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12:52 AM
Is it possible to know when a close vote was given? I have one on a recent question of mine, and I want to know if it occurred before or after I edited the question (due to a comment pointing out a mistake).
 
 
4 hours later…
4:49 AM
@user400188 Look at your reputation and it will show the time you got the downvote next to the -2 reputation change.
@user400188 Which question was it? I cannot see a downvote on either of your two recent questions.
 
5:02 AM
I wasn’t downvoted, I only got a close vote on the question (I guess they were nice)
 
Oops, sorry, yes.
 
So, is it the case that you can't know when a close vote occurred? (unless you watch it happen).

Also, it just occurred to me that this question would appear in the close vote queue, once it's given its first close vote (If I am understanding how that works properly).
 
The question did appear in the closed vote queue, but all that shows is the two reviews:
It doesn't show when the original close vote was.
 
Oh, ok then. Thanks for the help :)
 
 
2 hours later…
7:23 AM
Morning
 
 
2 hours later…
8:57 AM
I need some help regarding closed timelike curves
I need to build a time machine so that Angular was never invented
 
9:40 AM
@Slereah but how will you keep people from inventing it again?
 
"The error bars suggest a $3\sigma$ discrepancy between the two measurements" what does this statement mean? $\sigma$ means the standard deviation(error) usually but what should this mean when two measurements with two different errors are compared like this?
I don't think it's relevant but the measurements are of Hubble constant from different experiments. One is $H_0=74.0\pm 1.4$ and another is $H_0=67.4\pm 0.5$ in the same units
 
9:55 AM
@ManasDogra It means that you need to add/subtract more than three times their respective standard deviation to them to make them compatible - in your example, 74-3*1.4 = 69.8 and 67.4+3*0.5 = 68.9. It's not $4\sigma$ because subtracting 1.4 from the first value again lands you at 68.4, which is below the second value (hence they are compatible at $4\sigma$).
 
Thanks
 
 
2 hours later…
11:47 AM
@Slereah Whats angular?
 
12:05 PM
AngularJS, presumably
 
Why is the Hamiltonian density conserved? If you plug zero into both indices in eq. 2.30, you will get $\Theta_{00}$ to be conserved with time. Eq 2.31 shows that it is nothing but Hamiltonian density.
 
...why did you ask a question and then answer it yourself in the same message?
 
@ACuriousMind Sorry, I didn't word it correctly. The question is I don't understand why the density is conserved. It makes intuitive sense if Hamiltonian is conserved but why is the density also conserved?
 
@Yashas what do you mean "why"? Didn't that text derive its eq. 2.30 via Noether's theorem?
 
Yes, but it was surprising. I am accustomed to seeing Hamiltonian being conserved but was a bit surprised that to learn that the density is conserved too.
Why can't energy move from one place to another? Or could it not spread out or concentrate in one place? The density changes in these cases, right? But the total energy can still remain conserved.
 
12:34 PM
@ACuriousMind use the closed timelike curve to go back to the Big Bang and embed the message “don’t create Angular” in the CMB
either that or grandfather-paradox anyone who decides to invent it
but the former has the advantage of messing with cosmologists
”God exists, and He hates Angular”
 
@Yashas you need to look closer what the equations actually say
the conservation law is not $\partial^0\Theta_{00} = 0$, but $\partial^0\Theta_{00} = \partial^j\Theta_{j0}$
i.e. the claim is not that energy density is constant, just that its change with time is associated with a spatial flux of energy $\Theta_{j0}$
 
12:59 PM
Very embarassing. I picked out a single element instead of the sum. Still, the author says that density is conserved right below eq 2.32. Is that a typo?
 
yeah, that's a bad formulation - they mean "the integral of the Hamiltonian density over $V$ is conserved if there's no flux through its boundary $S$"
 
 
1 hour later…
2:12 PM
When we canonically quantize through $\Phi \rightarrow \hat{\Phi}$ what pictures are we going from and to?
e.g. is it the Heisenberg picture?
 
