Inverse Spectral Theorem - L10 - Frederic Schuller

[@lioneloddo]

La puissance intellectuelle de ce professeur est juste phénoménale. Ses cours sont une vraie performance.

[@ThomasDanielsen1000]

"If you want to scare little children, you say it's a C*-algebra homomorphism. If you want to calculate stuff you just remember what it means". Hahaha, good one!

[@benjaminwhiteside5176]

4:34 I think that might be just about the most perfect integral sign I've ever seen written on a blackboard.

[@rektator]

The sequence (f_n) @1:42:30 cannot be Cauchy in L^2(R), since that would imply that the pointwise limit f, which was assumed to be arbitrary measurable function, is an element of L^2(R), which cannot hold. But I think that you don't actually need it at any point, that the sequence (f_n) is Cauchy. It suffices that (\int f_ndP)(\psi) is Cauchy. It follows from the definition of \psi.

[@brendawilliams8062]

Professor Schuller. Thankyou so much for the lectures you willingly share with others. I like projective geometry. I can relate with interest to many lectures. You can get the message across so clearly. I hope these lectures will remain available for many years to come. Sincerely

[@lugia8888]

No problem

[@untwerf]

hello Mr scientist man, are you planning to film and upload any more lecture series in the future in addition to the QM and GR ones you have already kindly shared?

[@giannisniper96]

i hope so too!

[@pawej.szabowski2533]

Wow I love these lectures. Schuller is the excelent lecturer!

[@nickpreetic4670]

Such an excellent presentation of a difficult and important topic

[@AsamatBagatov135]

23:00 has to be the most \mathcal{L} L i have ever seen written with chalk

[@infinitjest96]

great lecture!

[@mathjitsuteacher]

At 1:14:39 I believe that he could just use the inequality he got to show that 1 is an upper bound for the norm of the operator. Then if you take f such that f(x)=1 for all x in IR and apply the operator to this f you get the identity. Finally since the ||f||_{infinity}=1 and ||id_H||=1 it follows that 1 is indeed the supremum.

[@nickpreetic4670]

at 49:01, the statement that the identity operator is not bounded is not correct. This is possible if one chooses different topologies on the Hilbert space but if this is not the case, the identity operator is bounded with operator norm 1.

[@jellyfrancis]

We need more please ❤

[@whig475]

Having some mathematical background, and having watched this video and the following one, it surprises me that he never (or at least very rarely) mentions eigenvalues/eigenvectors. Does his audience fully appreciate that self-adjoint operators are diagonalizable and that the eigenvectors corresponding to the (resultant) diagonal elements are orthogonal? As I understand his proof, much use is made of this latter fact. I am thankful for the chance to watch these videos; they are nicely done! I think he uses remarkably few examples in general, but I won't argue his teaching style.

[@JannikPitt]

He is talking about the infinite dimensional case, which is the case important in quantum mechanics. In infinite dimensions a self-adjoint operator doesn't have to have any eigenvectors at all. The spectral theorem he covers in this video is a generalisation of the finite-dimensional "every self-adjoint operator is diagonalisable with orthogonal eigenvectors".

[@whig475]

@@JannikPitt I understand what you are saying. But my point was just that he never mentioned the analogy between eigenvalues and spectral values at all (at least that was my recollection from 3 months ago). For me, that correspondence helped make sense of the integration--but if you have mastered the proof, then you are ahead of me! I only studied QM for a month or so, but I found it quite interesting. Some of the math I have been looking at was developed by physicists. Thank you for replying to my note.

[@benholder2766]

@@whig475 He deals with this case in the beginning of Lecture 8 when he talks generally about spectra. Perhaps that would have been the beginning of this lecture, but instead he had to do some "service" in the presentation of perturbation theory.

[@whig475]

@@benholder2766 Thanks Ben. Besides to thank the speaker, another reason I left my comment was in case it may be helpful to some other viewer--and your comment provides such a viewer with some direction! Cheers.

[@andrea.dibiagio]

At 1:41:00, is the sequence {f_n} actually Cauchy? I think if you let f(x) = x, which is clearly a measurable function, it's easy to see that {f_n} cannot be a Cauchy sequence. Besides, do you even need {f_n} to be Cauchy to prove that { (int fn dP)psi} is Cauchy in the Hilbert space?

[@sichengmao4038]

notice that the space L^2(R) here is actually L^2(R,mu_psi), the measure over it is the real-valued measure induced by P but not the common Lebesgue measure. In detail, because psi satisfies that "int |f|^2 d mu_psi < infinity", by the dominant convergence theorem it converges hence Cauchy. Same for the other question.

[@bishalkhadka1301]

Aditya Bhandari are you from Nepal