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Operators in QM

  1. Nov 16, 2013 #1
    An operator A defined by a matrix can be written as something like:

    A = Ʃi,jlei><ejl <eilAlej>

    How does this representation translate to a continous basis, e.g. position basis, where operators are not matrices but rather differential operators etc. Can we still write for e.g. the kinetic energy operator T:

    T = ∫∫dr dr' lr'><rl <r'lTlr>

    ? Or how would T be represented.
  2. jcsd
  3. Nov 16, 2013 #2
    I know there are some subtleties to shifting to a continuous basis, but basically yes. You change [itex]\Sigma[/itex]'s into [itex]\int[/itex]'s, and then all the same logic goes through. [itex]T[/itex] becomes a "matrix" with an infinite number of rows and columns, which you could think of as a function [itex]T(q_1, q_2)[/itex].
  4. Nov 17, 2013 #3
    The energy eigenstates form a countable basis. Would it be possible to do the above expansion in those rather than position eigenstates?
  5. Nov 17, 2013 #4


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    ^ The eigenstates of the Hamiltonian form a countable basis only if all the states are bound states, like in a harmonic oscillator. For example, the bound states of hydrogen atom form a discrete set, but the system also has a continous spectrum of scattering states which have positive total energy. The continuous spectrum is not a countable set.
  6. Nov 17, 2013 #5
    but either way you can expand any operator in energy just as well as position eigenstates right? or any other observables eigenstates..
  7. Nov 17, 2013 #6


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    ^ Yes, the eigenstates of any hermitian operator form a complete basis.
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