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I've been practicing with FRG techniques and I wanted to obtain the usual beta functions for Yukawa theory using the Wetterich equation. However, this has been more troublesome than I expected.

If I'm not mistaken, the Wetterich equation should read, for a Yukawa theory, as $$k\partial_k \Gamma_k[\phi, \psi, \bar{\psi}] = \frac{1}{2} \textrm{Tr}\left[\left(\frac{\delta^2 \Gamma_k}{\delta\phi \delta \phi} + R_{k,\phi}\right)^{-1}k\partial_k R_{k,\phi}\right] - \textrm{Tr}\left[\left(\frac{\delta^2 \Gamma_k}{\delta\psi \delta \bar{\psi}} + R_{k,\psi}\right)^{-1}k\partial_k R_{k,\psi}\right],$$ where $\phi$ is a scalar field and $\psi$ and $\bar{\psi}$ correspond to a Dirac fermion. I tried to consider the truncation, in Euclidean space, $$\Gamma_k[\phi, \psi, \bar{\psi}] = \int \frac{1}{2}(\partial_\mu \phi)^2 + \frac{m^2_k}{2}\phi^2 + \frac{\lambda_k}{4!}\phi^4 + i \bar{\psi} \gamma^\mu \partial_\mu \psi + i \bar{\psi} M_k \psi + i g_k \phi\bar{\psi}\psi \ \textrm{d}^d{x},$$ which I decided to try after taking a look at Eq. (5) of arXiv: 1308.5075 [hep-ph] (this paper is not central to the question, it is just the only reference I found working out a Yukawa interaction). For simplicity, I'm ignoring the running of the wavefunction normalizations $Z_{k,\phi}$ and $Z_{k,\psi}$ at this stage and setting them to $1$.

My problem is: when I take the functional derivatives of the EAA, I get $$\frac{\delta^2 \Gamma_k}{\delta\phi \delta \phi} = p^2 + m^2_k + \frac{\lambda_k}{2}\phi^2$$ and $$\frac{\delta^2 \Gamma_k}{\delta\psi \delta \bar{\psi}} = \gamma^\mu p_\mu + i M_k + ig_k\phi.$$

However, this means that $\psi$ and $\bar{\psi}$ are occurring nowhere on the Wetterich equation, and hence I'm not getting any running for the Yukawa coupling, for example, since that would be given in a term of $k\partial_k \Gamma_k$ with the form $\phi \psi \bar{\psi}$. I considered the possibility that my truncation is too violent and I should consider, e.g., a four-fermion term as well, but it seems to me that that would not recover the perturbative results, which can be calculated considering just the perturbatively renormalizable terms. What I am doing wrong?

I'm quite new to this formalism. Feel free to criticize any other mistakes you find in the previous expressions and to suggest relevant references.

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2 Answers 2

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In general, the Wetterich equation reads $$k \partial_k \Gamma_k = \frac{1}{2} {\rm Tr}\left[\left(\Gamma^{(2)}_k + R_k \right)^{-1} \cdot k \partial_k R_k \right],$$ where $\Gamma^{(2)}_k$ is a matrix of second-order derivatives of $\Gamma^{(2)}_k$ with respect to its arguments (here, $\phi$, $\psi$, and $\bar{\psi}$) and $R_k$ is the corresponding matrix of regulators. This is Eq. (4) in the paper you linked.

The original expression for the Wetterich equation missed an inverse on $\left(\Gamma^{(2)}_k + R_k \right)^{-1}$ before the edit. This inverse is important in general, because even if the regulator $R_k$ only couples to $\phi \phi$ and $\psi \bar{\psi}$ terms, the inverse involves all $9$ derivatives of $\Gamma_k$, not just $\delta^2 \Gamma_k/\delta \phi \delta \phi$ and $\delta^2 \Gamma_k/\delta \psi \delta \bar{\psi}$. This means that, in order to evaluate this inverse, in principle you will need to make your ansatz for $\Gamma_k$, evaluate all $9$ second derivatives, and then set $\phi$, $\psi$, and $\bar{\psi}$ to constant values, independent of space/momenta, so that you can actually evaluate the inverse. (The cited paper does not mention this point explicitly, but since they include a local potential $U_k(\phi^2/2)$ in their ansatz, they presumably take $\phi$ to be a scalar variable, and perhaps they set the $\psi$'s to be either zero or some other constant value). In this particular example, the inverse does simplify nicely when evaluated at zero fields (as shown for a simplified example below); the issue then comes from the fact that one needs to define the running fields in terms of derivatives of $\Gamma_k$ and differentiate the flow equation before evaluating at zero fields.