Why would there be a picture at all there?
you only have to decide what picture you're in once you talk about time evolution
 
That's what I was thinking
Okay thanks
What is meant by a 'dense subspace of the Hilbert space' in more physical terms?
Any time you see a commutator in QM, does that mean a Lie algebra is implicitely involved?
 
those are two very different questions :P
 
They are in neighbouring paragraphs in the book I'm reading lol
 
@DIRAC1930 you can think about a dense subspace as a subspace of vectors with which you can approximate any other vector arbitrarily closely (since the closure of the dense space is the whole space, all vectors not in the dense subspace are at least limit points of it)
 
2:21 PM
'Defining the commutator of two operators $A$ and $B$ on $\mathcal{H}$ is not as innocent as it may seem. The reason is that, in QM, observables are typically described by operators which are not defined on the whole Hilbert space $\mathcal{H}$ but only on a dense subspace. As a consequence, the composition of operators and hence also their commutator is not nessecarily defined'
Why aren't the operators defined on the whole space?
 
@DIRAC1930 the set of linear operators on a vector space is a Lie algebra with the commutator as the Lie bracket, I'm not sure what you mean by a Lie algebra being "involved" in that
 
And then after stating the the definitions of bi-linearity, antisymmetry and the Jacobi Identity, the book then says 'again we can summarize these properties by the statement that the commutator endows the space of linear operators on $\mathcal{H}$ with the structure of a Lie algebra'
 
@DIRAC1930 because there are no self-adjoint unbounded continuous operators defined on the whole space, cf. Hellinger-Toeplitz, but physical observables are often unbounded quantities like position or momentum
 
Is the following just a statement regarding the Complete Set of Commuting observables? 'Finally, the maximal abelian subalgebra provides the quantum numbers of the Hilbert space vectors'
 
CSCO is essentially just a different name for "maximal Abelian subalgebra", yes
 
2:34 PM
If we have two continuous symmetries, do we have to find the maximal Abelian subalgebra of all the generators combines?
 
in order to do what?
 
To classify the states
 
sure
but the Cartan subalgebra of $G\times H$ is just $\mathfrak{c}_G\oplus\mathfrak{c}_H$ if we denote the maximal Abelian subalgebra of $G$ as $\mathfrak{c}_G$, so it's not different from finding them invididually - you don't suddenly have to do something different for two symmetries
 
Why have you done the direct product of 2 algebras?
 
I was thinking in terms of the groups
$G$ is the group, $\mathfrak{g}$ is its algebra - the algebra of $G\times H$ is $\mathfrak{g}\oplus\mathfrak{h}$
 
2:39 PM
But where do you do the direct product of groups?
 
what did you mean by "having two symmetries" if not that you have one symmetry group $G$ and another symmetry group $H$ such that the total symmetry group is $G\times H$?
 
I meant like, for example, non relativistically, rotational symmetry and $SU(2)$ spin symmetry
 
that's not two different symmetries
spin is just rotation
 
I meant angular momentum and spin
 
still the same group
or rather, angular momentum and spin are not separately conserved, there's no two separate symmetries there
 
2:43 PM
Interesting
 
the whole point of spin is that it interacts with e.g. magnetic fields much like angular momentum does, and so you can convert one into the other
 
What are two different symmetries in non-rel physics
CPT and SO(3)
 
but to your original point: of course, in some concrete setting you can have that they are separately conserved, then you have two copies of the $\mathrm{SO}(3)$ rotation group as symmetries, with the full symmetry being $\mathrm{SO}(3)\times \mathrm{SO}(3)$
 
I'm confused
If I had to non relitavistic particles
Would SO(3)XSO(3) be a symmetry?
No that doesn't sound right
 
depends
symmetry is a property of your Hamiltonian/action
 
2:48 PM
$G\times H$ is the same as $G \oplus H$ right?
 
there is no $\oplus$ for groups
 
What is I had two Hamiltonians put together like $H_1 \oplus H_2$
Wouldnt the symmtery group be $G_1 \oplus G_2$ or something
 
again, there's no such thing as $\oplus$ for groups
it just doesn't mean anything
 
Yeah, that doesn't satisfy the group axioms
 
I don't know what that means - that thing simply doesn't exist, there is no definition of a direct sum for groups
 
2:50 PM
Can't we just define it
Eveything in math seems to just be someones definition thats caught on
 
and as what would you define it?
 
