Self adjoint and essentially self-adjoint operators - Lec 07 - Frederic Schuller

[@ivanphi]

I was wondering if there is any way to get the problem sheets to your classes?

[@jackozeehakkjuz]

We all wonder the same.

[@active285]

"A symmetric operator A"! 1:32:50 Fun in the theoretical physics lecture :)

[@thephysicistcuber175]

1:14:32 I actually found out that it's necessary too. Indeed if A is symmetric the following are equivalent: 1) A is self-adjoint 2) A is closed and ker(A*+i)=ker(A*-i)={0} 3) ran(A+i)=ran(A-i)=H Proof: 1)->2): Assume A is self-adjoint. Obviously A is closed. The second statement is a standard calculation, which is even performed in lecture 8 of this course. 2)->3) We employ the fact that the graph of a closed operator is closed the direct sum H+H. First of all it's easily seen that the closure of the two ranges have to be H. To show that the ranges are closed we prove that (A+i) and (A-i) are bounded below. I'll continue the proof with (A+i) only, and the proof for (A-i) is analogous. We have for x in D_A that ||(A+i)x||^2==-i+i+=||Ax||^2+||x||^2>=||x||^2, where we've used simmetry of A. Now if y_n is a sequence in ran(A+i) which converges to y in H then it's Ax_n+ix_n for some x_n. Since y_n is Cauchy and A+i is bounded below then it's easily checked that x_n is also Cauchy, and so is Ax_n=y_n-ix_n, difference of two Cauchy sequences. But then x_n->x and Ax_n->z and because the graph of A is closed this means that (x,z) belongs to it, and therefore Ax_n->Ax and (A+i)x_n->(A+i)x, concluding the proof of the closedness of ran(A+i). 3)->1) Go to 1:16:22, where it's proved in a slightly more general case. Incidentally you can substitute i above for any z with nonzero imaginary part (the proof is left as an exercise). This is not to be confused with the criterion for essentially self-adjointness, where both the closures of the ranges of (A+z),(A-z) are assumed, instead of the ranges themselves.

[@lugia8888]

ok nerd

[@Heuyy123]

@@lugia8888 😂

[@lucasdarianschwendlervieir3714]

I saw 10 students in the class from the reflection of the blackboard as he was cleaning it.

[@idiotstroszek]

good job

[@antoniolewis1016]

me too!

[@gastonbarboza3568]

And 5700 more students online

[@yuwuxiong1165]

At 1:02:38, I saw 5 more on the left side of classroom:)

[@thephysicistcuber175]

1:27:31 It seems from the proof that the hypothesis can be weakened to cl(ran(A+z))=H=ran(A+z*), where cl(.) is the closure operator and z* is the complex conjugate of z.

[@apoorvpotnis]

Reference: 40:30. I think that the conditions ker(A) = 0 and closure of ran(A) = H are not sufficient for an operator to be bijective. An operator is bijective if and only if it is invertible. We need the stronger condition that ran(A) = H, not just the closure. To see why, we can take H to be an uncountable, separable Hilbert space and let the domain D_A be a countable, dense subset, and take A to be the identity operator in D_A. We clearly see that closure of ran(A) = H, but A being a bijection (invertible) shall imply that a bijection exists between a countable and an uncountable set, a contradiction.

[@carlosleiva]

Does anyone know if there is a textbook for this course? It seems that the subjects flow directly from the Herr Schuller´s brain.

[@Bignic2008]

I think Prof Schuller personally probably pulls his knowledge and understanding from several sources, as everyone does. There are certainly books that follow a similar treatment of quantum theory. One well-known example is “Quantum Theory for Mathematicians” by Brian C. Hall. Another is “Mathematical Foundations of Quantum Mechanics” by John von Neumann. As is mentioned in the first lecture, the mathematics is essentially that of functional analysis, with a particular bias toward concepts that are relevant to quantum mechanics (self-adjointness, not focusing on bounded operators, the spectral theorem...) but if you follow any book on functional analysis closely, of which there are many, then you’ll get the same (likely broader but less focused) mathematical content.

[@theodorei.4278]

Is there a part of the exeprimental lectures of this series?

[@millerfour2071]

17:55, 27:33, 36:58, 50:25, 1:06:06

[@thephysicistcuber175]

1:38:00 Shuldn't there be an existential quantifier rather than a universal quantifier for z?

