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12:14 AM
we cannot decompose the $(1/2, 1/2)$ finite dimensional projective irrep of $\mathfrak{so}(1,3)$ into a direct sum $0 \oplus 1$ via Clebsh-Gordan business, right?
my actual question is: i don't see how I can read off from this expression that $\pi_{(1/2, 1/2)}$ induces the fundamental representation of $\text{Spin}(1,3)$
If i exponentiate $\pi(J_1)$, for instance, okay I get a tensor product of group elements $M_a \otimes M_b$, but what tells me that this is the fundamental representation?
i guess this one descends to a normal representation of $SO^+(1,3)$. And, by fundamental representation, I mean perhaps what is also called the "standard" or "defining" representation in which the representation is literally the action of matrices of $SO^+(1,3)$ on four dimensional vectors
 
 
6 hours later…
6:29 AM
okay i see there is a special change of basis you can make on the resulting generators in (34). i guess my question then is: is it obvious or intuitive that the (1/2, 1/2) irrep is isomorphic to the fundamental representation? Dimensionality certainly is not a final say as there are three irreps of dimension 4: (1/2, 1/2), (3/2, 0), (0, 3/2).
who introduced the $i$ notation in physics when dealing with lie algebras :cries:
 
 
1 hour later…
7:36 AM
this is like an abuse of notation right? in this answer
since we have taken such care to distinguish the two "copies" of $\mathfrak{su}(2)_\mathbb{C}$, it doesn't make sense to me to use the notation for tensor product representation of lie algebra representations $\pi_{(n,m)}(X)$. Rather we should use the explicit $\pi_{(n,m)}(X, Y)$ where $X \in \mathfrak{su}(2)_{\mathbb{C}_A}$ and $Y \in \mathfrak{su}(2)_{\mathbb{C}_B}$?
 
7:50 AM
okay i think i cleaned it up lol...so many subtleties D: and i don't understand them
 
8:05 AM
@ekardnam_ Since the discussion came up...I am finally planning to watch it. I am planning to go chronologically...
 
8:29 AM
the table of contents are looking real nice >:D
 
@Sanjana i think episode 4 is extremely good
the rest r good too. but some dialogue is laughable in episode 2
 
@RyderRude It's the oldest too...
 
it holds up, except for a part where theyre just traveling in a desert
it's two droids
that part doesnt hold up, but dont turn it off right then
becuz it gets really fun
the first time i tried watching it, i turned it off at that part
episode 123 r about politics and democracy
 
8:46 AM
@Sanjana Just watch it in release order
It's how must people do it
 
@SirCumference u can also try 45236. this is a self contained mini series
the rest can be watched anytime
i have watched it as 45236 @Sanjana
 
I wouldn't recommend skipping 1 tbh
 
i watched 1 later
 
It's a decent movie and sets up the rest of the prequels pretty well
 
ok then 451236
but i did 45236 and then 1789
 
9:00 AM
I guess you could. Though I'd just watch the originals first and then prequels for simplicity
Given that's how most people experienced it as it came out
 
yeah
in 45236, 2 and 3 are the flashback movies
i hated 2. i wish it cud b 45136
but 3 cant be understood without 2 i guess
theres also rogue one. can one put it in before 4
that would be cool
there is a scene continuity from rogue one to 4, but it wouldnt feel like a continuity because 4 was filmed decades ago
 
bruh how does anyone ever understand in full the finite dimensional representation theory of the Poincaré group
 
9:19 AM
my understanding is to constrct J_+ and J_, and get back J and K
but idk about the abstract stuff
 
9:41 AM
Thoughts about the mass renormalisation problem of naive treatment of quantum gravity. What if there is a ultraviolet cutoff in the spectra of momenta just enough to prevent all the powers of mass terms in the perturbation to blow up hence ensuring convergence?
 