Note that the ansatz for $\Gamma_k$ that you have written didn't denote which quantities were assumde to run with $k$ (in the original post, before editing), so just to compare to the cited paper's ansatz (Eq. (5)), $$\Gamma_k[\phi,\psi,\bar{\psi}] = \int dx~\left[ Z_k (\partial_\mu \phi)^2 + U_k(\phi^2/2) + Z_{\psi,k} \bar{\psi} i \gamma^\mu \partial_\mu \psi + i h_k \phi \bar{\psi} \psi \right].$$ It appears that you wish to explicitly truncate the authors' $U_k(\phi^2/2)$ at second order in $\phi^2$ (around $\phi=0$) and introduce an $i \bar{\psi} M_k \psi$. You will need to define how each coefficient is extracted from $\Gamma_k$ in order to compute your flow equations. For example, by defining $m^2_k = \lim_{p \rightarrow 0}\frac{\delta^2 \Gamma_k}{\delta \phi(p) \delta \phi(q)}\Big|_{\phi(p) \rightarrow \phi,\psi \rightarrow 0,\bar{\psi} \rightarrow 0}$, $M_k = \lim_{p \rightarrow 0}\frac{1}{i} \frac{\delta^2 \Gamma_k}{\delta \bar{\psi}(p) \delta \psi(q)}\Big|_{\phi(p) \rightarrow \phi,\psi \rightarrow 0,\bar{\psi} \rightarrow 0}$, $Z_k = \lim_{p \rightarrow 0}\partial_{p^2}\left[\frac{\delta^2 \Gamma_k}{\delta \phi(p) \delta \phi(-p)}\Big|_{\phi(p) \rightarrow \phi,\psi \rightarrow 0,\bar{\psi} \rightarrow 0}\right]$, etc. This will involve taking derivatives of the flow equation with respect to the fields in order to get $k \partial_k m^2_k$, $k \partial_k M_k$, etc. In doing so, it is again important to treat the inverse $\left(\Gamma^{(2)}_k + R_k \right)^{-1}$ carefully, since this will generate non-trivial terms in the flow equations for these quantities (using the identity $\frac{\delta}{\delta \phi} A^{-1} = -A^{-1} \frac{\delta A}{\delta \phi} A^{-1}$). See, for example, this paper for an example application to $\phi^4$ theory.

Edit: When multiple fields are involved, as in this case, or in a non-equilibrium problem, the inverse is a matrix inverse in the fields (when converted to momentum space). I'll sketch it for the cited paper's ansatz simplifed to $U_k(\phi^2/2) = m_k^2 \phi(x)^2/2$. In momentum space (assuming no mistakes...) $$\begin{array}{c c}\Gamma_k &= \int dp_1 dp_2~\Big[(Z_k p^2 + m^2_k) \phi(p_1)\phi(p_2)\delta(p_1 + p_2) + Z_{\psi,k} i \bar{\psi}(p_1) \gamma^\mu (p_1)_\mu \psi(p_2) \delta(p_1+p_2)\Big] \\ & ~~~~~~~~ + \int dp_1 dp_2 dp_3~ i h_k \phi(p_1)\bar{\psi}(p_2) \psi(p_3) \delta(p_1 + p_2 + p_3)\end{array}.$$ (I am not very familiar with dealing with fermionic fields, so you'll want to double-check my math here).

Then, the matrix of derivatives is $$\Gamma^{(2)}_k = \begin{bmatrix} \frac{\delta \Gamma_k}{\delta \phi(p) \delta \phi(q)} & \frac{\delta \Gamma_k}{\delta \phi(p) \delta \psi(q)} & \frac{\delta \Gamma_k}{\delta \phi(p) \delta \bar{\psi}(q)} \\ \frac{\delta \Gamma_k}{\delta \psi(p) \delta \phi(q)} & \frac{\delta \Gamma_k}{\delta \psi(p) \delta \psi(q)} & \frac{\delta \Gamma_k}{\delta \psi(p) \delta \bar{\psi}(q)} \\ \frac{\delta \Gamma_k}{\delta \bar{\psi}(p) \delta \phi(q)} & \frac{\delta \Gamma_k}{\delta \bar{\psi}(p) \delta \psi(q)} & \frac{\delta \Gamma_k}{\delta \bar{\psi}(p) \delta \bar{\psi}(q)} \end{bmatrix} = \begin{bmatrix} (Z_k p^2 + m_k^2)\delta(p+q) & \int dp_2~ i h_k \bar{\psi}(p_2) \delta(p+q+p_2) & \int dp_3~ i h_k \psi(p_3) \delta(p+q+p_3) \\ \int dp_2~ i h_k \bar{\psi}(p_2) \delta(p+q+p_2) & 0 & Z_{k,\psi} i \gamma^\mu p_\mu \delta(p+q) + \int dp_1~ i h_k \phi(p_1) \delta(p+q+p_1) \\ \int dp_3~ i h_k \psi(p_3) \delta(p+q+p_3) & Z_{k,\psi} i \gamma^\mu p_\mu \delta(p+q) + \int dp_1~ i h_k \phi(p_1) \delta(p+q+p_1) & 0 \end{bmatrix}$$

