I started reading about Conformal Field Theory a few weeks ago. I'm from a more mathematical background. I do know Quantum Mechanics/Classical Mechanics, but I'm not really an expert when it comes to Quantum Field Theory.

So while reading through several papers, some questions appeared and I would be more than happy if someone could answer them:

  1. First I'm refering to http://www.phys.ethz.ch/~mrg/CFT.pdf (which is a german document) Chapter 3.2.1 (starting at page 29). The main topic here is the Free Boson as an example of a CFT. In this chapter he tries to proof the conformal invariance of the quantized theory. He's doing this by constructing the generators of conformal symmetry. Now here comes the first question: Why does this show conformal invariance? In the classical case you show conformal invariance by showing that the integrand of the action functional doesn't change under a conformal group action (which seems reasonable). But why does constructing generators of conformal symmetry imply conformal invariance in the quantum case?

Next I'm refering to equation (3.2.26) in the same chapter. Here he states that the equation (3.2.26) $[L_{m},\phi]=z^{m+1}\partial_{z}\phi(z,\bar{z})$ proofs that the operators L_{m} actually do implement conformal transformations of the type $L_{n}=-z^{n+1} \partial_{z}$. Why is that so? Why does this proof that the L_{m} actually implement these conformal transformations $L_{n}=-z^{n+1} \partial_{z}$? Or better: What does he mean by "implementing a transformation"? (What's the defintion if you want). The equation $[L_{m},\phi]=z^{m+1}\partial_{z}\phi(z,\bar{z})$ kind of looks like an eigenvalue-equation (which of course it isnt). But if you think that there's no Lie-Bracket. It looks like $z^{m+1}\partial_{z}$ being an "eigenvalue" of $L_{m}$ of the eigenvector $\phi$.

  1. I'm now refering to the paper of David Tong: http://www.damtp.cam.ac.uk/user/tong/string/four.pdf (which is english). In Chapter 4.5.1 (on page 92) he adresses radial quantization. So he has a Theory living on a cylinder and maps this theory with the exponential map to the complex plane. So first of all, this seems a bit restrictive to me. I mean why should we only consider theories living on a cylinder as the general case? What I've heard is that that since you have a finite spatial direction on the cylinder you avoid singularities. But since I don't know Quantum Field Theory well enough I'm not so sure about this. N*ext it's not so obvious where the actual quantization appears in this process.* For me quantization means imposing certain commutation relations on operators/fields. And I don't see where this is happening here? Maybe it happens implicitly when you transform the energy momentum tensor from the cylinder to the plane by using the Schwartzian derivative??

I'd really be more than happy if someone can provide me with some detailed explanations. (because I'm really not an expert yet in this subject)

Thanks in advance!!


3 Answers 3


concerning the first question, related to the German text by Matthias Gaberdiel (greetings to him):

Closed algebra implies symmetry

It is enough to construct the generators of an algebra - in this case, conformal algebra - and calculate their commutators $[L_m,L_n]$ and so on. If the commutators are linear combinations of other generators, we say that the generators form a closed algebra. Now, you are right that we also want to use some "dynamical information about the theory". You wrote that the action should be invariant under the transformations that these generators generate - and you worry that the action - all the dynamical information about the theory - has been completely removed from the proof, right?

That's a good point but the dynamical information hasn't been removed because a particular generator (or, for a general basis of generators, a linear combination of generators) is the Hamiltonian that determines the dynamics itself. For conformal symmetry, it is $L_0+\tilde L_0$ that plays the role of the Hamiltonian. It generates translations of the cylinder or, equivalently (as we will discuss below), multiplicative translations in the radial coordinate. (Maybe additive shifts such as $c/24$ should be added on one of the backgrounds.)

Because $L_0+\tilde L_0$ is a linear combination of some generators and you can show that the set of generators is closed under the operation of taking the commutator, it proves that the whole algebra generated by this set of generators is a dynamical symmetry. In this case, the commutator $[L_m, L_0+\tilde L_0]$ is not strictly zero - so the generators $L_m$ don't commute with the Hamiltonian. Instead, the commutator is equal to another combination of the symmetry generators. But we still say that $L_m$ is a symmetry of the system and we know this situation from other contexts as well.