@Secret it cud be but it wud break lorentz invariance i guess
but i dont think there's any inconsistency with this theory yet
 
fqq
@SillyGoose of the three 4d irreps, (1/2,1/2) is the only one that contains 0+1 reps of the SO(3) subgroup, corresponding to the fact that a four vector reduces to scalar+3-vector under rotations
 
10:49 AM
if we know that the exponential of a matrix in the lie algebra is in the identity component and we also know that any matrix can be represented as $e^X$, does this mean that the matrices that "generate" the elements of the orthogonal group not in the special orthogonal group are just not in the algebra? if so, isnt this a problem? i thought the algebra should generate the entire group and for a compact group, it should only take one elements to do so?
 
11:45 AM
overleaf not working
oh well
 
123
Hello Everyone...
As per Newton's 3rd law action equal and opposite reaction. My question is that if we have only two people's in the universe. Person 1 applied 10N force on person 2 while person 2 applied 15N force on person 1.
What force is considered as equal and opposite reaction the lower one 10N , the higher one15N or 25N as total force?
 
@Relativisticcucumber what is "the Lie algebra" you're talking about here?
 
12:00 PM
@ekardnam_ i also just experienced this but its up now it seems
@ACuriousMind i think the first thing i mentioned is general but i am meaning the lie algebra of the $O(n)$
 
@Relativisticcucumber then by definition the Lie algebra of that group can only generate the connected component, i.e. SO(n), via its exponential map
 
@ACuriousMind right so i think im misunderstanding the idea of the algebra elements as generators of the group because it seems it doesnt generate all of $O(n)$?
 
they generate the identity component
 
@123 there will be two reaction forces, corresponding to the two action forces.
let's say person 1 punches person 2 and person 2 kicks person 1. then there r two reaction forces @123
 
okay i see.
 
12:07 PM
disconnected groups like O(n) are not really what Lie theory is about: write $\mathrm{O}(n) = \mathrm{SO}(n) \rtimes \mathbb{Z}_2$ and observe this is the (semi-direct) product of a connected Lie group and a finite group
 
the first is 10N on person 1 felt on the fist, and the second is 15N on person 2 felt on the foot
 
most of the statements in Lie theory will be about connected Lie groups
 
poincare group screams
but bleh i see
 
123
@RyderRude Thanks for the reply. Pls explain your answer
 
Poincaré is just $P_0\rtimes (\mathbb{Z}_2\times \mathbb{Z}_2)$, i.e. you do everything for the identity component and then you have to deal with space and time reversal separately, see also this answer of mine
 
123
12:11 PM
You are saying total action and reaction will be 25N? @RyderRude
 
i dont understand the question
 
123
F12 = - F21
 
this is correct..
 
123
What is F12 and F21 as per above equation in you example? @RyderRude
 
The total force applied by person 1 on person 2 is 25N in ur scenario, and same for vice versa
assuming the punch and the kick are anti parallel forces
otherwise, we have to add vectorially
so F12 = - F21=25N
 
123
12:17 PM
But person 1 punch 10N force
And person 2 kick 15N
 
yes but F12 is more than just the punch. F12 is the total force applied by person 1 on person 2
total force includes the punch and the reaction to the kick
 
123
Aaah... It means both antiparallel forces add as vector here.
 
the punch and the kick are anti parallel. the punch and the reaction to the kick are just parallel
so they add 10+15
 
123
I was confused because in kleppener and kolenkow numerical example they said person 1 has lower force so person 2 can not apply more force on him in tug war example
 
@Relativisticcucumber i think its up only for already logged in users
at least they say so on the official status page
 
123
12:26 PM
Aaahh... I see.. that was good explanation the punch and reaction to the kick are parallel. @RyderRude
 
In rel QFT, is it consistent to think of the Greens Function as having a pole at each of the excitations of the system (like it appears to be in non-rel QFT). The first pole is at $p^2=m^2$ (which is the first excitation of the system)...
 