Then, one would add to this the matrix $$R_k = \begin{bmatrix} R_{\phi,k} & 0 & 0 \\ 0 & 0 & R_{\psi,k} \\ 0 & R_{\psi,k} & 0 \end{bmatrix}\delta(p+q)$$

To calculate the inverse in practice, we need to choose functions $\phi(p)$, $\psi(p)$, and $\bar{\psi}(p)$ at which we wish to evaluate the inverse, since we cannot do it for arbitrary functions. Typically, one chooses momentum-independent values. If we had kept the full $\phi$ dependence of $U_k(\phi^2/2)$, we would typically set $\phi(p) = \phi$ (a scalar), but since we've truncated it, we could set all of $\phi(p) = \psi(p) = \bar{\psi}(p) = 0$, giving $$\Gamma^{(2)}_k + R_k = \begin{bmatrix} (Z_k p^2 + m_k^2) + R_{\phi,k} & 0 & 0 \\ 0 & 0 & Z_{k,\psi} i \gamma^\mu p_\mu + R_{\psi,k} \\ 0 & Z_{k,\psi} i \gamma^\mu p_\mu + R_{\psi,k} & 0 \end{bmatrix}\delta(p+q),$$ which can now be evaluated as a matrix inverse (the inverse of the momentum-dependence here is just another $\delta(p+q)$). For this particular choice, the inverse does end up just being 1 over the non-zero elements. If we had chosen constant values $\phi = \phi_0$, $\psi = \psi_0$, and $\bar{\psi} = \bar{\psi}_0$, the inverse would not be as trivial.

Now, for the choice $\phi(p) = \psi(p) = \bar{\psi}(p) = 0$, you'll notice that the dependence on $h_k$ dropped out when we set the fields to $0$. That does not mean $h_k$ would not flow in this model because, as I mentioned above, one needs to define $h_k$ in terms of derivatives of $\Gamma_k$, e.g., $h_k = \frac{1}{i}\frac{\delta^3 \Gamma_k}{\delta \phi(p_1) \delta \bar{\psi}(p_2) \delta \psi(p_3)}\Big|_{\phi = \bar{\psi} =\psi = 0}$, and then differentiate the flow equation with respect to these fields before setting $\phi = \bar{\psi} =\psi = 0$. The result should be a non-trivial flow equation for $h_k$ (coupled to other running parameters).

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  • $\begingroup$ I forgot to put in the inverse, but this one was just a typo. I was confident in the different expression I wrote due to Eqs. (1.1) and (2.12) on this paper, but I might have mistakenly assumed I could just add the RHS sides to obtain the full expression. Anyway, I think what I got wrong was that I was treating the inverses wrongly (I wasn't using the 9 derivatives, only the two I wrote down). Do you mind sketching how one would compute the inverses? I was just dividing by those expressions when getting to the trace integral in momentum space, + $\endgroup$ Commented Nov 18, 2021 at 4:25
  • $\begingroup$ since that procedure worked fine when there was a single field under consideration. It appears clear to me now that those other terms should enter as well, but I still can't see how that happens $\endgroup$ Commented Nov 18, 2021 at 4:29
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    $\begingroup$ The examples in Litim's paper appear to correspond to two separate cases, one with just bosonic fields and one with just fermionic fields, not a case with both types of fields in one action, which introduces cross terms. I will try to edit in a quick example of computing the inverse. $\endgroup$
    – bbrink
    Commented Nov 18, 2021 at 4:37
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    $\begingroup$ A follow up: as shown in the example inverse in my edited post, for this particular case the inverse simplifies nicely when evaluated at 0 fields. The $h_k$ dependence drops out of this inverse, but you will still get a non-trivial flow equation for $h_k$ because you'll need to take derivatives of the flow equation before evaluating at 0 fields, as mentioned at the end of the answer. $\endgroup$
    – bbrink
    Commented Nov 18, 2021 at 5:37
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For what it's worth, the Wetterich equation is an exact renormalization group equation (ERGE), which means that $\Gamma_k$ in principle contains all possible terms not forbidden by symmetry.

Even if we claim they are absent in the UV they get created during the RG flow of integrating out modes.

In particular there are also IR-irrelevant$^1$ terms with coupling constants of negative mass-dimension, such as e.g. the 4-fermion term that OP mentions.

To make contact with experiments, one have to impose appropriate renormalization conditions.

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$^1$ IR-irrelevant terms are usually called irrelevant terms. They are UV-relevant, i.e. they correspond to non-renormalizable terms in the Dyson sense.

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  • $\begingroup$ Notes for later: Replace mass dimension with inverse length dimension. $\endgroup$
    – Qmechanic
    Commented Mar 6, 2022 at 15:55

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