For the Ch. 27 in book QFT by Srednicki, in the modified minimal-subtraction renormalization scheme($\overline{MS}$), the residue for pole at $-m_{ph}^2$ is $R$, instead of one.

  1. However, I can not understand why we need only change the external leg as:

    • add a $R^{1/2}$ coefficient
    • replace the Lagrangian mass $m$ as $m_{ph}$

but we don't need to change the internal legs.

  1. Also, renormalization scheme means to determine the coefficient for counterterms, or say, determine the coefficient for $Z_i$ in Largranian: $$L=-\frac{1}{2}Z_\phi(\partial \phi)^2-\frac{1}{2}Z_mm\phi^2+\frac{1}{4!}Z_g g \phi^4\\=-\frac{1}{2}(\partial \phi_0)^2-\frac{1}{2}m_0\phi_0^2+\frac{1}{4!} g_0 \phi_0^4$$ according to the answer in Is the effective Lagrangian the bare Lagrangian?, I know that renormalized $m$ is not the physical mass, but the pole of propagator is. But I am confused that: since it seems like $m_0$ is just pole according to the second line of equation above, does this mean bare mass $m_0$ is actually the physical mass(the mass what we detected)? I think so because Srednicki says "bare parameters" must be independent of $\mu$" in Ch.28. But other book, such as p.323 Peskin, says bare mass is not not the values measured in experiments.

3 Answers 3


First, let me address the quick questions you have:

It seems like $m_0$ is just pole according to the second line of equation above, does this mean bare mass $m_0$ is actually the physical mass? And does the "physical mass" here means the mass what we detected?

No, $m_0$ is not the pole and it is not the physical mass. You can't just read off the pole from what appears in the Lagrangian.

Yes, the "physical mass" is the mass we can detect. If the theory have one-particle states, then the "physical" mass $m_{\text{ph}}$ is defined from the invariant of the momentum of the one-particle states $k^2=-m_{\text{ph}}^2$.

Now, let's clarify what is the pole and what are the differences between the renormalization scheme.

The pole:

In Srednicki's notations, the full propagator, defined via $\mathbf{\Delta}(x-y)=i\langle 0|T\phi(x)\phi(y)|0\rangle$ is given by the Lehmann-Kallen form (Eqn. 13.17): $$ \tilde{\mathbf{\Delta}}(k^2)=\frac{1}{k^2+m_{\text{ph}}^2-i\epsilon} +\int ds\rho(s)\frac{1}{k^2+s-i\epsilon} $$ Here the renormalized field $\phi(x)$ is assumed to satisfy Eqn. (13.2): $$\langle k|\phi(0)|0\rangle=1$$

Looking at the first term of the exact propagator, the important point here is that it has a pole at $k^2=-m_{\text{ph}}^2$. This is the pole we are interested in, which define the physical mass.

On the other hand, you know that the full propagator in our perturbation theory is given in terms of the self-energy $i\Pi(k^2)$ by: $$ \tilde{\mathbf{\Delta}}(k^2)=\frac{1}{k^2+m^2-\Pi(k^2)-i\epsilon} $$ where the self-energy can be computed by summing over the 1PI 2-point graphs.

One-shell scheme:

In the on-shell scheme, we want the renormalized mass $m$ to be the physical mass. That can be done by setting $m=m_{\text{ph}}$ and impose the certain conditions on $\Pi(k^2)$ such that the full propagator has a pole at $k^2=-m_{\text{ph}}^2$. This condition determines $Z_m$.

The second condition, $\langle k|\phi(0)|0\rangle=1$ is satisfied by requiring that the residue of the full propagator at the pole is $1$.

Minimal-subtraction Scheme:

In the (modified) minimal-subtraction scheme, all we care about is removing the infinities. $Z_\phi$ and $Z_m$ are chosen "minimally" such that the self-energy is finite.

