12:20 AM
Y'know it's been like a good year and a half since we've heard anything from Duffield
Wonder what ever happened to him

7 hours later…
7:14 AM
@ACuriousMind Oh yes!
Thanks

3 hours later…
10:22 AM
0

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...

3 hours later…
1:12 PM
Got the classic comedy bug
Bug disappears when I add a debugging console output and comes back without

those are fun

Tempting to just leave that console log in there and move on
Let a future person deal with it

the last ones I had were race conditions - adding the debugging slowed down the main process enough so that the deadlocks no longer occured

I suspect here it may be because the object is initialized and reused several times
I may need to check if it is reset properly

@Slereah I'm sure future-Slereah will love that :D

1:18 PM
@ACuriousMind If he gives me any lip I'll just start eating unhealthier food

a flawless plan

"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."
Not sure if I am exceptionally convinced

Maybe there are more exceptional things than those

Also, I have seen places like here where the state operator map is defined by the rule of getting states from operators which are valid in QFTs also

1:24 PM
What if the GUT uses an unexceptional group

@Slereah these notes look really great

@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 the focus word is local

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 $z=0$

If it's at 0 that sounds more than local, even

1:27 PM
"They are known as exceptional holonomy manifolds for this reason."
More of this exceptionality

@Slereah sounds like a classic case of wishful thinking to me :P

@ACuriousMind Don't want to live in some random universe

I mean, claiming that "having existence" is exceptional is pretty weird

Might be the worst nlab page I've seen
Although @ACuriousMind might have a better time reading it I suppose

exceptional means that there are lots of things that are not like it
but if there are many universes that don't exist...that doesn't make any sense at all

1:29 PM
🤔
It's not like Hegel's book was never translated

lol, it's literally just a Hegel quote

Although to be fair, Hegel's book is about as understandable in English as it is in German

I was just about to say that :P

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.

I learned more from those two pages on holonomy than I have anywhere else

1:41 PM
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?

I don't think people experience schadenfreude that often that it would be an everyday word
despite Germany's reputation
The Simpsons once pointed out that the opposite of Schadenfreude is sour grapes, and I don't think US people use it that often either

Thanks for the response 🤔

@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?

yes

What does it mean for an operator to act at $z = 0$
Aren't they defined on test functions
Is it on arbitrarily small support test functions

1:56 PM
technically it's $\lim_{z\to 0}\phi(z)\lvert 0\rangle$

If anything that's even less illuminating
2

if you want to think in terms of distributions, just smear the operator with some test function localized around $z=0$
but I've never seen anyone apply that pedantry to CFT - the mathematical phrasing there is in terms of operator vertex algebras, not distributions

"The traditional definition of vertex operator algebra (VOA) is long and tends to be somewhat unenlightening."
Oh no

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

2:22 PM
Is the assumption of unitarity somewhere taken in the state-operator correspondence?

unitarity of what?

of the CFT

physicists often use that word to mean several different but related things
what exactly do you mean?
are you asking whether the representation of the Virasoro algebra needs to be unitary?
are you asking whether the field operators are assumed Hermitian?
something else?

The CFT should believe that God has only one form

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

2:31 PM
@ACuriousMind If the Hamiltonian is assumed to be non-Hermitian...

yes, you need a Hermitian dilation operator for the correspondence to hold

When do we need this? I mean in which step of the proof of the map?

that's a bit hard to answer without knowing what proof you've seen :P

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)

2:52 PM
@ManasDogra unitarity is never obvious/easy in path integral formulations

Isn't it related to the reflection positivity thing in path integral formulations
I forget

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

Alarming

or, if already the dilation operator isn't unitary, you can't even begin to talk about "primary states2 at all

Is it mandatory?
I remember some examples of non-unitary theories that weren't that badly behaved
Dirac field around a black hole
Lack of unitarity representing particles falling into the singularity

2:57 PM
probably not, since non-unitary time-evolution can just mean your system is open
there's nothing inherently apocalyptic about that

Yeah I guess that's another way of having particles disappear :p
Less expensive
Experiment failed to be unitary due to electrons disappearing under the fridge

3:29 PM
0

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.

1 hour later…
4:39 PM
Alright, I eventually got ChatJax
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

As far as I know, given
$F_2(q,P)=qP+\varepsilon G(q,P)$
$F_2$ is the generating function, while $G$ is the generator
infinitesimal generator, of course

that's the terminology in the context of canonical transformations, yes
but that is not the meaning of "generator" in the context of gauge transformations or more general transformation

4:55 PM
So can't you make something like you do with momentum as the generator of translation?

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 $\mathrm{e}^{\mathrm{i}ft}$

Yes, this is what you get from my generating function

I don't know what you mean
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

Using the generating function $F_2$ as above you get
$p=\partial_q F_2=P+\partial_q G\implies\delta p:=P-p=-\partial_q G=\{p,G\}$
That is what you wrote above for infinitesimal transformations

yes, but what I wrote above has nothing to do with any $P$
I don't understand why you bring $F_2$ into this at all

5:03 PM
I was thought to think of infinitesimal transformations in terms, using the generating function of a canonical transformation
I think Shankar does this too, let me check

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

Maybe I don't know enough about this. Don't we associate a flow (canonical transformation) to every function in phase space?

you can associate a flow to every function, sure
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
Qmechanic has an overview here

The definition I know is "A coordinate transformation in phase space which jacobian matrix is symplectic"

the hell is a Kamiltonian

5:12 PM
But yeah you could also define it the geometric way as in your link
@Slereah The new hamiltonian after the canonical transformation

ah yes, that would be the definition as a symplectomorphism

Maybe I'm getting kind of insane doing stat mech but I thought all flows were canonical

as mentioned here, that's locally but not necessarily globally sufficient to have a generating function
@Feynman_00 the problem is imprecise terminology - in your notion of "canonical" (=symplectomorphism), that's true

Oh ok, I was going nuts about it

aaaanyway, I think this is distracting us from your actual question, sorry about that

5:18 PM
Don't mention it. It's a fun discussion

so, if I understand you correctly, you're wondering where the minus sign in front of the generator in QM comes from?

Yes, that is the problem. It should be the other way (the adjoint) as it turns out working out things without using generators
You could easily see that using path integrals

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!

Of course, even for momentum some use minus sign but that just depends on your convention
But all those conventions agree. I'll make an example

the minus comes essentially from the $H$ (or $\Lambda$) being placed in the second slot of the commutator
if we placed it in the first slot, we'd get a +

5:27 PM
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

it depends on what you think "a translation by $\delta x$" is
the first operator shifts by $\delta x$ in one direction, the other in the other

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
Yes, exactly

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 $\delta x$"

@ACuriousMind One could make that point also for momentum, couldn't they?

it doesn't matter in the end as long as your convention for what "being a generator" means doesn't switch in the middle of a derivation
@Feynman_00 yes, but for momentum the "correct" direction is not obvious, in contrast to time :P

5:31 PM
By the way, I suspected that this is actually connected to the fact gauge invariance is a bit different in QM

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
@ACuriousMind Sakurai does this

@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

5:41 PM
I am using the opposite convention in both cases

yeah, it makes sense that you get $\mathrm{e}^{-\mathrm{i}\Lambda t}$ then because I'm looking at notes that use what I wrote and get it without the -

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! :)

no worries, although I don't feel I helped much :P

5:56 PM
@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

6:50 PM
I like how nlab has become pretty ubiquitous in physics circles just by being one of the first result for a lot of physics searches