Jonathan Gleason
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 Oct 30 comment Is the density operator a mathematical convenience or a 'fundamental' aspect of quantum mechanics? @A.O.Tell Can you elaborate on what exactly you mean by "representing states of tensor factor subsystems"? May 30 comment Yukawa Coupling of a Scalar $SU(2)$ Triplet to a Left-Handed Fermionic $SU(2)$ Doublet This part of my question can be phrased as: given this statement of the problem, what is the correct form of the Yukawa coupling term? (I realize what I wrote down didn't make sense. The problem is, I don't know how to modify it so that it does make sense). May 30 comment Yukawa Coupling of a Scalar $SU(2)$ Triplet to a Left-Handed Fermionic $SU(2)$ Doublet This actually came from a homework problem (there was a five tag limit). The problem reads: "One method of generating neutrino masses that we did not discuss in class involves adding a Higgs field $T$ which is a triplet under $SU(2)_L$ with a Yukawa coupling to the left-handed lepton doublets. If this triplet is a complex field, then it is possible to assign a lepton number to this new scalar field such that the lepton number is conserved.". He doesn't actually write out the Yukawa coupling term, so we are left to figure this out on our own. May 20 comment The Faddeev-Popov Lagrangian In the Scholarpedia article, I am looking at the part that begins with "One more improvement was introduced by 't Hooft . . .". It almost seems as if the term I'm wondering about was inserted by hand by allowing a more general gauge fixing condition. With this more general condition, the relevant delta function contributes a nonzero term to the Lagrangian. If I understand this correctly, that's all well and good, but why the need for the more general condition? Is $\partial _\mu A^{\mu k}$ not sufficient? Does this not just complicate things further by introducing an extra term? May 3 comment Could a people do all sort of gymnastics movement in vacuum space? I don't understand your question. Could you try clarifying? May 3 comment QED BRST Symmetry With respect to (2), does this not imply that the Lagrangian is not invariant under the BRST transformation? Why would he ask me to show this if this is not in fact the case? May 2 comment The discontinuity of Electric Field "Why there exist the tangencial component?" -- Well, why not? Unless I know that the tangential component is 0, I can't just assume it is, even though that wouldn't affect this result. I am pretty sure Griffiths does show that the tangential component has to be continuous, i.e., in the notation above, $\mathbf{E}_{\text{tangential,above}}=\mathbf{E}_{\text{\tangential,below}}$. Thus, if the surface is a conductor (which by definition means the electric field inside it must be $0$), we do require that the tangential component is $0$. This is not necessarily true in general however. May 2 comment What areas of physics should a mathematician study to understand TQFT? As a mathematics student currently studying QFT, I can say that, if indeed any understanding of QFT at all is required for TQFT, you will need to know both special relativity and quantum mechanics pretty well. Nov 22 comment Angular Momentum Operator @atomicpedals ". . . why would I want to use $L^3$ instead of $L$;" First of all, be careful when you write $L^3$. $\mathbf{L}$ is a vector, so if you write $L^3$, I'm going to think you mean $(L_x^2+L_y^2+L_z^2)^{3/2}$ (because it doesn't make sense to cube a vector, although it does make sense to cube the magnitude of a vector), which is not your meaning. As for why would you work with $L_i^3$ over $L_i$: you wouldn't. I'm pretty sure this exercise is just for the sake of practice. Oct 28 comment Is the Lorentz group compact (and if not, is U(1)?) @RonMaimon There are only two connected one-dimensional manifolds: $S^1$ and $\mathbb{R}$. Both of them happen to be Lie Groups. $S^1$ has many names: $S^1$, $U(1)$, $\mathbb{R}P^1$, $\mathbb{T}^1$, $SO(2)$, $Spin(2)$, and the list probably goes on. All of these have different definitions, which happen to coincide for these small cases. To the best of my knowledge, there is really only the one name for $\mathbb{R}$. Also, there aren't "two versions" of $U(1)$. $U(1)=S^1$, period. The "other version" you're thinking of is just $\mathbb{R}$. It is not a "version" of $U(1)$. Oct 26 comment Does $E$ really equal $mc^2$? This should be closed. It's a double-post, and not a particularly high-quality question anyways. Oct 26 comment Is the Lorentz group compact (and if not, is U(1)?) I completely agree with BebopButUnsteady. Every physicist should know enough point-set topology to understand at least basic manifold theory. Oct 25 comment How must you spin the ball to make it alternate between 2 positions? SHM=Simple Harmonic Motion Oct 24 comment What is the definition of a 2-cocycle in Quantum Mechanics en.wikipedia.org/wiki/… In particular, "The set of isomorphism classes of central extensions of G by A is in one-to-one correspondence with the cohomology group H2(G,A), where the action of G on A is trivial." Oct 23 comment Defining a Riemannian manifold - made easy? @RonMaimon Still though, even if you could avoid this chicken-and-egg problem, in the end, ZFC is not powerful enough to talk about something like category theory, e.g., the notion of "the category of groups" (to the best of my knowledge) is not a set in the sense of ZFC, and is hence just an intuitive idea. Thus, to the best of my knowledge, if I want to work with categories, I have to make use of the "naive" notion of a set anyways. Oct 23 comment Defining a Riemannian manifold - made easy? "The mathematical stuff is very simple--- do not be intimidated! The picture is there, the symbols are just a baroque but necessary language to communicate the picture."+"The learning curve is only superficially steep--- after you solve even the simplest problems in GR, it melts away."--I couldn't agree more. Oct 23 comment Defining a Riemannian manifold - made easy? @RonMaimon I eventually decided you can't get something from nothing. You have to start somewhere. As a result, I wound up, more or less, accepting "naive ZFC" as given. If you're interested in how I came up with this "philosophy of mathematics" in more detail, you can read chicago.academia.edu/JonathanGleason/Papers/759541/…. I initially wrote it just for myself, so I apologize if it's not quite as "polished" as it should be. In any case, I certainly do not have the belief that this is the only correct way to think about things. Oct 23 comment Defining a Riemannian manifold - made easy? @RonMaimon Perhaps you misunderstood what I meant. I don't mean to question what you said regarding the mathematics, just your interpretation of it. For example, choice implies Banach-Tarski. This is fact. That this implication is philosophically dubious is an opinion. In any case, I have decided to "reject" any form of axiomatic set theory because of the philosophical "chicken and egg problem". That is to say, an axiomatic formulation of set theory requires, at least, first-order logic. However, you require at least the naive notion of a set in first-order logic. Oct 23 comment Defining a Riemannian manifold - made easy? @Peter4075 "Probably my ideal textbook would contain rigorous maths and pretty cartoons!"--I definitely don't disagree with this. IMHO, to obtain the best understanding, you need to understand both the rigorous formulation and what it all means intuitively. You use the intuition to solve problems and the rigor to make it precise. Both go hand in hand :). Oct 23 comment Defining a Riemannian manifold - made easy? @RonMaimon Fine. As long as we can agree that this (the philosophical dubiousness (is this a word?)) is just your opinion, and not mathematical fact.