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1929
bio website berkeley.academia.edu/…
location Berkeley, CA
age 24
visits member for 3 years, 4 months
seen 19 hours ago

Currently a graduate student in mathematics at the University of California - Berkeley.

Previously obtained a MASt. in Applied Mathematics from the University of Cambridge (2013), and a B.S. in Mathematics and a B.A. in Physics from the University of Chicago (2012).


Dec
5
comment Why quantum mechanics?
@EmilioPisanty Essentially, yes.
Dec
5
comment Why quantum mechanics?
See edit to question. Obviously, you'e going to have to appeal to experiment somewhere, but I feel as if the less we have to reference experiment, the more eloquent the answer would be.
Dec
5
comment Why quantum mechanics?
As an example of motivation that would satisfy me, the arguments that Weinberg makes in the first part of his first volume to motivate the introduction of quantum fields, while not a proof that nature has to be explained by a field theory, is more than satisfactory if all one seeks is justification to believe that a quantum field theory can be used to describe the universe.
Nov
8
comment What is the role of the vacuum expectation value in symmetry breaking and the generation of mass?
I don't quite see how the requirement that you not have a linear term requires us to make the substitution $\psi :=\phi -|v|$. For example, the trivial substitution $\psi := \phi$ doesn't have a linear term, but this can't be correct, because we don't see a massless particle and antiparticle pair, we see a Goldstone boson and a massive real scalar particle.
Oct
30
comment Is the density operator a mathematical convenience or a 'fundamental' aspect of quantum mechanics?
Perhaps I could rephrase my question as "Does there exist a quantum system that exists in reality which cannot be mathematically described by a pure state?". According to your answer, it seems that the answer is "Yes" and that an example is given by "open quantum systems". What exactly do you mean by this and how are they not described by pure states?
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.