3
$\begingroup$

TLDR: The question: Does it make sense to calculate scattering amplitudes using an off-shell renormalization scheme?

I expand a bit by using a theory of a single self interacting massive scalar. I consider both the bare and renormalized objects by subscripts $0$ or $R$. We have $\phi_0 = \sqrt{Z_{\phi}} \phi_R, m_0 = Z_m m_R$.

The connected bare (renormalized) momentum space Greens function reads \begin{align*} G_{c,0(R)}^{(n,m)}(p_1,...,p_n;p_1',...,p_m') = \prod_{i=1}^n\int d^dx_i~e^{-ip_i x_i} \prod_{j=1}^m \int d^dy_j~e^{ip'_j y_j} \langle\Omega| T \phi_{0(R)}(x_1)...\phi_{0(R)}(x_n) \phi_{0(R)}(y_1)...\phi_{0(R)}(y_m) |\Omega \rangle_c, \end{align*} where the subscript $c$ means that we take only connected diagrams. The full bare (renormalized) propagator $i\Pi_{0(R)}(p^2)$ is defined by \begin{align*} G_{c,0(R)}^{(1,1)}(p;-p') = (2\pi)^d \delta^{(d)}(p-p') i\Pi_{0(R)}(p^2). \end{align*} We have \begin{align*} i\Pi_{0(R)}(p^2) = \frac{i}{p^2 - m_{0(R)}^2 - \Sigma_{0(R)}(p^2) + i0} = \frac{i\mathcal{R}_{0(R)}}{p^2 - m_{ph}^2} + \text{terms non-singular as }p^2 \rightarrow m_{ph}^2, \end{align*} where $\Sigma_{0(R)}(p^2)$ is the sum over the all 1PI diagrams with two external legs, with the two external free propagators $\frac{i}{p^2 - m_{0(R)}^2 + i0}$ factored out. The pole mass $m_{ph}$ is defined to be the solution of \begin{align*} m_{ph}^2 = m_{0(R)}^2 + \Sigma_{0(R)}(m_{ph}^2) \end{align*} and the residue at this pole is given by \begin{align*} \mathcal{R}_{0(R)} = \frac{1}{1-\Sigma_{0(R)}'(m_{ph}^2)}. \end{align*} The LSZ formula reads \begin{align*} {}_H\langle p_1',...,p_m'| i\mathcal{T} |p_1,...,p_n\rangle_H &= \lim_{p_1^2,...,p_n^2,p_1'^2,...,p_m'^2 \rightarrow m_{ph}^2} \prod_{i=1}^n \frac{p_i^2-m_{ph}^2}{i \sqrt{\mathcal{R}_0}}\prod_{j=1}^m \frac{p_j'^2-m_{ph}^2}{i \sqrt{\mathcal{R}_0}} G_{c,0}^{(n,m)}(p_1,...,p_n;p_1',...,p_m') \\ &= \lim_{p_1^2,...,p_n^2,p_1'^2,...,p_m'^2 \rightarrow m_{ph}^2} \prod_{i=1}^n \frac{p_i^2-m_{ph}^2}{i \sqrt{\mathcal{R}_0/Z_{\phi}}}\prod_{j=1}^m \frac{p_j'^2-m_{ph}^2}{i \sqrt{\mathcal{R}_0/Z_{\phi}}} G_{c,R}^{(n,m)}(p_1,...,p_n;p_1',...,p_m') \\ &= \lim_{p_1^2,...,p_n^2,p_1'^2,...,p_m'^2 \rightarrow m_{ph}^2} \Big{(} \frac{\mathcal{R}_R}{\sqrt{\mathcal{R}_0/Z_{\phi}}} \Big{)}^{n+m} G_{c,amp,R}^{(n,m)}(p_1,...,p_n;p_1',...,p_m'), \end{align*} where $G_{c,amp,R}^{(n,m)}$ is the amputated Greens function.
Now an important assumption is that mass $m_{ph}$ is simultaneously the physical mass of the particles in the asymptotic state on the LHS of the LSZ formula and corresponds to the pole of the full propagator at $p^2 = m_{ph}^2$.
When using an on-shell scheme, where $m_R = m_{ph}$ and $\mathcal{R}_0 = Z_{\phi}, \mathcal{R}_R = 1$ everything is fine. The $\mathcal{T}$ matrix element is just the amputated Greens function with on-shell external particles.
However for a general off-shell scheme this is not the case. Moreover, in addition to the awkward factor $\Big{(} \frac{\mathcal{R}_R}{\sqrt{\mathcal{R}_0/Z_{\phi}}} \Big{)}^{n+m}$, the masses appearing in the propagators for a given diagram is $m_R$ while the $p_i^2 = p_j'^2 = m_{ph}^2 \neq m_R^2$.

The last line from in the above equation is clearly impractical for calculations. Am I therefore right to assume that calculating scattering amplitudes using an off-shell renormalization scheme is non-sensible?

However I do often see something like scattering amplitudes in MS scheme, how would one define then this amplitude or modify the LSZ prescription?

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.