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As we know, simple harmonic oscillator can be solved only by commutation relations between creation and annihilation operators, and the Hamiltonian expression. The spin energy is either solved only using commutative relations between spin operators in axes ($J_i$) and $J^{2}$. As an another illustration, for quantizing fields (such as Real Klein-Gordon scalar field) in QFT, one approach is to postulate canonical commutation relations between field and momentum operators. I already know quantum operators create Lie Algebra, and commutation relations are important in a Lie Algebra. However, I have a question: is it always true that commutation relations are sufficient to obtain eigenvalues and eigenvectors of a Hamiltonian in the Hilbert space? If it is true, why commutative relations are sufficient to solve a quantum mechanics problem?

heaven-of-intensity
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  • (+1) Another example is the Hydrogen atom: it can be solved using algebraic methods (first done by Paili if I recall). – AccidentalFourierTransform Mar 05 '16 at 19:31
  • @AccidentalFourierTransform yes. thanks for the remark – heaven-of-intensity Mar 05 '16 at 19:33
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    The commutation relations are not merely "important" in a Lie algebra, they define the Lie algebra completely. The question "why are they sufficient to solve a quantum mechanics problem" is ill-posed because surely there are problems that don't have enough symmetry that they are solved by knowing the Lie algebra of the symmetry alone. – ACuriousMind Mar 05 '16 at 19:41
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    @ACuriousMind is it always true? You know I want to know under what conditions commutative relations solve the problem. For example in continous groups or what? – heaven-of-intensity Mar 05 '16 at 19:48
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    What do you mean by "solve the problem"? Whether commutation relations solve it depends on what the problem is. – ACuriousMind Mar 05 '16 at 20:00
  • @ACuriousMind of course it is! To assure me like simple harmonic oscillator and other examples here, ALL eigenvector of THE Hamiltonian has been found – heaven-of-intensity Mar 05 '16 at 20:03
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    Congrats, I think you have just rediscovered a modern version of matrix mechanics. Too bad that Heisenberg had that idea some 90 years ago. :-) As for the practical usefulness: it's probably limited. Finding the complete algebra to a problem is (per "no free lunch" theorem) exactly as hard as solving the problem with other methods. There is no magic algorithm that can do this automatically for an arbitrary problem. – CuriousOne Mar 05 '16 at 21:13
  • @CuriousOne oh I am glad :) as I have understood from your previous comments, you said there is a Hamiltonian that its spectrum cannot be obtained only using commutative relations. Can you give me such a Hamiltonian? Thanks – heaven-of-intensity Mar 06 '16 at 07:40
  • I don't think that the spectrum of a general linear operator can be obtained exactly by any method, nor do I see a reason why that should be the case. Just because we are transitioning from classic Hamiltonians (all of which except for about a dozen are essentially unsolvable even for the single particle case!) to quantum mechanics doesn't mean things are getting easier. Are you expecting there to be an algebraic TOE? – CuriousOne Mar 06 '16 at 08:55
  • @CuriousOne I didn't talk about general linear operators rather self-adjoint operators that all observables are. I don't understand why you thought I am seeking a method rather seeking a counterexample (I think this is the case and more probable) or a proof that Hamiltonian eigenvector can be obtained solely by commutative relations. So, I am not seeking exact solutions. For example, we cannot solve exactly an arbitrary differentional equation but it has proven there is a unique answer on a closed interval. – heaven-of-intensity Mar 06 '16 at 09:11
  • I don't quite understand what you are looking for. Existence and uniqueness of a solution are very weak mathematical results and I am not even sure that uniqueness is even possible. I remember seeing non-harmonic oscillators with the same spectrum as the harmonic oscillator... so that kind of makes even the spectral theory weak. – CuriousOne Mar 06 '16 at 09:55
  • @CuriousOne I couldn't understand your purpose. You remembet what? Can you explain more? Is it empirical observation? – heaven-of-intensity Mar 06 '16 at 14:29
  • Is the existence of theory papers that construct different classes of Hamiltonians that have the same spectrum empirical observation? Yes. We had a question about that a while ago and people (including myself) dug up multiple papers, I believe. – CuriousOne Mar 06 '16 at 22:14
  • @CuriousOne I didn't mean by this empirical observations rather in a laboratory. But I would be glad if I can read the paper if they are availaible. Thanks – heaven-of-intensity Mar 06 '16 at 22:31
  • Not sure how I would use commutation relations in the laboratory... most of the time we are looking for spectra and symmetries, I guess. I would have to think about that some more. I think this is the post I was referring to: http://physics.stackexchange.com/questions/132688/is-the-harmonic-oscillator-potential-unique-in-having-equally-spaced-discrete-en – CuriousOne Mar 06 '16 at 22:50
  • @CuriousOne thanks. we didnt talk about commutative relation rather observing a system that its energies are the same as simple harmonic oscillator. Think about it too. – heaven-of-intensity Mar 07 '16 at 03:43
  • You don't need the commutation relations to solve the harmonic oscillator. The Schrödinger equation admits a perfectly fine solution as Hermite polynomials. – Slereah May 04 '17 at 09:28
  • @Slereah yup, but Schrödinger assumes $p=-i \nabla$ that is equivalent to the commutator of $X$ and $P$. – heaven-of-intensity May 10 '17 at 04:03

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If your Hamiltonian belongs to a Lie algebra for which you can solve the initial value problem in the corresponding group then you can use geometric quantization to solve the corresponding Schroedinger equation. This is because the solution of the Schroedinger equations is just $\psi(t)=e^{-itH/\hbar}\psi_0$, and $e^{-itH/\hbar}$ is an element of the group generated by the Lie algebra (in the appropriate representation on the given Hilbert space). Thus the problem is reduced to group representation theory.

Arnold Neumaier
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