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Logan M
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How to define transfinite derivatives of a function?
Speaking as a physicist, you can "regulate" $D^\omega \sin t$ by instead considering $D^\omega \sin \lambda t$. While not defined for $\lambda = 1$, for all $|\lambda| <1$ there is no difficulty in defining $D^\omega \sin \lambda t = \lim_{n\rightarrow \infty} D^n \sin\lambda t = 0$. For $|\lambda| > 1$ it is less well-behaved but can be considered as (complex) $\infty$ (there is some difficulty at integral multiples of $\pi/2$ but if more general perturbations are allowed this can be dealt with). I would not call this "robust" but it shows that your example is perhaps a degenerate case.
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Why is conformal invariance only possible for massless theories?
@AbdelmalekAbdesselam Your comment is fair. I was assuming the context of the answer, and to me as a physicist talking about "particles" implicitly means you have a weakly coupled limit with a discrete spectrum so that canonical quantization at least sort-of works, so in that context I would say that the theory is just not described by "particles" at all. Of course from the representation-theoretic perspective such a requirement is completely unnecessary, and as your answer shows, untrue for non-free CFTs.
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Why is conformal invariance only possible for massless theories?
As a physicist, this answer is the closest to what I would mean if I were to say that conformal invariance requires massless particles. The point is that the "mass" of a particle is a parameter of an irreducible representation of the Poincare group, but under the conformal group they are not irreducible; scale transformations send mass $m$ representations to mass $\lambda m$ ones for every $\lambda>0$. Unless you want to include representations for every $m>0$ (which we don't, at least in physics), the only way to get a scale-invariant mass spectrum is if all representations have $m=0$.
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Determinant of the "real part" of a matrix
@T.Amdeberhan See the comment on the OP's question on block diagonal matrices. Your formula predicts that $C(2)=3/2$ and $C(12) = 11+2^{-11}$, which violates the inequality with $n=2$ and $k=6$ because $(11+2^{-11})^{1/6} \approx 1.491 < 3/2$. Perhaps a better guess would be $C(n) = (3/2)^{n-1}$, which also matches the given numerical data for $n=1,2,3$ and does not seem to be in conflict with any of the bounds.
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Does quantum mechanics ever really quantize classical mechanics?
I am also not clear what you're asking. Naively, I'd think that you're looking for the phase space formulation of quantum mechanics, which was developed independently in the 1940s by Moyal and Groenewold. In particular, Moyal showed that one may formulate quantum mechanics via Wigner Quasiprobability Distributions, which are just functions on phase space. However, decoherence is still a bit subtle in this framework, so I'm not sure if that would answer your question.
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The amplituhedron minus the physics
@AllenKnutson You're welcome to edit as you see fit or post your own answer (for reference, I agree with everything you said in your comments). As I said above, the answer is based on what Nima explained to me. I tried to phrase it in a way that would be readable by mathematicians but keep it as close to the way he presented it at that time as possible. In my mind, the answer is already basically obsolete now that they have a couple of papers on the subject available which are quite readable.
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The amplituhedron minus the physics
added reference to the arXiv paper
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A gentle introduction to CFT
@Dilaton I agree that this question would not be accepted by Physics SE standards, but I don't think that immediately makes it acceptable on MathOverflow. MathOverflow is a site for professional mathematicians. While physicists and physics questions are sometimes welcome, we aren't the intended audience. As this question is right now, this is the wrong venue IMO (though there do seem to be some above who disagree). This question's closing on PSE may be an argument for creating a different Physics Q&A site or changing the rules there, but not for changing the rules here just to accommodate us.
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Higher dimensional Rubik's cube group
@JasonPioneer That seems correct to me. The full expression should be (assuming I haven't made a typo) $$\displaystyle (A_n \wr S_{2^n}) \times \left(\prod_{i=2}^{n-1} S_i \wr S_{{n \choose i} 2^{i}} \right).$$