For example, in special relativity, the angular momentum $J_{12}=J_z$ commutes with the energy $p_0$. However, the Lorentz boost generator $J_{03}$ doesn't commute with the Hamiltonian $p_0$: their commutator is proportional to $p_3=p_z$, a component of the momentum. It's nonzero but it's another symmetry generator. It's normal for symmetry generators that act nontrivially on time - such as the $J_{03}$ boost generator in relativity or $L_m$ in conformal symmetry - to have nonzero commutators with the Hamiltonian $p_0$ or $L_0+\tilde L_0$, respectively. What's important is that the commutator is another operator we know to be a generator of a symmetry, and the symmetry algebra is fully described by group theory - by the structure constants $f$ in $[L_m,L_n]=f_{mn}^k L_k$ - and doesn't need us to know any detailed dynamical information about the fields etc.

You proposed that one should be verifying that the action is invariant under symmetry generators. That sounds good except that the action is only good for a classical description - or a quantum description that is obtained by a direct quantization of a classical theory. Such a way to obtain a quantum theory is only smooth or useful if the quantum theory is "close enough" to a classical theory. The most general CFT, especially in 2 dimensions, is so strongly "quantum" that there is no natural notion of an action and classical degrees of freedom. One must directly work with the quantum operators, their commutators, and they don't have any natural or helpful classical limit. Take the Ising model CFT as an example. You will find lots of fields, such as spin fields and twist fields, whose (mass) dimensions are fractional numbers such as $1/16$, something that would be unthinkable in a classical theory: the whole dimension comes from quantum effects. That's why the structure of the CFT science tries to be as independent of classical concepts such as the action as possible.

Implementing a symmetry on operators

If you have a generator $G$ of a Lie symmetry, it (infinitesimally) acts on ket states $|\psi\rangle$ and bra states $\langle \varphi|$ as $$\delta |\psi \rangle = i \epsilon G |\psi \rangle, \quad \delta \langle \varphi| = i \epsilon \langle \varphi| G$$ If you also define the action of the generator $G$ on a general operator $M$ as $$\delta M = i \epsilon [G,M]$$ then you may prove that all matrix elements will be invariant under the symmetry, $$\delta \langle \varphi | M | \psi \rangle = 0$$ by the Leibniz rule. So the natural action of symmetry generators such as $L_m$ on operators such as $\phi(z,\bar z)$ is $$\delta \phi(z,\bar z) = i \epsilon [L_m,\phi].$$ So if the commutator of $L_m$ with some fields - operators - is the same as the appropriate $(n+1)$-st derivative of these operators, then the generators $L_m$ implement the symmetry whose infinitesimal form involves $\delta \phi\sim z^{n+1} \partial^{n+1} \phi$.

Your comments about "eigenvalues" are conceptually misguided because a defining property of an "eigenvalue" is that it must be a "value" - a $c$-number - but $\partial_z$ is not a value - it is an operation. (I avoided the word "operator" because $\partial_z$ is not an operator acting on the Hilbert space of the CFT; only operators such as $\phi(z,\bar z)$ and $\partial_z \phi(z,\bar z)$ or $L_m$ are operators acting on the CFT Hilbert space. Instead, $\partial_z$ itself is just a rule to produce one operator from another one. It would be an operator if the wave functions - state vectors - were equivalent to functions of $z,\bar z$ but in a two-dimensional CFT, they surely aren't.)

Why cylinder is important

Concerning the question based on David Tong's text (greetings to David!), the cylinder is important exactly because a CFT on a cylinder is exactly equivalent to a CFT on the infinite plane. If $w=\sigma+i\tau$ lives on a cylinder - with $\sigma$ being $2\pi$-periodic - and if $z=\exp(-iw)$, then the infinite cylinder will be fully mapped to the plane, in a one-to-one way.

I actually think that David is being very clear about it.