I would define $G \oplus H$ as $G \times H$
how does $G_1 \times G_2$ act on a Hamiltonian
The Hamiltonian would have to be of the form $H_1 \times H_2$ right?
 
Sure, then it exists and is a group. But you shouldn't do that because in general $\oplus$ is only used to denote the product in categories where the product and coproduct coincident, but the coproduct of groups is different from the product of groups.
@DIRAC1930 I think you're a bit confused about how groups and representations work
 
That notation is used for abelian groups, but it means direct product
 
There are essentially two main cases: Either you start with an abstract group and you try to construct representations, or you start with some space and some operators on it and you're looking for symmetry groups or whatever
I'm not sure in what sort of situation the question "How does this group act on a Hamiltonian?" arises
 
2:56 PM
Oh sorry, everytime I said group I meant representation
 
uhhhh
that, indeed, changes a lot :P
 
Lol
So a representation of $G\times H$ is $\rho(G) \oplus \rho(H)$ right?
 
although I'm not sure what a "symmetry representation" is, sooo...not everytime?
 
symmetry representation is just the representation of a symmetry group
 
Given a representation $(V_G,\rho_G)$ of $G$ and a representation $(V_H,\rho_H)$ of $H$, you get a representation of $G\times H$ as $(V_G\otimes V_H, \rho_G\otimes \rho_H)$
note that this is the tensor product, not the direct sum
 
3:01 PM
How exactly does this play a role in SU(3)XSU(2)XU(1)?
Actually I give up
 
I'm not sure what you mean
 
I've been trying to understand representation theory etc. for ages and I'm getting nowhere
 
maybe you should get comfortable with groups in basic QM before trying to understand the Standard Model :P
 
Whats the difference between a group and a representation in QM?
 
Why the "in QM"? Do you understand what the difference between a group and a representation in general is?
 
3:06 PM
Yes
A group satisfies some axioms
if it satisfies those properties
its a group
 
so then what issue do you have applying that knowledge to QM specifically?
 
Well explicitly written out, isn't the Hamiltonian operator a representation of the abstract H in a given basis?
 
I wouldn't say that
 
by representation I don't mean group representation, just representation in the colloqial sense
 
what's the "abstract H"?
 
3:09 PM
$H |\Psi\rangle = -\imath \partial_t |\Psi \rangle$
 
but that's not a thing
we often have an "abstract H" in our equations because whatever we're doing does not depend on the specifics of any particular H, but in each of these cases we still imagine this $H$ to be a concrete operator given on a single concrete space
there's a Hilbert space $\mathcal{H}$ and this $H$ is an operator on it, what does this have to do with "representation"?
(there are formulations of QM where we would talk about representations before even getting to a Hilbert space, but I don't want to bother you with that right now)
eh
if you're doing the free particle, it's $H\propto p^2$, no matrix in sight
 
I meant $H \propto p^2$ isn't the abstract H
 
but if you're on a finite-dimensional space, sure, it's given as a matrix because how else would you give someone an operator on a vector space?
 
$H$ is always concretely some operator on a vector space. it does something to vectors
 
What do groups act on?
Nothing
 
3:13 PM
what does $H$ have to do with groups?
there's no group here, it's just a single operator on some Hilbert space
 
Because how do find out what groups do in QM without defining the Hamiltonian first
 
I'm not following you
 
If I have a symmetry, I have $[H,O]=0$ right?
 
