New answers tagged

3

Actually... There is some evidence of CFTs in $d>6$. In [1] they construct a solution in AdS$_8$ implies the existence of a CFT in $d=7$. This is not a definitive answer because there are still some issues about the solution. One has to prove full nonperturbative stability and also there is a region in spacetime where the coupling becomes big and one has ...


4

I would say that the claim that there are no non-trivial CFTs in $d>6$ is just a speculation for which there isn't much evidence. The belief is that above six spacetime dimensions, the only unitary CFTs are simply free theories and that all non-trivial fixed points can be described by mean field theory. In addition to what you said, that there are fewer ...


5

Mass renormalization (in QED) does not involve virtual electron-positron pairs. It involves the emission and reabsorption of a virtual photon by an electron in motion. This is a quantum-mechanical interaction between the electron and the electron’s own electromagnetic field. The intuitive idea of mass renormalization actually can be explained without ...


2

It is very good and deep question. Let me try to answer and clarify some subtleties. In QED we have: $$ e^2_{phys}(\Lambda) = e_0^2\left(1-\frac{e_0^2}{6\pi^2}\ln \frac{\Lambda}{m_0}+\dots\right) = \frac{e^2_0}{1+\frac{e_0^2}{6\pi^2}\ln \frac{\Lambda}{m_0}} $$ $$ m_{run}(\Lambda) = m_0\left(1+ \frac{3e_0^2}{8\pi^2}\ln \frac{\Lambda}{m_0}+\dots\right) = m_0 \...


0

All theories are incomplete, but it does not mean they all need renormalizations. See, for example, this explanation here.


1

One way to understand the concept of a UV completion is in terms of fixed points in the space theories generated by the renormalization group. The RG flow, interpreted as a course-graining procedure, generates a vector field that is determined by the criterion that as we move along the vector field flow, we obtain theories that better and better describe the ...


0

Let us start from QED. In this theory, vacuum polarization in the lowest order is given by one-fermion loop correction to bare photon propagator (note that photon has zero mass). It is quite straightforward (or I can provide a derivation) that this correction modifies charge but not mass. For me, physical ground becomes clear after calculation. If you ...


1

The expression for $\lambda$ in terms of $\lambda_R$ is just perturbation theory. Given an expression for $\lambda_R$ in terms of $\lambda$ $$ \lambda_R = \lambda + \frac{\lambda^2}{32\pi^2}\ln\frac{s_0}{\Lambda^2}+... $$ we solve perturbatively, assuming that $\lambda$ can be written as a power expansion in $\lambda_R$. That is, we guess a solution of the ...


2

A more explicit way of writing this would be $$e^{-H'(s')}=\mathrm{Tr}_s \prod_{q=0}^{N'} T(s'_q;s^{B_q}_1,s^{B_q}_2,\dots s_9^{B_q})e^{-H(s)} $$ where $B_q$ denotes the blocks, $s_k^{B_q}$ is the $k$-th spin in the $q$-th block and $N'=N/9$ is the number of blocks (or of new spins)


1

That part follows from the condition that the residue at the pole should be $-i$ (or one, depends if you include the $i$ in the numerator of the propagator). In fact it's easy to see that defining $$\Pi(q^2) = \Pi_2(q^2)-\Pi_2(0)$$ one directly gets that, at the pole $$q^2=0 \to \Pi(0)=0$$ and so that the residue at the pole is $$\underset{q^2=0}{\text{Res}}\...


2

It looks like $T(s'; s_i)$ is shorthand for $T(s';s_1,s_2,s_3,s_4,s_5,s_6,s_7,s_8,s_9)$. This sort of abbreviation is pretty common.


1

(i) "The only tensors that can appear in $\Pi_{\mu\nu}$ are $g_{\mu \nu}$ and $p_\mu p_\nu$": this is because these are the only rank-2 tensors available in the problem. $p_\mu$ is a vector on which the problem depends and $g_{\mu \nu}$ is the invariant tensor. Using only this two tensors (one rank-1 and one rank-2), you cannot construct any rank-2 ...


