Case study: momentum operator - Lec09 - Frederic Schuller

[@ivanphi]

Just a comment for my future self: One can always find N such that eta=N' by simply defining N=\int_0^x dx eta. The integral is good because in a finite measure space all square integrable functions are also integrable. Then by the fundamental theorem of Lebesgue integral calculus, N is absolutely continuous and eta = N'.

[@jackozeehakkjuz]

@scuffed math Always.

[@abrlim5597]

3:29 is the Stone-von Neumann Theorem eventually proved in this series of lectures?

[@ANSIcode]

How much do the different self-adjoint extensions of the momentum on a closed bounded interval differ from each other? In terms of their spectra for example? Is there a way to tell which one is "physically correct" and what is the physical interpretation of the choices one has to make, or is such a choice also not unique in a meaningfull way? What about other domains, i.e. on what measurable subsets of the real numbers (or, more generally R^3) do meaningfull physical boundary conditions result in an essentially self-adjoint momentum? Is there literature on this? On what subsets can one define a physically meaningfull momentum by making additional choices?

[@ANSIcode]

What experiment would that be? How would you set up the system (one that can be described sufficiently well by the formalism covered here, i.e. a system equivalent to a particle in a bounded interval) and how would you measure its momentum?

[@gastonbarboza3568]

ANSIcode Google "self adjoint extensions of momentum on the interval" by Robert Helling

[@andrea.dibiagio]

These are interesting mathematical questions, but it's worth remembering that there is no such a thing as a half-line, or any way in which a quantum system is confined to an actual 1D region of space. You can approximate such a system by inducing huge potential barriers, but the system will never completely be confined, the domain will always be more 3 dimensional. The reason physicists introduce systems with lower dimension is that in certain situations exctiting the "transverse" degrees of freedom requires energies much higher than those available to the system. The wavefunction however will always "bleed" out in those directions. The only place where i know of literal 1D subspaces is in string theory, where if I understand correctly, people envision fermions propagatin *on* the string. In that case it looks like there would be ambiguities if the string is an open string.

[@mirceadolineanu9715]

Very interesting and revelatory lecture. But I have a question: would the conclusions change for a general first derivative operator? In the course, Mr. Schuller uses the momentum, -i d/dx, but the only points where the form has an explicit influence seem to be the symmetry and the fact that all the domains we can consider are in AC (set of absolutely continuous functions) in order to make the derivative work. Now, shouldn't a more general operator A(x) d/dx + B(x) go through exactly the same discussion as long as A and B are chosen such that the operator is symmetric?

[@tofu-munchingCoalition.ofChaos]

You will have problems for general A, B unbounded or if 0 is in the essential image of A.

[@SameerKumar-jf5mi]

Fascinating

[@ivanphi]

In the momentum operator introduced at the beginning cartesian coordinates are evident. If we want to look at quantum mechanics from a coordinate invariant point of view is it sufficient to consider the hilbert space as something on a "L^2(M)" (whatever that may be) with M a smooth manifold with connection? In such a case how would the momentum operator look?

[@lucasdarianschwendlervieir3714]

I would guess the answer would involve the construction of a 'hilbert fibre bundle' which means a fibre bundle whose base space is M and whose fibres are the hilbert space H. But again I havent really studied bundle theory so I wouldn't know.

[@giannisniper96]

exactly! and your connection allows you to exchange partial derivative with a covariant derivative, and it is not a tensor, which is coordinate independent. for any practical calculation tho, coordinates are essential. also no need for sophisticated bundles other than the tangent and the frame bundle

[@ActionPhysics]

can someone give example of a domain where momentum operator is self-adjoint

[@StephenCrowley-dx1ej]

Yes, the momentum operator −iħd/dx is self-adjoint on the domain of square-integrable functions L²(ℝ) that are absolutely continuous, vanish at infinity, and whose first derivatives are also square-integrable. This domain is denoted as H¹(ℝ) and ensures the self-adjointness of the momentum operator.

[@carlosleiva]

Does anyone know the textbook (s) of these lectures? I read across some and no one is so complete and rigorous.

[@taypangshiang7935]

I'm using Quantum Theory for Mathematicians by Brian C Hall. Its one of the recommended ones on his website

[@mauricesangermann1320]

Can you provide a link for the website, I havent been able to find it?

[@jessedaas6365]

@@taypangshiang7935 what website are you talking about? I can't find any where he recommends books

[@hamzaikanna1507]

I wish if there's some serie of exerciseS

[@jihongzhi]

Is it a mistake at 46:41, where he claims that the closure of {1} is the set of constant functions? This makes little sense to me, but also seems unnecessary for the following arguments.

[@ActionPhysics]

intuitively 1 function takes us to the constant 1 for any x , and the smallest extension for it to be open under appropriate topology on the function space ,can be collection of functions that takes us to constants not necessarily=1

[@maxd.9677]

I think you are right. In a Hilbert space A^perp perp is the closure of the span of A, and I think that is what he is using. Probably he forgot to put the span (or uses overline or curly brackets in some unusual notation)

[@jihongzhi]

@@maxd.9677 Ah yes, span makes sense. In fact as I have since learned, a Hilbert space is also a Hausdorff topological space, in which the closure of any singleton set is itself. So the claim in this lecture cannot have been correct as it stands.

[@thomaslamby1136]

I don't understand at 1:10:00 why if P** is a strict subset of P*, then P is not essentially self adjoint. I dont think that the "star-operation" is injective so we can't say "A different of B implies A* different of B*" ... I'm in fact pretty sure that is false because if it was true then not self adjoint would imply not essentially self adjoint since at that moment P strict subset of P* then P** strict subset of P* and then P** strict subset of P***. I presume then that we need to use that we have explicitely the A and the B : its P** and P*...

[@rifathkhan9995]

I got confused at this point too. In no where in his previous lectures he concluded an operator is not essentially self adjoint if the condition he mentioned holds. This seems to come out of the blue.

[@benholder2766]

I was confused here too, but maybe this is it? He proves (B extends A implies A* extends B*) at the beginning of Lecture 7: kzbin.info/www/bejne/aGOzkJuZj653btk If this holds true for "B properly extends A", then the result follows: P** is (properly) extended by P* so taking the adjoint of both sides we find that P** is (properly) extended by P***, i.e., the closure of P is (properly) extended by its adjoint and thus is not self-adjoint. It looks to me that his proof does indeed hold for "proper extension".

[@getrundelthd9807]

I don't really understand how a sobolev funtion can fullfill this boundary conditions. The sobolev space is a subset of L² which means, that the functions are really equivalence classes of functions. In these classes there are functions which differ only on a set of measure zero. So giving these boundary conditions is not correct in my opinion, because you could still choose every equivalence class because every equivalence class has a member that fullfills these boundary conditions. The only way to give such boundary conditions is the space H^1_0 which is the closure of C^\inf_0 which are sobolev functions which are 0 on the boundary

[@gustavomaycustodio5987]

isn't $\{ 1\}$ already closed since $L2$ is Haussdorff 47:06

[@andrea.dibiagio]

I think it's a slight abuse of notation and by {1} he means Span({1}), i.e. the space of constant of continuous functions, not just the specific function x \mapsto 1.

[@asaflevif]

That head on the corner made me crazy