$O$ is an operator in some Lie algebra
 
3:15 PM
no
$O$ is just an operator on a vector space
 
if continuous
 
it's just an operator on the Hilbert space
 
there's no Lie algebra here yet
 
I mean, the set of all operators on the Hilbert space is a Lie algebra, but I don't understand why we'd care about that at this point
 
i mean, there is a group action here already, i.e., $e^{-i H t}|\psi(0)\rangle = |\psi(t)\rangle$
 
3:16 PM
Okay, O is an operator on the Hilbert space
Okay, so thats what we mean by group action
 
i think i'm using that phrase right :P
 
If you write a commutator you are using a Lie algebra structure or else it is not consistent
 
what?
 
sure, as I said - the operators on any vector space naturally form a Lie algebra
but I'm not at the point where I can begin to understand what exactly @DIRAC1930 problem is
 
i guess, but there's not really an "or else" in there
 
3:18 PM
if we were to expand psi in a particular basis and H also, $e^{\dots} $ would be a representation of the time evolution operator on the space of wavefunctions in a particular baisis right?
 
no, that's not a representation
at least not in the sense of group and representation theory
whatever you do in a specific basis to evaluate an expression like $H\lvert\psi\rangle$ is...just a computation, it has no larger significance
 
But $\hat{X}: \psi(X) \rightarrow \psi(-X)$ is one representation of the parity operator acting on wavefunctions. Meanwhile $1 \oplus -1$ is another representation of the parity operator acting on $\psi(X)_S \oplus \psi(X)_A$ or something right? where S and A are symmetric or antisymmetric respectively
 
i'm not so sold on that being wrong. you have a 1D group of translations in time
with each possible Hamiltonian giving a different representation fo such
 
@Semiclassical but that has nothing to do with choosing a basis
 
no, it doesn't
 
3:24 PM
@DIRAC1930 True, I'm just not sure why you used "But" there ;)
parity is, abstractly, a $\mathbb{Z}_2$ group. You've given examples for representations of it
ahhh
 
i mean, you could define one Hamiltonian according to one basis and another according to another basis, and find they give different representations, but that's b/c they'd be different Hamiltonians
 
Because everyone was telling me i was wrong
lol
 
@DIRAC1930 well, what group do you think is being represented by the choice of $H$?
 
the choice of basis is a distraction
 
Thats right, the choice of basis is a distraction lol
 
3:26 PM
the reason your formulae for parity define a representation are because parity squares to the identity so the set $\{1,P\}$ is a group whose representation you cn give simply by writing down a formula for $P$
 
But how would we find that the group was $Z_2$ is we were given a system we couldnt visualize?
 
by knowing what our operators are doing
 
but the Hamiltonian doesn't square to the identity (usually :P), so there is no obvious group whose representation given a formula for it would define
 
if you don't have some way of mathematically describing what your operators are doing, you of course can't find a group structure
(besides just "group of Hermitian operators on a particular Hilbert space")
 
Oh I see I think. The Hamiltonian is the generator of a U(1) group
the group is U(1)
 
3:28 PM
you can indeed view all self-adjoint operators as defining a representation of $U(1)$ (or $\mathbb{R}$ in the unbounded case!) via their associated one-parameter groups
I'm just not sure why you'd want to do that
 
Because I don't know why we even mention groups and not just go straight to representations
$Z_2$ symmetry doesn't help you in any way
 
...how would you talk about representations without talking about groups?
 
Well don't we classify the representations and not the groups
 
we do both
 
But just knowing the group tells us nothing
 
3:30 PM
@DIRAC1930 the definition of a representation involves groups
 
sure it does
 
@DIRAC1930 depends on what you know about the group
 
If I just said to you, there is a $U(1)$ symmetry, that could either be time invariance, spatial invariance etc.
 
sure
 
sure? but you'd still know there's an associated conserved quantity
 
3:31 PM
Thats true
 
but e.g. knowing that every non-relativistic system has to carry a representation of $\mathrm{SO}(3)$ isn't "nothing"
 
That's true
When people say that there is an $SO(1,3)$ symmetry, everyone immediately thinks of the Lorentz transformations on a 4 vector.
Are the Lorentz transformations in the above example a representation of $SO(1,3)$ on the space of 4 vectors?
Just so I know I'm getting my terminology correct
 
@DIRAC1930 yes
 
So $SO(1,3)$ symmetry is a completely abstract notion
How would you find, for example the Little group without knowing the representation of SO(1,3)?
 
