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?
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asked Mar 5, 2016 at 19:12
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$\begingroup$ (+1) Another example is the Hydrogen atom: it can be solved using algebraic methods (first done by Paili if I recall). $\endgroup$– AccidentalFourierTransformMar 5, 2016 at 19:31
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1$\begingroup$ 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. $\endgroup$– ACuriousMind ♦Mar 5, 2016 at 19:41
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1$\begingroup$ @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? $\endgroup$– heaven-of-intensityMar 5, 2016 at 19:48
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1$\begingroup$ What do you mean by "solve the problem"? Whether commutation relations solve it depends on what the problem is. $\endgroup$– ACuriousMind ♦Mar 5, 2016 at 20:00
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1$\begingroup$ 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. $\endgroup$– CuriousOneMar 5, 2016 at 21:13
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1 Answer
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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.
answered May 4, 2017 at 9:22