12:49 PM
And the secondly, it seems that nothing out of the ordinary is happening comparing the rel to non-rel formalism. In rel QFT, for example with the boson field I will have $\hat{\Phi}=\sum_p \hat{a}^\dagger_p \psi^*_p + \hat{b} \phi_p$ where $\psi$ and $\phi$ refer to the plane waves of particles a and b. However, if I interpret $\hat{b}$ to be the creation and not annihilation operator for creating $-1$ particles with wavefunction $\phi^*$, I can combine this into
$\sum_i \hat{A}^\dagger \xi^*$ where $\hat{A}^\dagger = \{\hat{a}^\dagger ,\hat{b}\}$ and $\xi=\{\psi,\phi^*\}$
This allows transformation of particle a into particle b and the whole formalism seems to be the same
This appears to be alluded to at the end of L&L 3 QM when he briefly discusses 2nd quantization for rel systems
@bolbteppa What are your thoughts?
Now all that is left is to show is that transistions are supressed between a and b for low energies
 
There is no reason to interpret $b$ as a creation operator, the time dependence is that of what an annihilation operator should do so it should always be an annihilation operator, that same time-dependence argument is how you can recognize the other coefficient as a creation operator
Not sure about your notation is that an anti-commutator
 
No it just meant to be a set
 
Non-relativistically, $\psi(x) = \sum_i a_i \psi_i$ is a sum of annihilation operators, so if you were mimicking the non-relativistic case you'd need to interpret the creation operator as an annihilation operator
 
Yes so I have interpreted the anihilation operator for b particles as a creation operator for -1 b particles
 
L&L sec 14 start by trying to write the whole thing as a sum of annihilation terms and are forced to reinterpret the negative energy contributions as creation operators
 
1:03 PM
What if we have that the b particles are states moving backwards in time i.e. Stuckelbergs interpretation
By section 14 did you mean section 10? of L&L 4?
 
Sorry, section 11, eq. (11.1)
 
I was transforming 11.2 so I already have a and b particles
 
So what you are trying to do is interpret the second part of (11.1) as annihilation of particles running backwards in time via $e^{Et} = e^{-E(-t)}$
 
@DIRAC1930 i once read something similar but it was about Dirac sea
 
I am interpreting equation 11.2 as $a^\dagger$ creating 1 $a$ particle, and $b^\dagger$ creating $-1$ b particles
 
1:08 PM
Well this is not what we see in the non-relativistic case either, so no matter what you do things are looking different to the non-relativistic case
 
If you read the end of L&L 3, Second Quantization, he aludes to something similar
 
Above (11.1) they explain what the non-relativistic case looks like, you can see this looks different to (11.1) and (11.2), it doesn't make sense ot create $-1$ particles
I don't see how what he says in the NRQM book is saying this
I think (11.1) is your starting point, and it's impossible to make it look exactly like the NRQM case because of those negative energies, the only thing that makes sense is the 2nd quantization idea of raising and lowering numbers of particles in a given state
 
$\hat{A}_i(p)^\dagger$ = \{\hat{a}_p^\dagger$ , $\hat{b}_p \}_i, \xi_i(p)= \{\psi(p),\phi(p)^*\}_i$ therefore $\sum_i \sum_p \hat{A}^\dagger \xi^*$
 
It goes back to the matrix elements at the start of the NRQM book
I don't know what that notation means I can't read what the brackets mean when you first wrote it or there
 
One sec, let me write it again
Okay so I have $\hat{A}_i^\dagger (p)= \{\hat{a}^\dagger(p), \hat{b}\}_i(p)$
And the wavefunctions are $\xi_i = \{\psi,\phi^*\}_i$
So I can combine into one field operator $\hat{\Phi}= \sum_p \sum_i \hat{A}_i^\dagger(p) \xi_i^*(p)$
Dammit the first one is supposed to be $\hat{A}^\dagger_i (p) = \{\hat{a}^\dagger(p), \hat{b}(p)\}_i$
So $\hat{A}^\dagger_1 = \hat{a}^\dagger$, $\hat{A}^\dagger_2 = \hat{b}$
 
1:19 PM
But you have a $\psi$ and a $\psi^*$ in the same expansion, this is not what we see in the NRQM case
 
No because they are two different quantum numbers
The sum in the non rel case can be over any complete set of states
here I have labelled states by $p$ and by another quantum number $i$
 
It doesn't make sense and even if it did it's just relabeling all this unavoidable stuff, it's not going to change the meaning
 
@bolbteppa incidentally, is this volume 4?
 