ah
well, here's a thing: We often talk about "abstract" groups $G$, but how do you actually write down a group?
you usually do that by choosing some set of transformations on a space and saying "that's my group" - like with $\mathrm{SO}(1,3)$, we usually define this group as the group of Lorentz transformations
that we can mathematically treat this as an abstract group not necessarily bound to its representation on $\mathbb{R}^4$ doesn't mean you have to forget that it is the Lorentz transformations
you'll never be in the situation where you have an "abstract group" and don't know anything about it, not even a single representation
 
3:41 PM
well. you could be given a presentation of a group
 
true, but I can't think of a single example in physics where that would happen
 
yeah. physics examples have the virtue of being concrete
i wouldn't be shocked if there's some case where that can happen, but i can't think of one either
@ACuriousMind i guess there are some trivial cases, e.g., saying that "a non-relativistic system carries a representation of SO(3)" is arguably like that
you don't know which representation, so you only know the group
 
It is very annoying lol
 
but eh. in practice you definitely know which rep you're dealing with
 
@Semiclassical what I meant was that the way you explain to someone what $\mathrm{SO}(3)$ is you'll just give them the fundamental rep, even if in their application some other rep might matter
 
3:52 PM
yeah, true
 
I'm a big fan of the trivial representation
 
a fan of infidelity, i see
2
 
::snort::
 
@ACuriousMind even there, though, we do tend to get to SO(3) by looking at momentum/position operators and seeing what that implies for angular momentum operators
 
I don't get the joke
 
3:54 PM
the trivial representation is not a faithful representation
 
Oh
googles what faithful representation means
Okay
 
and if you're not faithful, you're unfaithful blah blah blah
 
I think I understand the standard model
 
@ACuriousMind so in that sense i do think we could start from so(3) as an algebra, see that we get SO(3) as a group by exponentiation, and only after worry about how this connects to 3D rotations
 
Only in the trivial case however
 
3:58 PM
(not that i'd say this is the standard way)
 
When people talk about group theory now with these random groups or whatever, are they not worrying about physical significance?
Wouldn't that be an unanswerable question if you just started off from the group?
Because SU(2) for example could lead you to isospin or non-rel spin
 
well, you might need to start from the group to figure out what the physical implications are
there's some pretty weird groups though
 
Does knowing $S_z$ commutes with $X,Y,Z$ help you in terms of physical significance?
 
@DIRAC1930 historically, we almost never start with a group and then wonder about physical significance
 
Do similar things happen with isospin?
 
4:02 PM
we develop the physics, and then in retrospect recognize that group theory can explain a lot of the structure we see
 
@DIRAC1930 yes, very much. it tells you that you can measure $S_z$ and $X$ independently
 
Does that happen with isospin or something similar?
Well there will always be one element of the Lie algebra that's diagonal so I assume it does happen
 
@Semiclassical yes, you could do that, but it's not a very physics-y way of thinking
 
yeah, fair
 
Do many string theorists care about physical significance anymore?
Except it looks like the people right at the top of the field
 
4:04 PM
depends on the string theorist
 
ask a string theorist :P
 
there are certainly some mathematical string theorists who care more about the mathematically interesting aspects than physical implications
 
Witten seems to get so much criticism but he at least seems to be trying to address the real world
 
but there's also plenty of string theory that tries hard to do model-building that results in realistic low-energy theories
 
So in QFT, do we postulate a symmetry, enforce that the Lagrangian transforms as a scalar under that group and see what happens?
 
4:08 PM
usually not
why would we postulate random symmetries?
 
maybe for effective field theory stuff
 
I rant about that conception of doing theory a bit here
 
there's not a lot of symmetries we usually worry about
 
I'm not sure why. Isn't SU(3) postualted?
 
that might be how it's taught now, but
 
4:09 PM
@DIRAC1930 the people who developed QCD had a lot of real-world collider data they were trying to explain
 
the reason we went to SU(3) was b/c of the evidence
(don't ask me to explain said evidence, mind)
 
and they figured out that a gauge theory with SU(3) explains the results
there's nothing about SU(3) that would make us postulate it, it just happens to be the symmetry group of the theory that turned out to actually model reality
of course, the success of such theories led theorists to more generally consider the properties of gauge theories with some symmetry group $G$
 
Interesting
 
but that's more than just postulating a symmetry - it's considering a specific class of physical theories, namely gauge theories, and there's an explicit action there
it's the action that contains the physics, not the mere notion of a symmetry group
 
You can derive the Maxwell Lagrangian just from symmetry principles right?
 
