17}\Psi^{\pm}_{\alpha} ~=~\Phi_{\alpha}+\int\!d\beta\frac{T_{\beta\alpha}^{\pm}\Phi_{\beta}} {E_{\alpha}-E_{\beta}\pm i0^{+}}.$$

The Lippmann-Schwinger equations (3.1.17) are not an approximation, and they are independent of the choice of wavepacket $g$.

--

$^1$ To simplify the discussion, we have taken the liberty to replace Weinberg's more general $\alpha$-integration with just an $E_{\alpha}$-integration. Here

$$\tag{3.1.4} \int \!d\alpha \cdots \equiv \sum_{n_1\sigma_1n_2\sigma_2\cdots}\int d^3p_1 d^3p_2 \cdots$$

Changing integration variable from $E_{\alpha}$ to momenta does not solve OP's problem, essentially because we still have to pick the branch of the pertinent square root that has positive real part, so that it doesn't bring us any closer in understanding negative energies.

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So basically you are suggesting that $g(E_\alpha)$, when analytically continued to $E_\alpha<m$ region, is always small in this unphysical region? But mathematically this seems to be a very strong assumption, I'm not sure to what extent it is justified. Besides, this will mean the conclusion derived from these argument is an approximate one instead of an exact theorem, and it's a bit unsatisfactory to me. –  Jia Yiyang Feb 6 '13 at 6:55
    
But we never have infinitely sharp wave packet so (3.1.12) remains an approximation. But I do like your point (5), it indeed offers more validity to Weinberg's derivation, so +1. If no other answers more reasonable than this come up I'll take it as the right answer. –  Jia Yiyang Feb 8 '13 at 4:55

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