All I am saying is that a CSCO is not just $\hat{p}$, it is also $\hat{T}$ where $\hat{T}$ has two possible eigenvalues of $1$ and $2$
 
Yes, but section 10-11 of vol 4
 
1:24 PM
@DIRAC1930 note that quantum.numbers are eigenvalues. u r expanding the field here, not the wavefunction
but yes, this rel field expansion is indeed onto a basis (but only in the linear algebra sense, not in the quantum sense)
 
The last sentence
This all works out
The question is if it works out in the next step
This must be what Stuckelberg was eluding to
 
No it doesn't, it's not making sense, that NRQM section does not fix how different fermions should relate to one another, there is nothing stopping us from making them commute or anti-commute, the fact that relativity forces what we interpret as two different kinds of particles as anti-commuting fixes this for us
 
Wait, my notation above is not for anticommutators, it is a set
I.e. $X_i = \{x,y\}_i$
$X_1 = x$ $X_2=y$
 
This is going all over the place, my comment is about the L&L quote from the pic you just sent
 
Ignore the first part, just the last sentence
All he is saying is that $\{\hat{A}^\dagger_i, \hat{A}_j \} = \delta_{ij}$ in my notation
In the other sentences
I am not changing anything. This is just a question about interpretation
Everything you said applies
 
1:36 PM
He is saying that we should take $a_i$ and $b_j$ to anti-commute rather than commute, $\{a_i,b_j\} = \{a_i,b_j^{\dagger}\} = ... = 0$, where $\psi_a = \sum_i a_i \psi_i$ and $\psi_b = \sum_i b_i \chi_i$ are non-relativistic wave functions for two different types/species of identical particles
 
Oh, ignore the L&L 3 stuff in that case. I was just focussing on the bosonic case at the moment
 
Right, well that paragraph has nothing to do with the relativistic bosonic case, it doesn't apply to this KG thing you're trying to do
 
All I'm doing is interpreting $\hat{b}$ to be a creation operator for $-1$ b particles
If I can do that, then everything I wrote above is just reshuffling terms
The question about whether I can do that is ambiguous due to the time dependence
However, maybe the backward time interpretation fixes this
 
I don't know what you're doing, but look at (11.1), look at the 2nd term, it looks like it has an annihilation operator in front of it, the only problem is the sign in front of the energy, if that was a $-$ sign the whole expansion would be as in the NRQM case which is written in the paragraph above in an unlabelled equation, what is wrong with (11.1)
 
That is a good point
Okay, so starting at 11.2, if we just interpret $b$ to be a creation operator for $-1$ b particles moving backwards in time, everything works
 
1:48 PM
No, (11.2) does not have any 'moving back in time' issues at all, only 11.1 admits a 'moving back in time' interpretation
 
Yes, that is not in contradiction with what I am saying
 
Well I don't know what you're saying but it looks like you're trying to hide something wrong in notation, and this 'creating $-1$ particles' thing which doesn't make sense to me
 
I will have to think about it more
 
Below 11.1 they trace the time dependence leading to the creation and annihilation interpretation back to the end of section 2 which traces back to the transition elements from section 13 of vol. 3
Trace (11.1) back to that section 13 thing and you'll see it's unavoidable
 
2:39 PM
I think you're right about the going backwards in time thing is incorrect
However creating $-1$ b particles is consistent with the matrix elements time dependence
Creating $-1$ b particles is just linguistically different from annihilating $1$ b particle
Mathematically it is the same
I need to think about this some more
 
3:05 PM
I'm just going to assume everything I said above is incorrect
 
3:27 PM
Does anyone have access to the S Matrix paper by Heisenberg translated to English?
 