So the analysis of the CFT defined on a full plane in general, and its behavior near the $z=0$ origin in particular, is totally equivalent to an analysis of a CFT defined on a cylinder in general, and in the limit $w\to -i\infty$ in particular. The two problems are exactly equivalent, exactly because of the conformal symmetry. The cylinder has a periodic spatial coordinate but this periodicity is not postulated for ad hoc reasons. It's postulated because if you write $z$ in the radius/phase form, $$ z= \exp(-i\sigma+\tau), $$ then the points with $\sigma\sim \sigma+2 \pi$ are identified with one another. The coordinate $\sigma$ is periodic. This is the basic fact of the exponential function - or its inverse function, the logarithm, if viewed as a function of a complex variable. And the exponential conformal map is very useful which can be seen if you follow what David is doing with that. You could ban the exponential function because you don't like it (or you think that other functions are being discriminated against) - but then you couldn't learn much of the CFT calculus because the CFT calculus largely depends on this clever exponential conformal map.

Because a small piece of any two-dimensional world sheet - regardless of the topology - looks like the flat plane and because the flat plane is equivalent to the infinite cylinder, the infinite cylinder is important for the understanding of local physics of the CFT at any Riemann surface - of any topology.

It is just true that I may just describe the infinite plane in coordinates such that one of them is periodic. This is what makes the analysis of the states defined on the cylinder - closed string states - automatically useful for the analysis of any properties of CFT, including its operators on the plane. In fact, the states of a closed string - obtained by quantizing the CFT on a cylinder - are in one-to-one correspondence with the local operators $\phi_K(0)$ at the origin (or any other point), because of the very same conformal map from the plane to the cylinder.

Now, you ask, where is the quantization?

Many of the formulae would work for a classical (non-quantum) conformal field theory, too. However, there are is no Hilbert space of "states" of a closed string, obtained from the quantization. So many of the interesting things, including the state-operator correspondence discussed two paragraphs above this one, only arise in the quantum theory. Pretty much all the objects such as $H$, $T_{\rm cylinder}$, and so on that David lists on page 86 or almost any other page are operators, so he deals with a quantum theory.

In some equations, David surely also uses commutators, to prove that it is a quantum theory, but it's not necessarily page 86 or another page you could find that has no commutators on it. ;-) But your complaint that David doesn't play with commutators of some fields exactly on some page where you would expect it surely not a sensible complaint, is it?

I am pretty sure that if you listen carefully, you will also understand that the commutators of operators in a CFT may be obtained from the OPEs, the operator product expansions. Just place two operators $T_k(z)$ and $T_l(0)$ to two nearby points $0$ and $z$ and calculate their product. The product will typically include a singularity that diverges as $z\to 0$ - as the two operators are very close to each other. (The singularity will be visible in any sensible expectation value.) The coefficient of $1/z$ or $1/z^2$ or $1/z^4$ - the leading singularity - is either a $c$-number or another operator. From that operator, you may determine the commutator of Fourier modes of $T_k$ and $T_l$ expanded over the cylinder, and so on.

Quantum mechanics has many effects that you wouldn't encounter in classical physics. For example, as you correctly mention, it influences the transformation from the cylinder to the plane, and so on. However, I don't know what to do with questions such as "And I don't see where this is happening here?" What should be happening here? Well, what's happening is probably something else than what you were expecting to be happening - but that's the very reason why you're trying to learn new things from David Tog, isn't it? If you were only learning old things that you knew, you would be wasting your time.

The things that you have to learn to understand two-dimensional conformal field theories are not "the same things" that you already learned for a generic quantum field theory in a generic flat space (such as a four-dimensional one). It is a new topic with new special features such as the exponential maps, OPEs, state-operator correspondence, and so on, and you shouldn't insist that the physics of OPEs has to be composed of the same insight that you already knew from QED in $d=4$. It is not the same thing - if it were the same thing, people wouldn't teach it twice.

So I would suggest that you ask about some particular statements that David makes and that you don't understand. A necessary assumption is that you actually try to listen what David is saying, instead of trying to force him to say things that you wanted to hear in the first place. ;-) When you switch to this mode of learning, the discussion could become a little bit more constructive. At any rate, I assure you that David is talking mostly about quantum mechanical systems, so all observables are operators on a Hilbert space that can get multiplied and whose expectation values may be calculated. The previous sentence could help if you misunderstood every single formula in David's lectures that includes an operator - which would be pretty much every formula.

