7

The Quick Answer The big yellow book (namely Di Francesco et al) that the OP quotes, largely obscures the distinction between what I call (b) and (c) below. If the OP is just interested in deriving the result in the fastest way he can Taylor-expand in $\delta$ the quantities $w(z+\delta/2)$, etc., and take the limit $\delta\rightarrow 0$. E.g., $$ w(z+\...


7

When both "space" ($x$) and "time" ($y$) directions are periodic, the Laplacian on torus with coodinate $z=x+iy$ has a normalized zero mode $$ \varphi_0(z) = \frac 1 {\sqrt{{\rm Im}(\tau)}} $$ (Here $\tau$ is the modular parameter defining the torus.) As $$ -\nabla^2 \varphi_0=0. $$ the zero mode means that the Laplace operator is not 1-1 and so prevents ...


5

The representations that satisfy the usual unitarity bounds are the only representations that Are irreps of the Lorentzian conformal group, which is the universal cover $\widetilde{SO}(d,2)$ of $SO(d,2)$. Are unitary Have positive spectrum of energies. Note that the Hamiltonian $P^0$ is a part of the conformal algebra and hence its spectrum is fixed by ...


5

This OPE is not trivial, it just doesn't have singular terms. Suppose there is a quasi-primary operator $\mathcal{O}$ with weight $(h,\bar h)$ appearing in the right-hand side. We can compute the coefficient with which it appears by looking at three point function $$ \langle T(z)\bar T(\bar w)\mathcal{O}(x,\bar x)\rangle=\frac{f_{T\bar T\mathcal{O}}}{(z-w)^{...


5

In my lectures the space of states is not assumed to be a Hilbert space, i.e. to have a positive definite scalar product. Actually it is not assumed to have a scalar product at all. So you cannot write $\langle v|$ and the short answer to your question is no. Actually you do not need a scalar product for computing OPEs or anything else. Nevertheless, you ...


5

I understand that any quantum lattice model at the critical point which can be described by a massless relativistic quantum field theory has emergent conformal invariance. This is not always true. For example, free quantum electrodynamics in $(2+1)$d is scale invariant and relativistic but not conformal. However, there is a proof that in $(1+1)$ dimensions, ...


4

Using momentum conservation and onshell conditions, $$ p_1+p_2+p_3=0, \qquad p_1^2=p_2^2=p_3^2=\frac{4}{\alpha'}, $$ it follows that for any $i\neq j\neq k$, \begin{equation} \begin{aligned} 2p_i\cdot p_j &= (p_i+p_j)^2-p_i^2-p_j^2 \\ &=p_k^2-p_i^2-p_j^2\\ &=-\frac{4}{\alpha'}, \end{aligned} \end{equation} so this relation is really just ...


4

If you understand the key content of the Renormalization Group theory RGT), there is no need they should have the same Lagrangian, (well, in a context of Statistical Mechanics it would be more appropriate to speak in term of Hamiltonian, but this is a side remark). The explanation of the existence of universality classes provided by RGT is based on the fact ...


4

Re second question: we can formulate quantum mechanics in coordinate representation or equivalently in momentum representation. So do we live in position space or in momentum space? The answer is that it doesn't matter. Some questions are easier to answer in position space, some in momentum space. Everyday life is more easily addressed in position space. ...


4

In these types of problems, it is very useful to work on a generic spacetime with metric $g$. For instance, we can promote the usual free-field Lagrangian $$S=\int\mathrm{d}^dx\,\mathcal{L}=\int\mathrm{d}^dx\,\partial_{\mu}\phi\,\partial^{\mu}\phi$$ to one on a general curved spacetime as $$S=\int\mathrm{d}^dx\,\mathcal{L}=\int\mathrm{d}^dx\,\sqrt{g}\,g^{\...


4

The $G$-WZW model depends not only on the group $G$, but also on a number $k$ called the level. The symmetry algebra is an affine Lie algebra, and it also depends on $k$. Both $SU(2)$ and $SO(3)$ have the same affine Lie algebra, and the central charge is $$ c = \frac{3k}{k+2} $$ where $k\in \mathbb{N}$ for $SU(2)$ and $k\in 2\mathbb{N}$ for $SO(3)$. It ...


4

The critical dimension of the superstring is $d=10$, fixed by the conformal anomaly. The total central charge for the world sheet conformal theory is therefore \begin{equation} c=d\left(1_\text{scalar}+\frac{1}{2}_\text{fermion}\right)=10 \times\frac{3}{2}=15. \end{equation} From the spacetime point of view we want to compactify on a manifold \begin{equation}...


4

The four-dimensional $\mathcal{N}=4$ SYM contains scalars, fermions and gauge fields and under gauge transformations, the fields and their covariant derivatives transform in a covariant way. Gauge-invariant operators are obtained by taking the trace of a product of such covariant fields at the same spacetime point, i.e $$\begin{align} \mathcal{O}(x) = \text{...


4

The constant term is needed because on a compact manifold with periodic boundary conditions there is a zero mode in the spectrum of the laplacian. This is easier to see on the circle, where $\frac{d^2}{dt^2}f = \lambda f$ has the periodic constant solution $f(t) =c$. This makes the operator not-inveritble. One should instead work in the space orthogonal to ...


3

You are very close to discover the so-called ''lightcone bootstrap'' or ''analytic bootstrap.'' There is a lot of literature about it, I'll post some refs. at the bottom. I'll first rewrite your idea in a more precise way. Let us consider two operators $O_1$ and $O_2$ in a $d$ dimensional conformal field theory. In particular let us write down their four ...


