# Are the matrix elements of $S$-matrix Lorentz invariant?

The *whole point* of QFT is that it is a framework that allows you to define Lorentz (co)variant scattering amplitudes. In fact, under some general hypothesis it is the *only* framework with that property. The expression in the OP is not *manifestly* Lorentz covariant, although it turns out to be, after a very cumbersome analysis. See ref.1 for a detailed discussion.

Alternatively, the Lorentz covariance of the scattering amplitudes follows immediately from the Lorentz invariance of transition amplitudes, say, as given by the LSZ formula, e.g. $$ \langle p_{1},\ldots ,p_{n}\ {\mathrm {out}}|q_{1},\ldots ,q_{m}\ {\mathrm {in}}\rangle =\int \prod _{{i=1}}^{{m}}\left\{{\mathrm {d}}^{4}x_{i}{\frac {ie^{{-iq_{i}\cdot x_{i}}}\left(\Box _{{x_{i}}}+m^{2}\right)}{(2\pi )^{{{\frac {3}{2}}}}Z^{{{\frac {1}{2}}}}}}\right\}\prod _{{j=1}}^{{n}}\left\{{\mathrm {d}}^{4}y_{j}{\frac {ie^{{ip_{j}\cdot y_{j}}}\left(\Box _{{y_{j}}}+m^{2}\right)}{(2\pi )^{{{\frac {3}{2}}}}Z^{{{\frac {1}{2}}}}}}\right\}\langle 0|{\mathrm {T}}\varphi (x_{1})\ldots \varphi (x_{m})\varphi (y_{1})\ldots \varphi (y_{n})|0\rangle $$ for a scalar field theory in four spacetime dimensions. In turns, the correlation function $\langle 0|{\mathrm {T}}\varphi (x_{1})\ldots \varphi (y_{n})|0\rangle$ is Lorentz invariant, which is manifest when written as a path integral, $$ \langle 0|{\mathrm {T}}\varphi (x_{1})\ldots \varphi (y_{n})|0\rangle\sim \int \varphi (x_{1})\ldots \varphi (y_{n})\ e^{-I[\phi]}\ \mathrm d\varphi $$ where $I$ is the action, a Lorentz scalar.

All the elements in the above formulas are manifestly Lorentz covariant, and so it immediately follows that so are the $S$-matrix elements. The case of arbitrary spin is handled similarly, although the details are more involved. Again, see ref.1 for the explicit construction.

FWIW, the $S$-matrix can also be written directly as a path integral, sidestepping the need of ever introducing the LSZ formula. See e.g. ref.2, where we can read, for example, (p.11)

Comparing [...] with [...] we can say that to get the $S$-matrix in the momentum representation one is to calculate the Feynman functional integral [...]. This observation shows that the $S$-matrix is defined by means of a functional integral in a more elegant manner than the evolution operator. [...] This is specially attractive in the case of field theory because [it] becomes manifestly Lorentz covariant [...].

**References.**

Weinberg -

*Quantum theory of fields, Vol.1. Foundations*.Faddeev -

*Introduction to Functional Methods*, from*Methods in Field Theory*, edited by North Holland. See also this PSE post.

1)

A normalization condition for one-particle states $| \vec p \rangle$, which is Lorentz invariant is

$\langle \vec p | \vec q \rangle = 2 E_p (2 \pi)^3 \delta^{(3)} (\vec p - \vec q)$

(The factor $2$ is conventional)

However, if we write

$\delta^{(3)} (\vec k) = \int \frac{d^3x}{(2 \pi)^3} exp (i \vec k \cdot \vec x)$

we can state

$\delta^{(3)} (\vec 0) = \frac{V}{(2 \pi)^3}$

Hence

$\frac{\delta^{(3)} (\vec 0')}{V'} = \frac{\delta^{(3)} (\vec 0)}{V}$

So, the normalization in your post is invariant as well.

2)

The S-matrix element $\langle f|S| i \rangle$ for $n$ asymptotic momentum eigenstates is given by the LSZ (Lehmann-Symanzik-Zimmermann) reduction formula which for scalar quantum fields $\phi (x)$ states

$\langle f|S| i \rangle = [i \int d^4 x_1 exp (-i p_1 x_1) (\Box_1 + m^2)] \cdot \cdot \cdot [i \int d^4 x_n exp (i p_n x_n) (\Box_n + m^2)] \langle \Omega | T \{ \phi (x_1) \cdot \cdot \cdot \phi (x_n)\} | \Omega \rangle$

where $-i$ in the exponent applies to initial states and $+i$ to final states

The LSZ formula is constructed from Lorentz covariant fields, hence the S-matrix element is an invariant.