However, I can't explain you all other details about David's text (and not even all effects of quantum mechanics - because pretty much everything in the text is quantum mechanical) unless you say exactly what's your problem. I would have to take 107 of his pages, inflate them by a factor of 10, and you could still end up being dissatisfied because your dissatisfaction could have some totally different causes. ;-)

  • $\begingroup$ Hey. Thanks for your response. It really helped me a lot. However one thing still seems to be a bit unclear for me: I do understand that $\delta\phi = i\epsilon[L_{m},\phi]=i\epsilon z^{m+1} \partial_{z} \phi$. However the $z^{m+1} \partial_{z}$ seems to appear on a classical level as a generator of conformal symmetry (or a generator of the Witt Algebra). This confused me since we're working here in the Quantum-case. Compare for example Page 21 of the paper written by Prof. Gaberdiel. Greeting and thanks again. $\endgroup$
    – mr_conf
    Commented Apr 13, 2011 at 9:16
  • $\begingroup$ Quick Remark: I've posted another question here: physics.stackexchange.com/questions/8540/… since I didn't know how to add a second question. It's in some way related to my previous questions (at least the references are the same). Maybe if you have time, I'd be more than happy if you could adress these questions as well. $\endgroup$
    – mr_conf
    Commented Apr 13, 2011 at 9:18
  • $\begingroup$ @Lubos: There seems to be a typo. $\delta \langle \varphi| = i \epsilon \langle \varphi| G$ seems to missing a minus sign, right? Is the action of operators through its commutator or the adjoint action have any significance in the light that the group action after exponentiating become something like $ \phi -> g\phi g^{-1}$. Group action by conjugation seems to be used in all transformations n Physics. Why is this true? $\endgroup$
    – user7757
    Commented May 31, 2014 at 17:50
  • $\begingroup$ I guess that you are right about the $-i$, there should probably be some $g^\dagger$ along with $-i$, too. ... Conjugation is the right way to conjugate operators because operators are maps from the Hilbert space to the same Hilbert space, so they live in a bilinear space, and the transformation of this adjoint/bilinear representation of any group is given by conjugation! It's omnipresent in physics because all observables (measurable quantities) are given by operators in quantum mechanics. $\endgroup$ Commented Jun 3, 2014 at 11:31

First of all, It is not enough to have a representation of a Lie algebra on a Hilbert space of a quantum field theory in order to affirm that the theory is "symmetric" with respect to this algebra. We speak about the "symmetry" only in the case where the correlation functions of the theory verify so-called Ward identities. The Ward identities can be formulated without referring to a classical action or to any processus of quantization although a beginner may not have this impression upon the first reading of review articles. ( Indeed, in the literature the Ward identities are often "derived" from a classical action and the path integral, but in reality this "derivation" is not more than a chain of plausible arguments motivating why the Ward identities should have the form which they have). The true logic of things is indeed different: the Ward identities must be postulated at the beginning and any theory verifying them is called symmetric. To be clearer, let me give an example of a scalar field theory in two-dimensional Minkowski spacetime. The postulated Ward identity for the translation symmetry is $$\langle 0\vert\Phi(x_1+a)\Phi(x_2+a)...\Phi(x_n+a)\vert 0>=\langle 0\vert \Phi(x_1)\Phi(x_2)...\Phi(x_n)\vert 0\rangle, \qquad (Ward)$$ where $\Phi(x)$ is an operatorial distribution (to be smeared with smooth compactly supported functions on the Minkowski space) and $\vert 0\rangle$ is the vacuum vector. Of course, this form of the translational Ward identity can be motivated from a classical action and path integral but this is not important for us. The identities $(Ward)$ are in fact simply postulated and any theory verifying them is by definition called translationally symmetric.

