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I am going through the basics of renormalization in QFT from Schwartz's book and I'm confused about what is the measurable coupling constant. enter image description here

In the $\phi^4$ theory, to compute this $2\rightarrow2$ matrix element, Schwartz introduces a cut-off $\Lambda$ and the matrix element is then given by $M_2(s)=-\lambda-\dfrac{\lambda^2}{32\pi^2}\ln(s/\Lambda^2)$ for some energy scale $s$.

He then proceeds to renormalize and defines a renormalized coupling $\lambda_R$ by $$\lambda_R=-M_2(s_0) \tag{15.65} $$ at some reference energy level $s_0$, which means that $\lambda_R$ is an observable, hence physical. He then proceeds to find an expression for $M_2(s)$ that does not depend on the cut-off, but on $s$ and $s_0$; we're predicting cross-sections at one energy level with respect to a cross-section at $s_0$.

My questions follow:

(1) We could define another $\lambda_R'=-M_2(s_0')$ (which is again physical and measurable) and we could find $M_2(s)$ with respect to $s$ and $s_0'$. I think the physics of this is that our predictions are relative to an arbitrary energy level. What I don't understand is what is the physical parameter describing the strength of the interactions of the theory*.

(2) Before I reached the renormalization part of QFT, I thought it was the parameter in the form of $\phi^4$ in $\mathcal{L}_{int}$ but now I don't know what it is. I suppose that in QED (having seen its Lagrangian), the analogous quantity to $\lambda$ there is proportional to the electron charge, $e$. So, in that case, we have $e_R, e(s_0), e(s_0')$, etc. But then when everybody's talking about the value of the electron's charge, everybody agrees with a particular value. At other times, people say that it changes with the energy scale. So, what is the electron's measurable, physical charge?


*Is it that we just pick a reference energy level and define the renormalized coupling (or charge) at that level, but the actual parameter that describes the strength of the interactions changes with the energy? If so, this would answer (2).

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  • $\begingroup$ "Is it that we just pick a reference energy level and define the renormalized coupling (or charge) at that level, but the actual parameter that describes the strength of the interactions changes with the energy?", Yes, see physics.stackexchange.com/q/436173 $\endgroup$ – MadMax Jan 12 at 16:35
  • $\begingroup$ @MadMax Please find it here: physics.stackexchange.com/q/607205 Thanks! $\endgroup$ – TheQuantumMan Jan 12 at 18:06
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Yes, OP's interpretation is correct. Call the bare coupling constant $\lambda_0$, the renormalised coupling at the renormalisation point $\lambda_R = \mathcal{M}(Q^2)$. Indeed, the choice of the renormalisation point is arbitrary, so we could equivalently define $\lambda'_R$ at $Q'$, and this gives rise to renormalisation group methods, however it is necessary to choose one such $Q$ and stick with it for future theoretical calculations. $\lambda_R$ is then in principle calculable from an observable quantity, since the relation $\lambda_R = \mathcal{M}(Q^2)$ is defined to hold to all orders of perturbation theory, but it will involve a sum of formally infinite powers of $\lambda_0$ (this is fine of course, since the bare parameters are nothing more than unphysical in the interacting theory). As such, we can then use the value of $\lambda_R$ to make further predictions about observables measured at energy scale $Q$.

However, the interaction coupling $\tilde{\lambda}$ at a different energy scale $\tilde{Q}$, will typically vary from $\lambda_R$ through a non-trivial relation $\tilde{\lambda}(\tilde{Q}^2) = f(\lambda_R, m)$ such that $\tilde{\lambda}$ and $\lambda$ agree at $Q$ - this is the effective coupling or running coupling, which can be used to calculate observables at any energy scale.

Simply put, the procedure is: Measure an observable matrix element from a physical process at an energy scale $\rightarrow$ define the renormalised coupling at that energy scale $\rightarrow$ calculate coupling at a different energy scale using the beta function$^\dagger$ $\rightarrow$ make predictions about observables at this new energy scale.

This implies, for example, that the electron charge is energy-dependent (the oft-quoted value is usually at $Q^2 = 0$), and that weakly coupled theories at low energies can become strongly coupled at high energies (e.g. quantum electrodynamics) and vice versa (e.g. quantum chromodynamics) - this dependence, again, can be discerned from the beta function.

$^\dagger$The beta function describes the dependence of the coupling on the energy scale $\mu$, defined by $$ \beta(\lambda) = \frac{\partial \lambda}{\partial\log\mu} = \mu\frac{\partial \lambda}{\partial\mu}$$

(This is conventional, see for example the Callan-Symanzik equation). As mentioned previously, the physics should not depend on the choice of the renormalisation scale or scheme, so the beta function only depends on $\mu$ through $\lambda$.