4:14 PM
depends on what you mean by that
it's pretty much the only invariant Lagrangian you can build for a U(1) gauge field in 4D
but who told you you needed a gauge field? Symmetry can't tell you that, especially not because gauge symmetries are unphysical in the sense that you can't see the local symmetry in any measureable quantity
 
Yes I read your post you linked here
Do we usually start with a classical field theory and then quantize it?
It seems like finding symmetries of classical fields is easier
 
100% yes - we essentially don't know how to build a QFT except by quantizing a classical field theory
 
Okay, so i'm thinking about this in the completely wrong way
 
i do think we sometimes 'postulate' symmetries in order to go from an unsolvable problem to an approachable one
but that's b/c we want to get some useful toy model
 
the closest to "building a QFT without building a classical theory" I know is the approach Weinberg takes in his QFT books
because he constructs the fields out of c/a operators only after doing a lot of heavy lifting
 
4:22 PM
that's one reason why QFT feels more mysterious to me than QM
 
but still, he ends up with something that's clearly also an action for a classical field theory, so...
 
it's easy to come up with simple QM models which lack a classical analog
 
there are some CFTs that lack classical actions
 
Is Landaus book good for classical field theory
Or does it it not go into the stuff I want to learn?
 
or only have them "locally" or whatever, I don't understand what's going on there fully - the thing with 2d CFT is that having the Virasoro algebra gives you so much information you don't need an action to build a quantum theory
 
4:24 PM
QFT at present is much more tied to classical field theory than QM is tied to classical mechanics
oh, speaking of things involving 2D cft
there was a talk at my uni (which was recorded, i couldn't see it in person) talking about getting 'simple' models of quantum gravity via averages of CFTs :)
which...bizarre, but also neat
e.g. this paper by the presenter + Ed Witten: arxiv.org/abs/2006.04855
 
What's the best book on classical field theory that's short?
Do these gauge groups pop up there too?
 
you can have classical gauge theories, sure - EM is one, after all
but non-Abelian classical gauge theories are rare
 
classical CD, oof
 
So non-abelian gauge theories are really only used in QFT?
 
i imagine there's some weird examples you could come up with in statistical mechanics
something something biaxial liquid crystal
(though i don't think those have ever been observed)
 
4:31 PM
I had a course on classical field theory that had a whole chapter on non-abelian gauge theories but I imagine that part was really quantum but didn't really make a difference
 
well, there is stuff like the gauge theory of the falling cat
that's SO(3), I think
 
@ACuriousMind You can even make it $E(8)$, but then the cat complains
 
if the cat complains, you can't make it that :P
 
@DIRAC1930 GR is a non-Abelian gauge theory
 
just like you just can't stand up if there's a cat lying on you
it's natural law
 
4:40 PM
@ACuriousMind I've done it
I'm not letting a cat boss me around
 
y-y-y-y-you monster!
 
@ACuriousMind What did you do your PhD in?
 
funny you should say that
 
@DIRAC1930 I don't have a PhD
 
I don't know what gauge groups arise in normal people physics but I know condensed matter physics has a lot of weird shit in there
 
4:47 PM
biaxial nematics would have non-abelian topological defects
too bad i don't think they've ever been seen
 
You should write a small monograph summarizing everything
 
"small"
 
As in just the bare essentials
Starting from classical field theory and ending with gauge theory
 
@DIRAC1930 is it gonna be one of those monograph that starts with "Consider a $\infty$-topos"
@ACuriousMind What about the non-parity term
 
@Slereah which one?
 
4:52 PM
The one that's like uuuuh
$\varepsilon_{abcd} F^{ab} F^{cd}$
Or some garbage like that
 
do you mean $F\wedge F$ that's just topological?
 