Now what is the relation with the unitary representation of the translation group? Well, let $U(a)$, $a\in \mathbb{R}^2$ be such representation, i.e. $U(a)$ are unitary operators on the Hilbert space $H$ of the theory such that $U(a)U(b)=U(a+b)$. If it moreover holds \begin{equation} U(a)\vert 0\rangle=\vert 0\rangle \qquad(1) \end{equation} and $$ U(a)\Phi(x)U^{-1}(a)=\Phi(x+a)\qquad (2)$$ then we see easily that the Ward identities $(Ward)$ are satisfied for each $n$. Thus we observe that the existence of the unitary representation of the translation group such that the condition (1) and (2) are satisfied yields the translational symmetry of the quantum field theory in the sense that the Ward identities are verified. By the way, the condition (2) is often rephrased by saying that the translation transformation $x\to x+a$ is "implemented" by the unitary operator $U(a)$ acting on the Hilbert space. The infinitesimal version of the relation (2) reads $$i[P,\Phi(x)]=\partial_x\Phi(x)$$ where $U=\exp{(iaP)}$ and $P$ stand for infinitesimal (Lie algebra) generators of the translations. In this sense must be understood the Gaberdiel formula (3.2.26) where not just translational but all infinitesimal conformal transformations are implemented.

I shall not embark here on a detailed description of the Ward identities for the conformal symmetry, because it is not a subject of your question (they can be found e.g. in the founding BPZ paper in NPB, 1984) Let me just mention that the story is in this case a bit more complicated, because the conformal symmetry is softly broken by the non-invariant vacuum so the Ward identities are "anomalous". Anyway, let me just finish the first part of my answer by saying, that the conformal field theory is simply the theory which satisfies the BPZ conformal Ward identities

Concerning the transition from the cylinder to the plane: Perhaps the best way to start the exposition is to mention the Eguchi-Ooguri reformulation of the BPZ conformal Ward identities (NPB, 1987). Eguchi and Ooguri work in the Euclidean picture and they couple dynamical fields of a CFT to a non-dynamical background gravitational field (Riemannian metric $g_{ab}$) on the world-sheet. In particular, they postulate that a field theory is called conformal only if its correlation functions change in a particular way if we replace the background metric $g_{ab}$ by $e^{\sigma}g_{ab}$, where $\sigma$ is an arbitrary function on the world sheet. This means that the functional derivatives of the correlation functions with respect to the Weyl factor $\sigma$ must have a particular form and the corresponding quantitative expressions of this fact may be called the Eguchi-Ooguri Ward identities. The upshot is that the standard BPZ Ward identities can be derived from Eguchi-Ooguri Ward identities. All this means that if we know correlation functions in a given gravitational background $g_{ab}$ we can calculate them also in the "Weyl equivalent" background $e^{\sigma}g_{ab}$. This observation is very useful since we may study some quantitative aspects of the CFT theory in one background and other aspects in some other Weyl related background. The principal example of this situation is just the transition from the cylinder to the plane. The correlation functions of CFT theories in the natural flat Euclidean coordinates on the plane have very nice analytic properties (the so called fundamental Borcherds OPE relation has the nicest possible form) while the theory on the cylinder is more useful for the construction of the Hilbert space of the theory and e.g. questions related to the representations of the Virasoro algebra. Let me add a few technical details on this point.

Consider the Euclidean metric on the Euclidean plane $ds_1^2=dx^2+dy^2$, and the flat metric on the cylinder $ds_2^2=d\phi^2+d\rho^2$, where $\rho$ is the coordinate along the axis of the cylinder and $\phi$ is the angle "coordinate" around it. Although $\phi$ is not defined globaly, the following map relating the plane and the cylinder is globally defined : $$x=e^\rho\cos{\phi}, \quad y=e^\rho\sin{\phi}.$$ Using this transformation, we observe that the cylinder can be parametrized by the plane coordinates $x,y$ in which the metric $ds_2^2$ becomes $$ds_2^2=\frac{1}{x^2+y^2}(dx^2+dy^2).$$ We observe that the Weyl factor $\sigma=-$ln$(x^2+y^2)$ considered in the Eguchi-Ooguri formalism has naturally emerged and, due to the Eguchi-Ooguri Ward identities, there is a one-to-one relation between the correlation functions of a CFT theory on the plane and of its "cousin" CFT theory on the cylinder.

