I had a -10 serial upvote reversal today. I was a bit puzzled since I had not seen much activity on my account recently. It seems that yesterday, someone upvoted two of my answers on neutron stars and the algorithm decided this was serial upvoting.
Is upvoting two answers, even in quick successio...
"In non-conformal field theories, a typical local operator creates many different states."-got this line from Tong's string theory lecture notes.
But isn't what's special to CFTs is there exists a unique local operator corresponding to a state and that the other way round is valid even in generic QFTs? And if that is the case how can a local operator create different states in QFT let alone in CFT?
"Since nature (reality) is exceptional in that it has existence, it is plausible that it is the exceptional structures among all mathematical structures – such as the exceptional examples in the classification of simple Lie groups, the exceptional Lie groups – that play a role in the mathematical description of nature, hence in physics and specifically in phenomenology."
@ManasDogra I think what Tong means is that in a general QFT, you can't reduce all operators creating states to those living at the origin: To get all QFT states, "the local operator" $\phi(x)$ (i.e. a quantum field) has to act at $x=0$ as well as at all other values of $x$
in contrast, in a CFT you can get the entire Hilbert space just from the operators at $z\to 0$
so, in a QFT, $\phi(x)$ can be associated with many different states for different values of $x$, but in a CFT, it is uniquely associated with the state it creates at $z=0$
What about this where the OP writes the equation for only the operator to state map and not the other way round?
I have seen the same thing in for example Becker Becker Schwarz page 67. They seem to give not only the half of the correspondence but the half which actually isn't special to CFTs in particular.
Pardon the interruption, but I have a question about the German language, if you guys don't mind... is the word "schadenfreude" a common everyday word, or has the internet just turned it into some sort of meme, by looking for its opposite etc?
@ACuriousMind Just to be sure about an obvious point...Do we need many local operators acting at $z \to 0$ to get the full Hilbert space? I mean, we get a particular state from a particular operator, so we need many local operators to build the Hilbert space right?
the idea is this: The non-rigorous way of thinking about quantum fields in 2d CFT just as operator-valued holomorphic functions $\phi(z)$ gives rise to operator product expansions (OPEs) that tell you how to multiply two of them (in terms of the holomorphic functions it's a bunch of residue theorem applications etc.)
these OPEs are really everything you need to know for a CFT
and so you formalize the CFT data in terms of some algebra of operators $A$ that has a multiplication rule that maps two operators to a formal Laurent series (i.e. a series of terms that are operators "divided" by powers of $z$)
this contains now all the OPE data without ever claiming anything is a holomorphic function (or a distribution), and so all the annoying details with that have been defined away
here's a nice answer by Connor Behan discussing "unitarity" in a CFT context
note that there is at least one other notion of "unitarity" in a CFT context, namely that of the unitarity of the Virasoro algebra for certain values of the central charge
The path integral kind of proof..The one given in Polchinski vol. 1...
page 66
Should I share a screenshot ?
In short one takes the fields on the outer circle of a disk and the limit of the path integral over the inner circle defines the local operator...
Here in 59:00 Shiraz Minwalla seems to skip the question for the time being.
I was talking of unitarity in the sense it was said in that video (although what "unitarity" means isn't mentioned explicitly but used by a student to ask the question I was asking)
yes, it is (Connor's answer above mentions that, too)
in the Hilbert space formalism it's much easier to see: If you don't have unitarity, then the descendant states of the primary states will have negative norm sooner or later
I'd like to ask for an example in which the minimum energy principle is used (thermodynamics). However i've never seen question in which examples are required so i got the doubt that it's not allowed to ask for examples.
The generating function $F_2(q,P)=qP+\frac{e}{c}\Lambda(q)$ generates a gauge transformation, right?
