#### [SOLVED] Counterterm Lagrangian and Renormalisation?

I am going through the notes on QFT by M. Srednicki (online here), and I am having a hard time to understand the renormalised Lagrangian.

Consider a Klein-Gordon scalar field, with an interaction term e.g. $V(\phi)\propto \phi^3$. When proving the LSZ reduction formula, Srednicki argues that in order for it to make sense, the field must obey the following relations:

1. Let $|\Omega\rangle$ be the exact ground state. Assuming that $a(p)|\Omega\rangle=0$, we must have $\langle \Omega |\phi(x)|\Omega\rangle=0$. To this end, we redefine $\phi\to \phi+\text{const.}$ (shift the field).

2. Let $|p\rangle=a^\dagger(p)|\Omega\rangle$ be a one-particle state. To ensure a correct normalisation of such a state, we must have $\langle p|\phi(x)|\Omega\rangle=\exp(-ikx)$. Just like before, we redefine $\phi\to \text{const.}\ \phi(x)$ (rescale the field).

After we redefine the field, we end up with something like $\phi(x)\to A\phi(x)+B$. Srednicki states that the Lagrangian becomes something like $L=Z_1 (\partial \phi)^2+Z_2 m^2 \phi^2+Z_3 g \phi^3+Z_4 \phi$.

First question: we had two constrains, so how do we end up with four renormalisation constants $Z_i,\ i=1,2,3,4$. That is to say, we should have $Z_i=Z_i(A,B)$, where $A,B$ are the aforementioned "shifting" and "rescaling" constant $\phi(x)\to A\phi(x)+B$. As there are two constants, there should be two (linear) relationships between the four $Z_i$. Is this right? Is this even important at all? Why is this never discussed?

Second question: next we study the dynamics through, say, a Path integral. As usual, we define

$$Z[J]=\int D\phi\ \exp\left[i\int \mathrm d x\ L_0+L_1+J\phi\right]$$

where $L_0$ is the free Lagrangian and $L_1$ is "everything else":

\begin{aligned} L_0&=(\partial \phi)^2+m^2\phi^2\\ L_1&=Z_3 g\phi^3+(Z_1-1)(\partial \phi)^2+(Z_2-1)m^2\phi^2+Z_4 \phi \end{aligned}

We get the usual $Z_0[J]$ in terms of the propagator, and treat everything else as perturbations:

$$Z[J]\propto \exp\left[i\int \mathrm dx\ L_1\left(\frac{\delta}{\delta J(x)}\right)\right] Z_0[J]$$

I understand the necessity of treating the cubic and linear terms as perturbations, but why don't we treat the $(Z_1-1)(\partial \phi^2)$ and ($Z_2-1)m^2\phi^2$ terms exactly? These terms are quadratic in the fields, so they can be included in the propagator, simply by following the usual steps while taking into account these multiplying constants.

I know that we usually use the counterterms to "absorb" infinities, but this feels a bit like cheating: we can solve a problem exactly, but we don't because we know that we will soon need some degrees of freedom to avoid troubles... There probably is something that I am not getting right, and I would really appreciate if any of you can lead my thoughts in the right direction.

Finally, there is a technicality that is annoying me a fair bit. The field is an operator, so when dealing with the exponential in the path integral, we should be careful, as in general $\exp(A+B)\neq\exp(A)\exp(B)$, we should use Baker-Campbell-Hausdorff instead. This means that, in the path integral, we should not write

$$\exp\left[i \int\mathrm dx\ L\right]=\exp\left[i\int \mathrm d\ x L_1\left(\frac{\delta}{\delta J(x)}\right)\right] Z_0[J]$$

because $\exp[S_0+S_1]\neq\exp[S_0]\exp[S_1]$. Anyway, as both $\phi$ and its momentum commute with their commutator, BCH should be fairly easy to implement, as we would only need the first correction $\exp(A+B)=\exp(A)\exp(B)\exp(-\frac{1}{2}[A,B])\exp(...)$

I would like to apologise if my notation is sloppy (obviously, I'd be glad to make it more precise if asked to). Also, there might be some constants missing (as the $\frac{1}{2}$ factors in the Lagrangian), but this is not really important here.

#### @TotallyRhombus 2015-07-15 23:31:37

First I'll address the last question you mentioned regarding commutativity in the path integral. One benefit of the path integral formulation is that the integrand behaves like an ordinary complex or Grassman number---if you're computing the trace of a product of matrices in Einstein notation, you can rearrange terms in the product as long as you keep track of index contractions and whether the numbers anticommute. In a sense, the operator character is encoded in the measure $\mathscr{D}\phi$ on the space of fields (the range of indices that you sum over).

Now your second question regarding counterterms. The reason quadratic counterterms are not included in the propagator ultimately has to do with the process of renormalization, which I will review briefly for clarity. In free theory, the propagator coincides with the two-point correlation function. In an interacting theory, we keep the propagator of the 'closest' free theory, and the two-point correlation function has perturbative corrections. The question is, what small parameter do we expand in when computing the perturbative corrections? Ideally, this parameter will be a small dimensionless ratio of two different energy scales. However, these physical quantities do not appear immediately as parameters in our model ('bare' coupling constants, masses, field normalization, etc.). We need to relate the model parameters to a small number of measured/fixed physical quantities first, like pole masses and known scattering amplitudes, and then use perturbation theory to relate the initial set of reference quantities to other `nearby' quantities (like scattering cross sections at energy scales comparable to the reference energy).

After you relate your model parameters $\mathbf m$ to some reference quantities $\mathbf x$ (e.g. fixed scattering amplitudes and pole masses), you can make new predictions $\mathbf x'(\mathbf x)$ once you measure $\mathbf x$ experimentally. The relationship between the reference amplitude $g_R$ and $g$ starts at linear order in $g$: $g_R=g+\mathcal{O}(g^2)$. After inverting this order-by-order to obtain $g(g_R)$, any new physical prediction $\mathbf x'$ will be expressed as a power series (or asymptotic series) in $g_R$. In particular, the two point correlation function will equal the propagator plus $\mathcal{O}(g_R)=\mathcal{O}(g)$. Of course the exact 2-point function has higher order corrections that in general can change the locations of poles, and hence masses, but the zeroth order term is always just the propagator, the 2-point function of free theory. The terms $(Z_i-1)$ keep track of contributions that are only important at order $g_R$ (equivalently order $g$) and higher, and hence do not appear in the propagator. [In other words, you can think of $Z_i$ as formal power series in $g_R$, with 0th order term equal to 1, where the important information is in the form of the coefficients of the expansion, not in any function that could be associated with the series.]

Finally, your first question. Constraint 2 on the normalization of single-particle states is subtle, because you must ensure that the masses of particles are fixed as well (your density of states remains on the same momentum-space hyperboloid). Imposing this constraint requires the $Z_2$ parameter. The $Z_4$ and $Z_3$ parameters are constrained by $\langle\phi\rangle_\Omega=0$, and the additional parameter can be absorbed into the interaction strength $g$. Typically $g$ is constrained by measuring a scattering amplitude.

#### @AccidentalFourierTransform 2015-07-16 13:07:30

Path integral: I must say that I had the feeling that in the path integral, the field was (somehow) different as when canonically quantizing. Nevertheless, it is still not to be thought of as a function, but rather a distribution, right? Some subtleties remain after all :P

#### @AccidentalFourierTransform 2015-07-16 13:08:01

The other answers: yay! you made it really clear for me now. Thankyou very much!!!