 
3 hours later…
6:03 PM
This might be a question without a satisfying answer, but is there any reason to expect the double sum here in the definition of a general polarization state? In all other instances I'd expect a "general vector of type X" to just be a single linear combination of the basis. I see that the $\lambda'$ sum only goes over the indices of the metric, but I couldn't justify if asked why there should be a double sum here, is there something I'm missing?
The context here is QED, this section leads into definition a quantum state of a general polarisation
 
 
3 hours later…
9:17 PM
@Charlie first, note that there's a typo there and it should be $\epsilon^\mu(\vec p,\lambda')$; second it'll turn out that this choice means that $\sum_\lambda \alpha_\lambda a_\lambda(p)$ generate a state of momentum $p$ and that polarization $\zeta$ there corresponding to the $\alpha$, i.e. the coefficients are defined to work well with the modes $a_\lambda$, not to give a nice expression for $\zeta$
 
10:02 PM
@fqq hm okay. and by "contains" is what is meant that $\frac{1}{2} \otimes \frac{1}{2} \subset \frac{1}{2} \otimes \overline{\frac{1}{2}} \equiv (\frac{1}{2}, \frac{1}{2})$ and by Clebsh-Gordan business this subset of the image of the irrep reduces to $0 \oplus 1$?
i am abusing notation and using the symbol usually used for the irrep itself as a symbol for the image of the irrep
@Sanjana are you thinking of pursuing a physics phd :0
 
@SillyGoose I don't know what's going on with your notation but all that fqq meant is that the $(1/2,1/2)$ rep of the Lorentz group decomposes as the $0 \oplus 1$ rep of the subgroup of rotations $\mathrm{SO}(3)\subset \mathrm{SO}(1,3)$
 
Let $\frac{1}{2} \otimes \overline{\frac{1}{2}}$ be the $(1/2, 1/2)$ irrep of the restricted Lorentz group where $\overline{\frac{1}{2}}$ is the conjugate representation of $\frac{1}{2}$.
 
Every few months I somehow come back to the topic of asymptotic states and realise that I still don't understand anything about it
 
consider the image $\frac{1}{2} \otimes \overline{\frac{1}{2}}(\mathfrak{so}(1,3))$. Based on fqq's statement, I guess I want to show that $\frac{1}{2} \otimes \overline{\frac{1}{2}}(\mathfrak{so}(3) \subset \mathfrak{so}(1,3)) = \frac{1}{2} \otimes \frac{1}{2} (\mathfrak{so}(3))$
 
I have no idea what your notation is supposed to mean
 
fqq
10:12 PM
@SillyGoose what does that mean? you cannot "let A be B" if A and B are already well defined. also if by \frac{1}{2} you mean the 2d rep of SU(2), it's self-conjugate
 
The representation (1/2,1/2) of SO(1,3) induces a representation of SO(3) (because every representation of a group is a representation of all of its subgroups). The only claim here is that that this induced representation of SO(3) is $0\oplus 1$
 
is $\pi_{(m,n)}$ not equal to $\pi_m(A_i) \otimes \mathbb{I} + \mathbb{I} \otimes (\pi_n(B_i))^*$?
i.e., the tensor product of representations of an $\mathfrak{su}(2)_\mathbb{C}$ irrep with a conjugate irrep
if $(1/2, 1/2)$ meant $\frac{1}{2} \otimes \frac{1}{2}$ immediately, then this would just be isomorphic to $0 \oplus 1$ which cannot be true
 
I don't know where you got that conjugate from
@SillyGoose why can it not be true?
you have to be careful with the notation here, again
the numbers in $0\oplus 1$ refers to the rotation algebra $\mathfrak{so}(3)\subset\mathfrak{so}(1,3)$
 
oh... well I mean to refer to the whole algebra
 
the numbers 1/2 in your tensor product refer to the summands in $\mathfrak{so}(1,3)_\mathbb{C}\cong \mathfrak{so}(3)_\mathbb{C}\oplus \mathfrak{so}(3)_\mathbb{C}$
 