6

As @knzhou noted, we first Wick-rotate so $\color{blue}{k}\in\Bbb R^4$ is Euclidean. Then $\int_{\Bbb R^4}f(\color{blue}{k}^2)d^4\color{blue}{k}=2\pi^2\int_0^\infty f(\color{red}{k}^2)\color{red}{k}^3d\color{red}{k}$, where $\color{red}{k}\in[0,\,\infty)$ is the radius of $\color{blue}{k}$, and the proportionality constant is the solid angle in $4$ ...


5

In general, such cutoffs only have meaning when you Wick rotate to Euclidean signature. In that context, $k$ is a vector in $\mathbb{R}^4$, and the condition is that its magnitude is less than $\Lambda$.


0

Recall that $c=26$ is needed to guarantee the vanishing of the string theory $\beta$ function at linear order in $\alpha{´}$, then if you start with the the construction of a wolrdsheet CFT with $c \neq 26$ then the trace of the energy momomentum tensor no longer vanishes and one should derive the dependence of the new effective action on the zero modes of ...


2

on the one hand $$ T(s_1 s_2) = T(s+\delta s) = T(s) + \delta s \frac{dT}{ds'}\bigg|_{s=s'} + O(\delta^2)$$ while on the other hand $$ T(s_1 s_2) = T(s) T(1+\delta) = T(s) \left[ 1+\delta \frac{dT}{ds}\bigg|_{s=1}+O(\delta^2)\right]$$ where we used $T(1)=1$. Comparing we get the equation $$ s \frac{dT}{ds} = T \frac{dT}{ds}\bigg|_{s=1}$$ which can be solved ...


1

Here's my shot at an explanation—would love anyone's feedback. In the above, I argued that since we normalized the fields to create a single particle state in the free theory (and satisfy the canonical commutation relations), we should expect them to do so in the interacting theory as well. As far as I can tell, this is just wrong. It assumes that the ...


0

One can always write down a mean field solution to a field theory. The question is whether this solution is locally/perturbatively stable. Its analogous to local minima of the free energy. If the upper critical dimension ($d_c$) of a field theory can be correctly identified, then for $d > d_c$, generally, mean-field solutions tend to be stable due to the ...


0

The argument is easier to follow from the continuous RG point of view (a la Wilson), though it can be adapted in the block spin picture. I'll take the former point of view here. The RG procedure gives rise to a flow of the Hamiltonian, $H_s$, with $H_{s_0}$ the initial Hamiltonian, and $\partial_s H_s = R(H_s)$ (to connect with the OP notations $b=e^{s}$). ...


0

The scaling relation must hold for arbitrary scaling factors $b$ and thus it must also hold for the specific choice $$b = |t|^{-\frac{1}{y_{T}}} .$$ Inserting this into the scaling relation you get the above form. I would say this is somewhat similar to parametrising a function of two arguments $f(x,y)$ via setting $y=y(x)$ and effectively getting a function ...


1

Let us first remark that the RG equation that you quoted, \begin{align} \left[-p \frac{\partial}{\partial p}+\beta(\lambda) \frac{\partial}{\partial \lambda}+\left(\gamma_{m}(\lambda)-1\right) m \frac{\partial}{\partial m}+d_{n}-n \gamma_{d}(\lambda)\right] \tilde{\Gamma}^{(n)}(p ; m, \lambda, \mu)=0, \end{align} is a linear partial differential equation for ...


1

While saying that the temperature and the magnetic field are relevant parameters is standard, it is somewhat sloppy, and confusing at first. What is really meant is that both the temperature (more precisely $\beta J$ for the Ising model) and the magnetic field (or $\beta h$ for the Ising model) have non-trivial projections onto the two distinct relevant ...


0

Well, dimensional regularization has issues with the representation theory for spinors, Clifford algebras, gamma matrices, gamma-five & the Levi-Civita symbol, cf. e.g. this & this Phys.SE posts and links therein.


0

You cannot have true renormalization group (RG) fixed points in finite systems. You need infinitely many degrees of freedom in order to restore the system to its original scale---as you identify in your example, you would be missing terms in a finite system. Think of it as a kind of reverse Hilbert Hotel problem: you have a hotel with infinitely many guests, ...


1

The natural answer to your question is provided by lattice QCD. Like the classical action of QCD (with massless quarks) the lattice (Wilson) action has no dimensionful parameters, only a dimensionless coupling constant $g$. Masses of hadrons are extracted from the exponential decay of correlation functions, so they are expressed in units of the inverse ...


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