Operators in QM

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  • #1
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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.
 

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  • #2
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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].
 
  • #3
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The energy eigenstates form a countable basis. Would it be possible to do the above expansion in those rather than position eigenstates?
 
  • #4
hilbert2
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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.
 
  • #5
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but either way you can expand any operator in energy just as well as position eigenstates right? or any other observables eigenstates..
 
  • #6
hilbert2
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^ Yes, the eigenstates of any hermitian operator form a complete basis.
 

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