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  • $\begingroup$ So, I guess the renormalized coupling is what all physicists implicitly agreed on using as "the" coupling and then the running coupling is the one that gives a measure of the strength of the interactions? $\endgroup$ – TheQuantumMan Jan 12 at 17:03
  • $\begingroup$ This would imply, for example, that the electron's charge is actually energy dependent and physically changes with energy, even though we still say that the electron charge is $~1.6*10^{-19} Coulomb$ but we mean that it's just the value at a universally accepted energy level. Another concequence is that there are energy levels that can turn some weakly coupled theory into a strongly coupled theory above that energy level. Is my assesment correct? $\endgroup$ – TheQuantumMan Jan 12 at 17:03
  • $\begingroup$ @TheQuantumMan good observations; updated $\endgroup$ – Nihar Karve Jan 12 at 17:17
  • $\begingroup$ Thanks. So, in essence, there's no such thing as "the" electron change (as a fixed value) then? Only the electron charge as the agree upon value and the electron charge as in being the effective electron charge that's energy dependent? $\endgroup$ – TheQuantumMan Jan 12 at 17:22
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    $\begingroup$ @TheQuantumMan yes, nicely done $\endgroup$ – Nihar Karve Jan 12 at 17:27
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This issue also confused me for quite some time and I present now my current understanding. I am still not quite sure if my understanding is completely correct and I am very thankful for corrections. I refer as an example to the QED coupling constant. Some points

  • The parameter in the bare Lagrangian $e_0$ is formally infinite, i.e. it diverges in the limit where you remove the regulator, e.g. $\Lambda \rightarrow \infty$ for cutoff or $\epsilon \rightarrow 0$ for dim. reg. But this is not a problem, since $e_0$ is not measurable! Note that the theory should be seen as being defined with the regulator and only the observables, such as cross sections, should be finite in the limit where you remove the regulator.

  • In renormalized perturbation theory one redefines the parameters of the theory, e.g. $e_0 = Z_e e_R$, where all the singular behaviour (when removing the regulator) is absorbed in these $Z$-factors, such that renormalized quantities, such as $e_R$ are finite. This procedure is of course not unique, in other words there are many different renormalization schemes that remove the divergences but give you different $e_R$. Keep in mind however that you know that certain quantities are independent of the renormalization scheme, namely the observables such as cross sections. This is easy to see, since the actual Lagrangian is the same, you just made a redefinition of the parameters. Keep in mind however that at finite order in perturbation theory such observables do depend on the scale, it is only independent when summing all orders. To get a prediction one should choose the scale such that the convergence of the series is as good as possible and this is usually acomplished by choosing the scale around the process energy.

  • Now the bare coupling is not an observable, and $e_R$ depends on the renormalization scheme. But nevertheless we can of course measure the electromagnetic coupling in experiments and in fact one can define a renormalization scheme in which the renormalized parameters, such as the electron mass $m_R$ and coupling $e_R$ correspond to the real physical values. Such a scheme is called a on-shell renormalization scheme. As an example consider the full QED vertex function $$ -ie_0 \Gamma(p,p') = -ie_0 \left[ \gamma^{\mu} \Gamma_1(q^2) + ... \right], $$ where $q^2 = (p'-p)^2$ is the momentum transfer, or the scale. The factor $\Gamma_1(q^2)$ is UV divergent. In an on-shell scheme we define $Z_e = [\Gamma_1(q_0^2)]^{-1}$ at some reference scale $q_0^2$. This removes the UV-divergence in $\Gamma_1$ but we still retain a dependence on $q_0^2$, i.e. $e_0 = [Z_1(q_0^2)]^{-1} e_R(q_0^2)$.

  • Now I tell you that in this on-shell renormalization scheme the renormalized coupling is the actual real coupling. But $e_R(q_0^2)$ still depends on $q_0^2$. But in fact it should be this way! The electromagnetic coupling is actually dependent on the momentum scale of the process from which it is measured. In particular the familiar result $$ \alpha(0) = \frac{e_R(0)^2}{4\pi} \approx \frac{1}{137} $$ is the coupling measured in a low energy process, or equivalently in non-relativistic setting. In contrast, at the scale of the $Z$-boson mass $\sim 90 GeV$ one measures $$ \alpha(90~GeV) \approx \frac{1}{127}. $$ This dependence is intuitively explained by the screening of charges particle-anitparticle pairs that are created from the vacuum. So to be precise the statement "the electromagnetic coupling is an observables" is technically wrong. More correctly one could say "the electromagnetic coupling at a certain energy scale is an observable".

  • For the mass it is a bit different. In the on-shell scheme we define $m_R = m_{\text{pole}}$, where $m_{\text{pole}}$ is the location of the pole of the propagator. This quantity is independent of the process energy and it can be identified with the electron rest mass in QED. In contrast with the coupling the rest mass does not depend on the energy of the process at which it is measured.

  • If you are not using an on-shell scheme, such as minimal subtraction, one can not identify the parameters in the Lagrangian with the real physical quantities. Again the coupling is scale dependent and now also the mass is scale dependent. Minimal subtraction is sometimes better since one has a better control over the convergence of the perturbative series. Furthermore in QCD the full quark propagators do not have poles at the "physical quark mass" (whatever that means), since in the real world free quarks do not exist. So the physical significance of the on-shell scheme is lost in QCD and in addition to being impractical.

  • I really recommend https://arxiv.org/abs/1901.06573. It provides a great overview over the topic.

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  • $\begingroup$ On your last point, this definition seems suitable, but as I understand, in other schemes $m_R\neq m_{pole}$. So, what would be the reason behind not defining the two to be equal? Furthermore, the mass of the electron that's "agreed upon" (in that it's the effective mass at a specific energy scale $s_0$, $m_{eff}(s_0)=m_R$) is $m_pole$ in the on-shell scheme, but how would other schemes agree on the same numerical value? Is it just that in their scheme, they just defined $m_{eff}'$ at some other energy scale $s_0'$ so that $m_{eff}'(s_0')=m_R'=m_R$? $\endgroup$ – TheQuantumMan Jan 12 at 17:30
  • $\begingroup$ @TheQuantumMan I made some edits to my post, since I was quite unprecise on the mass part. I hope this answers your question. $\endgroup$ – jkb1603 Jan 12 at 20:57

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