10:20 PM
i mean if I take literally that $(1/2, 1/2)$ is supposed to represent a single tensor product of irreps of $\mathfrak{su}(2)_\mathbb{C}$, then from the normal theory of angular momentum $(1/2, 1/2) \equiv \pi_{1/2} \otimes \pi_{1/2} \cong \pi_0 \oplus \pi_1$. where have I gone wrong
 
@SillyGoose you've assumed that the two $\pi_{1/2}$ are representations of the same algebra
 
@ACuriousMind i know $\mathfrak{so}(3)$ and $\mathfrak{su}(2)$ are isomorphic, yet people always write $\mathfrak{su}(2)$ for some reason
 
they aren't - one of them is the representation of the left summand, the other is the representation of the right summand
 
what is the convention for distinguishing the two algebras? I read a math stack that there is a reason for treating one as a conjugate irrep
let me find it
 
and then there's a third $\mathfrak{so}(3)$ here, which is the rotation algebra as a subset of $\mathfrak{so}(1,3)$, which is neither the left nor the right summand
@SillyGoose what do you mean? When we write $(s_1,s_2)$, the $s_1$ is the rep of the left summand, the $s_2$ the rep of the right summand
 
10:22 PM
6
Q: Conjugate Representations of Lie Algebra of Lorentz Group

Edward HughesI'm trying to understand the Lie algebra of the Lorentz group and am almost there, but am stuck at the final hurdle! It's easy to prove that $$\frak so(1,3)^\uparrow_{\mathbb{C}}=sl(2,\mathbb{C})\oplus sl(2,\mathbb{C})$$ by considering generators. Indeed $\frak so(1,3)^\uparrow$ has generators...

@ekardnam_ maybe it makes it explicit that one is looking for irreps of $SU(2)$, not of $SO(3)$
 
@ekardnam_ well, the usual convention is to call the algebras by the name of the associated simply-connected Lie group (i.e. SU(2)), I just wanted to stress here that one of these at least is the literal subalgebra $\mathfrak{so}(3)\subset \mathfrak{so}(1,3)$
 
and if i follow the conventions set out in the math stack, I get the appropriate irreps
 
@SillyGoose i mean there is no difference between $\mathfrak{su}(2)$ and $\mathfrak{so}(3)$
@ACuriousMind oh i see, was wondering if there was a reason people mostly write su and i guess thats it
 
@ekardnam_ well yes $\mathfrak{su}(2) \cong \mathfrak{so}(3)$ but $\mathfrak{su}(2)$ is of no relevant until you want to identify the Lie algebra of $SO(3)$ with the Lie algebra of its universal cover
at least that is my impression
@ACuriousMind i would much prefer to understand this business without talking about conjugate irreps. but I did do that way the first time around and got the incorrect explicit irreps
 
@SillyGoose What exactly is the question you want to answer?
 
10:26 PM
what is the explicit definition of $\pi_{(m,n)}: \mathfrak{so}(1,3) \rightarrow \mathfrak{gl}(V)$ and the conventions underlying (if any) this definition
 
anyhow saying that $(1/2, 1/2)$ is $0\oplus 1$ has some sense no? very roughly $(1/2, 1/2)$ is the four vector, which you can get a scalar (the spin 0) from (the norm) and some other bits which transform in the spin 1 I guess
however you get stuff which ends up mixing the left handed and right handed parts then
 
@SillyGoose you can just look this up on Wiki no?
 
yes...and on the wiki they use the conjugate representation
 
where?
 
fqq
@SillyGoose is the explicit definition not just the first formula in the picture you just posted?
 
