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  Schwartz's proof of LSZ formula

+ 3 like - 0 dislike

I have trouble understanding the proof of the LSZ formula given in Matthew Schwartz, Quantum Field Theory and the Standard Model. For reference, this is in section 6.1, equations 6.18-6.19. The author claims that:

$$\langle{\Omega}|T\left\{[a_{p_3}(+\infty) - a_{p_3}(-\infty)]\ldots[a_{p_n}(+\infty) - a_{p_n}(-\infty)]\\
\quad \times [a^\dagger_{p_1}(-\infty) - a^\dagger_{p_1}(+\infty)][a^\dagger_{p_2}(-\infty) - a^\dagger_{p_2}(+\infty)]\right\}|\Omega\rangle\\
= \left[i\int\!\mathrm{d}^4x_1\,e^{-ip_1x_1}(\square_1 + m^2)\right] \ldots \left[i\int\!\mathrm{d}^4x_1\,e^{ip_nx_n}(\square_n + m^2)\right]\\
\quad \times \langle\Omega|T\left\{\phi(x_1)\ldots\phi(x_n)\right\}|\Omega\rangle$$

based on the relation derived earlier:

$$i\int\!\mathrm{d}^4x\,e^{ipx}(\square + m^2)\phi(x) = \sqrt{2\omega_p}[a_p(+\infty) - a_p(-\infty)]$$

The $\phi$ appearing under the time-ordering symbol in the first equation come from the $\phi$ under the integral in the second equation. I do not understand this step. Indeed, in the l.h.s of the first equation, the time ordering bears upon the time in the individual annihilation operators, while in r.h.s, the time ordering bears upon the time variables in the $\phi$. I do not want to discuss the issue of putting the integrals outside of the time-ordering symbol, which the author claims he will address later. What I wonder is whether one is allowed to change the meaning of the time-ordering symbol as described above.

For instance, consider the simpler case where you have some time-indexed symbols $x_t$, where $t$ is discretized (taking it to be half-integer) for simplicity. Then, we have, for instance:

$$T\left\{(x_1 - x_{-1})(x_{1/2} - x_{-1/2})\right\} = x_1x_{1/2} - x_1x_{-1/2} - x_{1/2}x_{-1} + x_{-1/2}x_{-1}$$

Now, suppose that there is a symbol $\Delta_t$ such that $x_{t + 1} - x_t = \Delta_{t + 1/2}$. Then, the discrete analog of the integral expansion Schwartz did to prove the LSZ formula is (well, I think):

$$T\left\{(x_1 - x_{-1})(x_{1/2} - x_{-1/2})\right\}\\
 = T\left\{(\Delta_{-1/2} + \Delta_{1/2})\Delta_0\right\}\\
= \Delta_{1/2}\Delta_0 + \Delta_0\Delta_{-1/2}\\
= T\left\{(x_1 - x_{-1})(x_{1/2} - x_{-1/2})\right\} + [x_{1/2}, x_0] + [x_0, x_{-1/2}]$$

The two results do not coincide because of the commutators. For this reason, I am not sure how Schwartz goes from equation 6.18 to equation 6.19.

asked Sep 1, 2017 in Theoretical Physics by maharishi (25 points) [ revision history ]
edited Sep 1, 2017 by maharishi

Under the symbol of T-ordering you may commute variables since the T-ordering will make them in the right order by definition. Maybe this helps.

Thanks for this hint. Unfortunately, if I take Schwartz's formula (or my simplified version) literally, the commutators are outside of the $T$ operator so that I cannot get rid of them.

Assuming that asymptotically the field behaves as a free field, you can expand \(\phi\) in terms of the creation and annihilation operators \(a_p,a_p^{\dagger}\) as standard, cf. Eq.(6.7) of Schwartz. Now, Fourier transform  \(\partial_0 \phi\), that is, derive Eq.(6.7) in time, multiply both sides by \(e^{ip \cdot x}\)and integrate over \(d^4 x\) . Then you can invert the free field expansion to write the \(a_p,a_p^{\dagger}\)  operators in terms of \(\partial_0 \phi\), and plugging this in the first equation of your question you get the second. For a detailed derivation, cf. section 5.2 of Maggiore's book on QFT, particularly Eq.(5.17--18). If you wish, I can write a detailed derivation tomorrow.

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