Maybe?
 
you can add it to the Lagrangian but it doesn't do anything classically
 
Peskin talks about it
 
@ACuriousMind When did you start learning everything you know?
Was it in your spare time after university?
 
4:55 PM
@DIRAC1930 no - I studied physics at uni
 
Yes but you know way more than a standard undergraduate course
I'm assuming you learnt all of gauge theory youself
 
well, for one thing, I didn't just do an undergraduate course - I did a Bachelor's and Master's course (Master and PhD are two entirely separate stages here)
for the other, Heidelberg is pretty atypical in that it is feasible (and recommended for people interested in theoretical physics!) to take QFT in the third year of undergraduate/Bachelor
 
In my 2nd course on QFT, I studied non-abelian gauge theories etc., however as you can tell, it was way to advanced for me so I learnt nothing
 
I essentially just took every math and theoretical physics course I could, I don't have anything special I did to learn what I know
 
I think I'm just dumb
 
5:03 PM
@DIRAC1930 welcome to the club
:P
 
haha lol
 
Everyone is part of it except Acuriousmind :P
 
well, you seem to lack a lot of prerequisites- we did proper theoretical mechanics (so also Lagrangian and Hamiltonian mechanics and basic classical field theory) in our first year and proper quantum mechanics in our second year
I don't think a lot of places do that to the extent we did
 
@ACuriousMind I never learned classical field theory and was thrown straight into the pool of QFT and drowned
 
Me too
 
5:05 PM
@ACuriousMind what book would u recommend for self study?
 
I can't recommend any books because I didn't learn this stuff from books :P
 
@ACuriousMind Any free lecture notes?
:P
 
the only textbooks I have are more specialized stuff
 
@ACuriousMind QFT for the gifted curious mind I suppose :P
 
@MoreAnonymous Here's the notes of the course I took
5
 
5:08 PM
@ACuriousMind Thank u so much ... I feel the genius flowing in me
 
the Weinberg book is probably the most "elementary" book on QFT I've really read, and I wouldn't recommend that as a starting point to anyone
 
I think my problem is that I hate doing long calculations
 
not because it's bad but because it's very different in approach from what the rest of the QFT world does :P
 
That looks like the way two of my courses were done
 
I would probably recommend Schwarz for QFT
 
5:11 PM
I recommend giving up on physics and living a worthwhile remaining life
:P
 
I think the issue is that most QM books can be self contained because the math is a lot easier
QFT assumes so much background knowledge
 
Honestly there are too many ways to think of QFTs ... Personally I like the many body approach
Its similar to QM
 
Every physical theory has a bunch of different formalisms
GR and QFT are particularly horrendous
I'm not sure what's the worst GR formalism, but special mention to that paper I found once that was "GR from quaternions"
 
6:08 PM
would anyone know of situations where the Wigner-Brillouin perturbation series is clearly advantageous to the usual Rayleigh-Schrodinger series? Methinks in degenerate (or nearly degenerate) perturbation theory for large Hilbert spaces but that just reflects my ignorance of applications of the WB scheme.
 
7:01 PM
An entertaining little experiment:
 
fqq
7:59 PM
@DIRAC1930 most QM books and courses just skip all the hard maths. Functional analysis, $C^*$ algebras etc are definitely not easier than representation theory
 
most intro books stick to being 'formally correct' without attempting to be rigorous
e.g. expressions like $\psi(x)=\langle x|\psi\rangle$ and $\langle x|x'\rangle=\delta(x-x')$
within the standard construction, these expressions don't really mean anything: mathematically there's no such thing as an $|x\rangle$ ket
you can make that rigorous by including a rigging for your Hilbert space, but i doubt most physics people care about that
myself i tend to treat them as mnemonics
 
 
2 hours later…
10:28 PM
What do people think of The Road to Reality by R. Penrose?
'Any reader who is not too concerned with the full details of the formalisms
that I shall be presenting is recommended simply to ignore this issue
completely. (Most experts would do the same—until the moment comes
when they have to write articles or books on the topic!)' lol
 

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