In this sense would it be right to take $\Lambda(q)$ as the infinitesimal generator of gauge transformation in QM? One would get a unitary transformation $\hat{U}=\exp(\frac{ie}{\hbar c}\Lambda)$ But it turns out that the unitary trasformation acting on the states is $\hat{U}_\Lambda=\exp(-\frac{ie}{\hbar c}\Lambda)$ i.e. the adjoint
@Feynman_00 the meaning of "generating" in "generating function for a canonical transformation" is not the same as that in "generator of a transformation"
canonical transformations are about coordinate changes, but the kind of transformations that have "generators" via the exponential map aren't about coordinates
In general, a "generator" of a classical transformation with infinitesimal changes of the generalized coordinates $\delta q,\delta p$ is a function $f(q,p)$ with $\delta q = \{q,f\}$ and $\delta p = \{p,f\}$
the quantum version of that is replacing the Poisson brackets with commutators
the corresponding finite transformations are given a) classically by the flow of the Hamiltonian vector field of $f$ and b) quantumly by the exponential $\mathrm{e}^{\mathrm{i}ft}$
your generating function contains $q$ and $P$, implying you're thinking about a transformation between two sets of Darboux coordinates $q,p$ and $Q,P$
the statement that "momentum is the generator of translation" is just that the finite version of $\delta q = \{q,\epsilon p\} = \epsilon$ is $q \mapsto q+\epsilon$, i.e. translation
that doesn't make a lot of sense to me because a transformation is only a canonical transformation if it is a symmetry, i.e. leaves the Hamiltonian invariant
but the notion of a generator of a transformation via the Poisson bracket/commutator is much more general and not restricted to symmetries
but that won't be a "canonical transformation" for every function
what exactly a canonical transformation is in mathematical terms is a bit annoying because different authors use the term "canonical transformation" differently
so, how did you get from "the generator of a gauge transformation is $\Lambda$" to "the gauge transformation should be $\mathrm{e}^{\mathrm{i}\Lambda\epsilon}$" without the minus sign?
because we also often say "the Hamiltonian is the generator of time translation" and the time evolution is $\mathrm{e}^{-\mathrm{i}Ht}$ with a minus, too!
After acknowledging momentum as the generator of translation, some define translation as $\hat{U}=\exp{i\frac{\hat{p}\delta x}{\hbar}}$ others as $\hat{U}=\exp{-i\frac{\hat{p}\delta x}{\hbar}}$ but those are actually two different operators
One shifts $|x\rangle$ to $|x+\delta x\rangle$ (and thus the wave function from $\Psi(x)$ to $\Psi(x-\delta x)$, multiplying by the bra). The other does the opposite
depending on the direction of one's x-axis and whether or not one is thinking about this transformation as active or passive and even whether one thought carefully about this or not, one will pick one or the other of these as "the translation by $\delta x$"
I'd argue the only consistent usage would be that all generators generate their finite transformation with the minus sign
but I can't really think of a case where it would matter that you pick the other version for some operators as long as you don't change your choice for these later on
What i mean is that the kinetic momentum is gauge invariant in classical mechanics, while the kinetic momentum operator is not (its average is, though). On the other hand, in classical mechanics canonical momentum is gauge dependent, while in QM we decide to keep it unchanged as the generator of translations
@Feynman_00 it's not a decision - it is always the canonical momentum that generates translation in its canonical position
that's just what the Poisson brackets tell us
there's nothing magical about kinetic momentum that would make it always generate translations - the thing that generates translations in $q$ is the thing $f$ with $\{q,f\} = 1$
My phrasing was bad. What I meant is that what you get by applying the gauge transformation to canonical momentum is not anymore the generator of translation
i.e. canonical momentum after gauge transformation
I don't think that has anything to do with the sign
for gauge transformations, there is simply an additional sign choice in its classical definition, namely whether a gauge transformation does $\vec A \mapsto \vec A + \nabla \Lambda, \phi \mapsto \phi - \frac{1}{c}\partial_t \Lambda$ or with the signs flipped
Yes, that is fine. The easiest way to prove it would be using the path integral propagator and I get exactly that minus sign
Oh wait
I want to transform the states i.e. kets not the wavefunction
I should use the minus sign :P
In the same fashion you use the minus sign to have a positive shift in position ket
To make a parallelism with position and momentum, I was thinking in terms of the shift of the wavefunction, that is opposite to the shift of a position ket
@ACuriousMind I think I understood. This discussion was very helpful, thank you! :)
@ACuriousMind Oh, you did. I'm the one who's not helping himself. I should be studying structure of matter and stat mech and I'm chatting about minus signs in QM :P
I'd like to ask for an example in which the minimum energy principle is used (thermodynamics). However i've never seen question in which examples are required so i got the doubt that it's not allowed to ask for examples.