3

OK I think I know what is going on. It's all about primes. Consider an active spacetime transformation: $$ x^{\mu} \mapsto x'^{\mu}(x)$$ $$g_{\mu \nu} (x) \mapsto g'_{\mu \nu} (x') = g_{\alpha \beta} (x) \frac{\partial x^{\alpha}}{\partial x'^{\mu}} \frac{\partial x^{\beta}}{\partial x'^{\nu}}$$. (the transformation of the metric tensor follows ...


3

The stress tensor is not unique, whether it's defined using Noether's theorem or defined by varying the action with respect to the metric. The non-uniqueness of the Noether-derived version is familiar. This answer addresses the non-uniqueness of the other version. The stress tensor $T^{ab}$ is defined by $$ \delta S\propto \int d^Dx\ \sqrt{g}\,T^{ab}(x)\...


3

Probably you are already familiar with the reviews of conformal bootstrap, but this is the standard list David Simmons Duffin's TASI lectures 1602.07982 Slava Rychkov's EPFL lectures 1601.05000 Recent numerical bootstrap review 1805.04405 These links don't discuss the inversion formula (except sec IX of 3. very briefly), but the notations and all basic ...


3

Let me redo the calculation, while explicitly writing OPE coefficients. Let $(A_n(z))_n$ be a basis of operators at $z$. We use the two OPEs $$ Y(y)Z(z) = \sum_n c_n(y,z) A_n(z) $$ and $$ X(x)A_n(z) = \sum_m d_{m,n}(x,z) A_m(z) $$ We end up with the result $$ X(x)Y(y)Z(z) = \sum_{m,n} c_n(y,z)d_{m,n}(x,z) A_m(z) $$ where the coefficient of $A_m(z)$ is $\...


3

There's no need to go crazy: you just have to work out $$\frac{\partial}{\partial x^\nu} \frac{x^\mu}{|x|^2} = \frac{1}{|x|^4} \left[ x^\mu \frac{\partial |x|^2}{\partial x^\nu} - |x|^2 \frac{\partial x^\mu}{\partial x^\nu}\right] = \ldots$$ This is a concrete $4 \times 4$ matrix, and there's nothing strange about computing its determinant.


3

TL;DR: The conclusions surrounding eq. (2.19) are not true in general. First of all, Ref. 1 makes it clear on top of p. 19 that $T^{\mu\nu}$ denotes the metric/Hilbert stress-energy-momentum (SEM) tensor, not some other SEM tensor. Now it is true that if $\epsilon=\epsilon^{\mu}\partial_{\mu}$ is a Killing vector field (KVF), then the current (2.19) ...


3

When you expand the normal ordered term, you have \begin{align} :\partial \phi(z) \partial \phi(0): &= :[ \partial \phi(0) + z \partial^2 \phi(0) ]\partial \phi(0): \\ &= : \partial \phi(0) \partial \phi(0): + z : \partial^2 \phi(0) \partial \phi(0): \\ &= : \partial \phi(0) \partial \phi(0): + \frac{z}{2} \partial \left( :\partial \...


3

Indeed as said in the comments, when people say "the free boson has central charge $1$" they mean, "the theory with just one free boson has central charge $1$." It is an example of "synecdoche" I believe... Anyway, the reason why we consider $\partial\phi$ as a field rather than $\phi$ is because, as you noticed, $\phi$ has conformal dimension $h=0$, so its ...


3

For starters, several conformal transformations, e.g. the special conformal transformations, take finite points $p\in M$ to $\infty\notin M$, which technically violates OP's suggested definition. For Euclidean space $\mathbb{R}^n$, the conformal compactification $\overline{\mathbb{R}^n}$ is simply the real projective space $\mathbb{P}_n(\mathbb{R})$, which ...


3

Of course, the free energy on the cylinder is not a measurable observable if you're given the theory on the infinite plane. But one can measure other observables which are proportional to the central charge, such as the two-point function of the stress-energy tensor. There are situations where that expression is an observable. If you have a one-dimensional ...


3

The character of a representation of the Virasoro algebra depends on the whole structure of the representation, not just of the conformal dimension $h$ of the highest-weight state. For a Verma module, the character is simply $$ \chi_{c,h}^\text{Verma}(\tau) = \frac{q^{h+\frac{1-c}{24}}}{\eta(\tau)}$$ as you wrote. However, for generic $c$ there are two ...


2

The path integral measure is not one-loop exact in general. In the example of the anomaly for the chiral symmetry, the path integral measure for the fermions are one-loop exact because the integral is gaussian on the fermions $\psi$ and $\bar\psi$ (with the gauge field $A_{\mu}$ fixed, as well as any scalar in the Yukawa coupling). The anomaly comes from the ...


2

For 3-manifolds which are circle fibrations of 2-manifolds, this follows from the fact that any such partition function may be computed as trace of an operator on the Hilbert space of the 2-manifold. More generally however, we need to understand how to cut up our 3-manifolds and compute the partition function of the pieces. Using a Heegaard splitting for ...


2

In QFT, the reduced density matrix can be written as, \begin{equation} \rho = \frac{e^{-\beta H}}{\mathrm{Tr} (e^{-\beta H})} \end{equation} where $H$ is the modular Hamiltonian often used in QFT literature while it is called entanglement Hamiltonian in the condensed matter theory literature. See the first few minutes of this talk (https://www....


2

I don't think it's the case that all $n>0$ terms vanish, because the mode expansion of $\phi$ has a zero mode $\phi_0$. Its expansion is \begin{equation}\phi \left(z,\bar{z}\right) = \phi_0 - i\pi_0 \log\left(z\bar{z}\right) +i \sum_{n\neq 0} \frac{1}{n} \left(a_n z^{-n} + \bar{a}_n \bar{z}^{-n}\right)\end{equation} Computing $\langle:\phi^n:\rangle$ ...


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