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
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:
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 $Y \in \mathfrak{su}(2)_{\mathbb{C}_B}$?
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?
@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
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?
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?
@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 $O(n)$?
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
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
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
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)...
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
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
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
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
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
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
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)
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
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
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
@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 $\zeta$
@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 $\frac{1}{2}$.
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 $\frac{1}{2} \otimes \overline{\frac{1}{2}}(\mathfrak{so}(3) \subset \mathfrak{so}(1,3)) = \frac{1}{2} \otimes \frac{1}{2} (\mathfrak{so}(3))$
@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 $0\oplus 1$
the numbers 1/2 in your tensor product refer to the summands in $\mathfrak{so}(1,3)_\mathbb{C}\cong \mathfrak{so}(3)_\mathbb{C}\oplus \mathfrak{so}(3)_\mathbb{C}$
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
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
I'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_ 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 $\mathfrak{so}(3)\subset \mathfrak{so}(1,3)$
@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
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
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
@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?
@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 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
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?
I'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...