The renormalized mass $m$ is not the physical mass. The pole is at $k^2=-m_{\text{ph}}^2$, not $k^2=-m^2$

And since the field has not been normalized "properly", the residue at the pole is $R=|\langle k|\phi(0)|0\rangle|^2\neq 1$.

  • $\begingroup$ Thank you so much for your wonderful answer! But I am confused that if bare mass, e.g. $m_0$, is not the physical mass (something we detect), why we still say bare parameters must be independent of $\mu$, as in Ch.28 by Srednicki? $\endgroup$ Mar 29, 2020 at 17:24
  • 1
    $\begingroup$ @MerlinZhang The intuitive picture is like this: The bare mass is the (physical) mass when there were no interaction ($g\rightarrow 0$). In the presence of the interaction, the free particle becomes "dressed" particle with physical mass $m_{ph}$. The difference between $m_0$ and $m_{ph}$ depends on $g$. So $m_0$ is some parameter we started off with. It shouldn't depend on anything. $\endgroup$
    – JF132
    Mar 29, 2020 at 18:07

Point 1.
In the renormalized propagator the pole mass $m_{ph}$ is physical and independent of any subtraction scheme used to set the finite parts of the counterterms. In the on-shell subtraction scheme the finite parts of the counterterms are chosen so that $m_{ren} = m_{ph}$. In minimal subtraction $m_{ren} \ne m_{ph}$.

In minimal subtraction the lagrangian mass parameter is $m_{ren}$ and the LSZ formula has to be corrected by a $R^{-1/2}$ factor for each external particle because of the normalization of the field $R^{-1/2} \phi (x)$. The Green's function appearing in the LSZ formula already accounts for all of the interacting terms.

Point 2.
The bare Lagrangian is written in terms of bare (unrenormalized) fields and couplings, which are infinite. That is why of the renormalization procedure.

You write:
$\phi_0 = \sqrt{Z_\phi} \phi_{ren}$
$m_0 = Z_m m_{ren}$
$g_0 = Z_g g_{ren}$

The mass you detect is the pole mass, that is $m_{ph}$, which is related to the renormalized mass $m_{ren}$ as $\Sigma_{ren} (m_{ph}) = m_{ren} - m_{ph}$, where $\Sigma_{ren} (m_{ph})$ is the sum of all of the $1PI$ (one particle irreducible) graphs including the counterterms.


Here we will try to answer OP's title question, which seems spurred by the following paragraph above eq. (27.6) in Srednicki's book:

Finally, in the LSZ formula, each external line will contribute a factor of $R$ when the associated Klein-Gordon wave operator hits the external propagator and cancels its momentum-space pole, leaving behind the residue $R$. Combined with the correction factor of $R^{−1/2}$ for each field, we get a net factor of $R^{1/2}$ for each external line when using the $\overline{\rm MS}$ scheme. Internal lines each contribute a factor of $(−i)/(k^2 + m^2)$, where $m$ is the lagrangian-parameter mass, and each vertex contributes a factor of $iZ_g g$, where $g$ is the lagrangian-parameter coupling.

Srednicki is trying to say that the internal/amputated part of the connected correlator function in the LSZ formula can be calculated as a sum of amputated Feynman diagrams that are built from free propagators $\Delta=1/(k^2 + m^2)$, vertices $ig$, and counterterms as usual. Srednicki mentions indirectly the counterterms $Z_g-1$ for the vertex, but for some reason he does not mention the kinetic counterterm $Z_{\phi}-1$ and the mass counterterm $Z_m-1$, but of course they are also there.

The internal free propagators can typically not be re-summed into full/exact propagators a la this Phys.SE post; in particularly if they are part of a 1PI subdiagram, cf. my Phys.SE answer here.

On the other hand, the external legs are already re-summed into full/exact propagators $${\bf \Delta}~=~\frac{1}{k^2 + m^2-\Pi}~\simeq~\frac{R}{k^2 + m_{\rm ph}^2}$$ (which carry the residue factor $R$) in order to fit into the LSZ formalism.


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