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Poisson brackets, commutators, transformations

  1. Sep 13, 2009 #1
    Hi all,

    I've taken a two-course undergrad QM sequence and have been reading Shankar's Principles of Quantum Mechanics. There is some reference to the similarity between the Poisson bracket in Hamiltonian mechanics and the commutator in QM. E.g.

    [tex]\{x, p\} = 1[/tex] (PB)
    [tex][x, p] = i \hbar[/tex] (commutator)

    At various points in the book it seems to me that Shankar suggests that this commutation relation is a very fundamental thing. However, I don't understand where it comes from. I do understand the derivation of the poisson bracket relation. Why does the canonical commutation relation hold? An equivalent question, I guess, is why the momentum operator in the x basis is
    [tex]-i \hbar \frac{\partial}{\partial x}[/tex]
    or why there is the Fourier-transform relationship between position and momentum space. I don't see why these things must be true either. Ditto for the angular momentum operators. Also, why the parallel between the classical poisson bracket and the QM commutator? What is the fundamental connection between position and momentum that transcends classical and quantum mechanics?

    In Shankar there's a section on generators of infinitesimal canonical transformations in Hamiltonian mechanics, stating that

    [tex]Q = q + \varepsilon \frac{\partial g}{\partial p}[/tex]
    [tex]P = p - \varepsilon \frac{\partial g}{\partial q}[/tex]

    gives an infinitesimal canonical transformation and that [tex]g[/tex] is the generator of the translation. If the Hamiltonian is invariant under the transformation generated by g, then g is conserved. E.g. [tex]g = p[/tex] is the generator of translations; therefore, translational symmetry gives conservation of momentum. I'm OK with that. However in QM infinitesimal transformations seem to be given by

    [tex]T(\varepsilon) = I - \frac{i \varepsilon}{\hbar} G[/tex]

    where the operator G is the generator of an infinitesimal transformation. Where does this factor of [tex]\frac{i}{\hbar}[/tex] come from?! Shankar doesn't justify this form for infinitesimal transformations in QM.

    I can follow the classical derivations of the fact that the momentum, angular momentum, and Hamiltonian are the generators of infinitesimal translations, rotations, and time translations respectively, and how this leads to the corresponding conservation laws. However, I don't understand why this is true in QM too. Which means I don't understand the justification for the Schrodinger equation itself, I think.

    In looking at related stuff on Wikipedia I see a lot of references to "symplectic spaces" and stuff. From what class or textbook or website would one learn about such things?

    Thanks for any help!
     
  2. jcsd
  3. Sep 13, 2009 #2
    Try L. E. Ballentine "Quantum mechanics. A modern development". Perhaps it will answer some of your questions.
     
  4. Sep 13, 2009 #3
    First of all, the relation is pretty fundamental. Dirac constructed his formulation of QM by this very principle: "Classical poisson bracket -> commutator, multiplied by i.hbar". The technical name for the procedure is canonical quantisation, and is very important in many presentations of quantum field theory.

    Regarding the infinitesmal translations in QM: I've not seen the reciprocal factor of hbar before, but as far as I can tell it doesn't affect anything important. What is important is the factor of i in front of the infinitesmal epsilon. In QM symmetry operations are associated with unitary operators, so that transition probabilities are preserved. (The only exception is time reversal, which is an anti-unitary operator- don't worry about it.) If G is a hermitian operator, then you'll see that multiplying T by its adjoint gives the identity operator if you take second order quantities in epsilon to vanish. So symmetry transformations are generated by hermitian operators- which, remember, are observables.
    Also, the Schroedinger equation is a postulate of QM. It can't be rigorously justified, but Schrodinger spotted the formal analogy and it worked.
     
  5. Sep 14, 2009 #4
    Thanks for the replies. Looks like there's a copy of Ballentine's book in my school's library, so I'll have a look at it.
     
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