What is then the radial quantization? In my opinion, the word quantization is a misterm and one should rather speak about a "radial reconstruction". What I mean is the following: The BPZ Ward identities are usually formulated on the complex plane where they take particularly simple form. Once we have an Euclidean conformal field theory on the plane (i.e. the correlation functions which verify the BPZ Ward identities) we would like to know whether there exist a Minkowski version of this theory with its Hilbert space and with its operator valued distributions in such a way that the mean values of those distributions in the vacuum yield (upon a suitable Wick rotation) the original Euclidean solutions of the BPZ Ward identities. In this respect we have to stress that the Minkowski CFT lives always on the cylinder! This means that, working on the plane, we cannot take as the Euclidean time to be Wick rotated a flat coordinate $x$ or $y$ but the true Euclidean time is instead: $\rho=\frac{1}{2}$ln$(x^2+y^2)$ which is just the logarithm of the radial polar coordinate on the plane. The true "space" of the Minkowski quantum field theory is any circle on the Euclidean plane centered in the origin. The "radial reconstruction" is then a way to construct the Hilbert space of the Minkowski version of the theory, the operator valued distributions and the Virasoro generators directly working on the plane without performing the coordinate transformation to the cylinder. The outcome is the usual quantum theory stuff with its Hilbert space, operators etc so it looks like a quantization but in reality the starting point is not some classical story but it is also quantum albeit in the Euclidean sense. Let me conclude by a warning that the radial reconstruction does not always work, i.e. not every solution of the Euclidean BPZ Ward identities leads to a Minkowski quantum field theory. The so called Osterwalder Schrader positivity conditions imposed on the Euclidean correlation functions have to be also verified for this purpose.


I'm not really qualified to answer, but I think I can give a hint on your first question.

First, you have to think of the field $\phi$ as an operator. In particular, I tend to think of $\phi$ as a linear combination

$$ \hat\phi(z,\bar z) = \int dz d\bar z\ \phi(z,\bar z) \Psi^\dagger(z,\bar z)$$

of operators $\Psi^\dagger(z,\bar z)$ that create particles at positions $(z,\bar z) = (x+iy,x-iy)$ (where $x$ and $y$ are the independent coordinates in the plane). The construction in terms of normal modes is similar, except that you use a different basis - not positions but modes (mainly because creating a particle that it localized at precisely one point has its mathematical problems).

EDIT: The following is not very accurate.

Now, what does it mean for the operator $\mathcal L(\phi, \partial\phi)$ to be invariant under a symmetry operator $T$? It simply means that both operators commute,

$$ [T,L(\phi, \partial\phi)] = 0$$

Now, if the symmetry operator $T=L_n$ fulfills the commutator relation in question, and the Lagrangian as a function is invariant under conformal transformations, then it follows that the Lagrangian as an operator commutes with $T$. The reason is that the commutator relation for $L_n$ makes $[L_n,·]$ act on the Lagrangian operator exactly like the corresponding conformal symmetry acts on the Lagrangian function. (I think this can be shown by using the above picture of $\phi$ as an operator.)

  • $\begingroup$ Hi Greg Graviton, there are glimpses of right ideas in your answer. Apologies but your first formula is conceptually incorrect. It is not true that $\phi(z,\bar z)$ in a free boson CFT is bilinear in creation operators. Instead, the right expansion is a sum (not product) of terms that only depend on $z$, and terms that only depend on $\bar z$. Also, the Lagrangian of the CFT is obviously not invariant under conformal symmetries (it shifts!) - only the overall action is - as long as we treat the system classically. But quantum mechanically, the CFT is not just about the "classical action". $\endgroup$ Commented Apr 7, 2011 at 16:04
  • $\begingroup$ @Luboš Motl. Thanks for your comment! Indeed, the bilinearity is nonsense. But are you sure that it's a sum of $z$ and $\bar z$ separately? Treating $z$ and $\bar z$ as independent coordinates, I would now think that the creation operator is $\Psi^\dagger(z,\bar z)$. $\endgroup$ Commented Apr 7, 2011 at 17:25
  • $\begingroup$ Dear @Greg, look e.g. at the first blue-rectangled equation at physics.thetangentbundle.net/wiki/String_theory/bosonic_string/… - the $\alpha$ are the Fourier modes of $X$ on the cylinder, or, equivalently, the Laurent modes of $X$ in the plane around the origin, and they create "quanta of $X$" on the world sheet. They're still modifying a single-string Hilbert space, its excitation, of course, not the second-quantized multi-string Hilbert space. $\endgroup$ Commented Apr 7, 2011 at 17:36

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