[@Nooofa51]

Thank you for this great lecture professor I have a question for something you say: That “The Spectra is real if and only if A is self-joint” in 13:48 .. so I know if an operator is self-adjoint then the Spectra is real, but the other side!!!?.. Please answer me or any one know the answer .. i need reference or proof for it ..

[@piercingspear2922]

U need to watch his next lecture on spectral theorem and perturbation theory

[@mashedpotatoez99]

2 years too late, but hopefully helpful for future viewers (and probably you as well). Well, first of all, the statement as written is false; the implication “if the spectrum of A is real then A is self-adjoint” is the wrong one. A simple counterexample is to consider a 2x2 matrix with real entries on the diagonal, but the off-diagonals being not complex conjugates of each other. But anyway, this was probably just a slip of the tongue, as the correct statement is that “for a *symmetric* operator on a Hilbert space, it is self-adjoint if and only if its spectrum is real”. For reference, one can take Reed-Simon Volume I, Theorem VIII.3 (which was essentially covered in this lecture): *Theorem VIII.3 (The Basic Criterion for Self-Adjointness):* Let A be a symmetric operator on a complex Hilbert space H. Then, the following statements are equivalent: (a) A is self-adjoint (b) A is closed and ker(A*+i) and ker(A*-i) are both equal to {0} (i.e A*+i and A*-i are injective) (c) the ranges of A+i and A-i are H (i.e these operators are surjective) With this in mind, suppose A is a symmetric operator such that its spectrum is real. Since the spectrum is real, it follows that i and -i are not in the spectrum. By definition this means they belong to the resolvent, which by definition means A+i and A-i map the domain D(A)->H bijectively and the inverse is bounded. Ok so in particular A+i and A-i are surjective maps D(A) -> H. Hence, condition (c) is satisfied and thus A is self-adjoint. Some (personal )remarks: Reed-Simon Volume I is a wonderful reference, as are their other volumes, for a lot of the material covered in these lectures. For this lecture specifically, they introduce unbounded operators using their graphs, and frankly I think is the way to go because it really motivates the definitions and terminology (closable, closure, closed), and furthermore if we go all the way back to basics, then a function in set theory is really defined in terms of its graph so this just ties together all the concepts nicely. Of course, proving the equivalence of the graph-based definition with the version given here using adjoints is straight-forward, so maybe this was just a time issue in the lecture that this aspect was not elaborated on and only an off-hand comment @25:50 was made. Also, Reed-Simon Volume II, Chapter X is a wonderful resource; right in the first section they develop Von-Neumann’s theory of deficiency indices (which was not elaborated here for time issues).

[@ANSIcode]

I'm stlightly confused as to the various criteria. For example I found the criterion: A densely def. operator A is self-adjoint if and only if: A is symmetric and ran(A+i)=H=ran(A-i). This seems a much better result than the one quoted here. Can someone clarify or refer to a comprehensive list of theorems?

[@active285]

That's indeed a result he uses in the proof of the corresponding theorem. Just take a closer at the implication in the last line (and carefully listen Mr. Schuller's explanations). In order to conclude ψ = ρ it is necessary that they (at least) lie dense in H.

[@jimnewton4534]

Is this the correct video to follow kzbin.info/www/bejne/paWVZmaCnd1lldE, I think I missed something. Does it really go from Integrable functions to Adjoint operators?

[@jackozeehakkjuz]

Yes, he does, indeed.

[@seanziewonzie]

Jim Newton From what I gather, it seems that he was surprised that his students were not familiar with measure theory, so he gave them a crash course as soon as he learned about this and then continued on with the functional analysis he was already teaching.

[@jimnewton4534]

That's great that he did the section on measure theory. I never really understood it until watching that video.

[@seanziewonzie]

Yep. Perhaps some students would learn better with more discussion about examples, but I personally don't need examples to learn. I get confused by all the objects swimming in the dense notation: if I don't recall what each piece of the theory does clearly, I can't understand. Dr. Schuller is the only professor I've seen so dedicated to reaffirming "remember, this operation takes these objects here to these objects here" and other things like that. A real appreciation for the intricate details of mathematical structure. For example, I know I would have been confused if I learned QM from a source that handwaved the details about symmetric vs. essentially self-adjoint and just said "the spectra will be real!". And all these rants he makes about notation- the bra and ket notation and the f(x) notation within integrals- they are always such "aha!" moments for me because I simply cannot roll with unclear notation.

[@somasundaramsankaranarayan4592]

This is a reflection of myself

[@ChrisRossaroGG]

Has someone else though of Mickey Mouse's alien friend when \eta\beta was written?