10:31 PM
@fqq yes, and it is using the conjugate representation
 
the (A2) on Wiki specificially does not have the $(\dots)^\ast$ you have
 
that's in terms of sl(2,C)
look at (A2) in the section I linked
 
@ACuriousMind is $\mathfrak{sl}(2, \mathbb{C})_{\mathbb{C}} \cong \mathfrak{sl}(2, \mathbb{C}) \oplus \mathfrak{sl}(2, \mathbb{C})$?
 
well i can try to do some explicit computations of (A2), but i'm pretty sure those are the formulas i started with in the first place and got the incorrect irreps from, but perhaps my $J_i$ and $K_i$ were defined differently or something than in the wiki
wait what
 
10:34 PM
@ekardnam_ yes
 
@ACuriousMind great im not completely crazy afterall
@SillyGoose anyhow the fact that the representations of $\mathfrak{g} \oplus \mathfrak{h}$ are tensor producs of representations of the individual summands should be a pretty general fact no?
 
@ekardnam_ i am okay with taking this fact for granted
 
so whats the problem exactly?
 
@SillyGoose from your screenshot, it looks as if you're using the mathematician's convention that the algebras are anti-hermitian
since you have $\mathrm{i}\sigma_i$ and not just $\sigma_i$ as image of the representation map
 
after you noticed $\mathfrak{so}(3, 1)\otimes \mathbb{C} \cong \mathfrak{sl}(2, \mathbb{C}) \oplus \mathfrak{sl}(2, \mathbb{C})$
 
10:37 PM
this will only lead to suffering if you try to compare this with any remotely physics-related text :P
 
it should all folllow
 
bleb i thought i was following the physics convention oh no
 
@ACuriousMind nothing more true has ever been said
@ACuriousMind its difficult even to compare different physics related texts
 
@SillyGoose you're not - the $K_i$ are the boost generators and in physics convention, they famously turn out to be anti-Hermitian, but your $K_i$ are Hermitian
 
oh god
 
10:39 PM
I'm fairly sure you have no conceptual problem here, you just need to pick one convention and then compute everything properly
 
does srednicki use the physics convention...or did i just somehow manage to turn the physics convention into the math conveniton at some point lol
hm wait so is this related to making the choice to totally antisymmetrize the lorentz algebra generators by raising an index
 
going to sleep anyhow
goodnight or day or anything else
 
nightz
 
@SillyGoose goodluck with the convention hell
 
XD thank you
 
fqq
10:44 PM
@SillyGoose yes, Srednicki uses the physics convention
 
oh no but srednicki also uses the opposite metric that i use...
okay well i'll just try to stick with some convention lol
okay wait so to get this straight, in general we exponentiate a lie algebra element like $e^{tX}$ where $X \in \mathfrak{g}$ and $t \in \mathbb{R}$. In physics, we choose to actually compute this as $e^{-i (iX) t}$ so we multiply generators by $i$ and exponentiate like $e^{-it (-)}$
the benefit being that this turns skew-symmetric generators (say, of $SO(3)$ ) into Hermitian generators
 
no, the minus is on the $X$ - we exponentiate generators $T$ as $\mathrm{e}^{\mathrm{i}Tt}$
 
oh
okay so the physics convention would write the generators of $\mathfrak{su}(2)$ as $\frac{1}{2} \sigma_i$. then, mathematicians would write this as $-i\frac{1}{2} \sigma_i$?
 
and moreover physicists would exponentiate by $e^{it(X_{phys})}$ and mathematicians would exponentiate by $e^{t(X_{math})}$
hm, wait but in the convention i have used for normal quantum mechanics, I write a clockwise rotation around the $z$ axis as $e^{-it\frac{1}{2}\sigma_z}$. Is this another choice of convention?
 
11:01 PM
remember that the "right" direction in math is usually counterclockwise, so the minus is because clockwise corresponds to negative angles
 
ohhh okay so this sign comes in from actually just writing the paramter $t = -\theta$
okay i shall write this down for future me :P thank you v much
 
11:18 PM
This is interesting pirsa.org/08050029
 
11:41 PM
@ACuriousMind Do you think the dressed particle approach will help me with my confusions?
 
@DIRAC1930 I have no idea what "the dressed particle approach" is
 

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