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QM commutators

  1. May 12, 2004 #1
    Does anyone know of any tables that show the commutation relations of all QM opeartors? Any information would be appreciated.
     
  2. jcsd
  3. May 13, 2004 #2
    You don't need a table, they can all be derived simply from

    [tex][x,p]=i\hbar[/tex]

    That's the beauty of physics and the thing that distinguishes it from Botany or stamp collecting.

    You may need a couple more for relativistic quantum physics and spin, but since you didn't specify I am assuming you mean non-relativistic QM.
     
  4. May 16, 2004 #3

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    Yes indeed. There are some important commutation relations involving rotations and boosts.
     
  5. May 16, 2004 #4
    I was looking for non-relativistic, just wondered if such a thing existed. The matrical rep of the Serret-Frenet formulae is very neat because it makes them very easy to remember. Now that I've looked at all the classical QM relations together at once I see that, as Slyboy stated a table is not really necessary, because they are all so similar. Thanks everyone.
     
  6. May 16, 2004 #5

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    I can't resist adding-

    But if you had been looking for relativistic commutation relations, one way to sum up the relations for generators of rotation J and generators of boost K are that they form the algebra SO(3,1), or equivalently SU(2) x SU(2).

    Tell that to your friends to impress them. :approve:
     
  7. May 17, 2004 #6
    but perhaps you should read a book on Lie algebras first, just in case any of them know what you are talking about. :biggrin:
     
  8. May 18, 2004 #7
    I've read about 20 books on Lie Algebras and Superalgebras - I know more math than physics.
     
  9. May 18, 2004 #8

    Tom Mattson

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    Well, you'd also need to know that [itex][x_i,x_j]=0[/itex] and [itex][p_i,p_j]=0[/itex].

    Yes, and you need the commutators for spin even in nonrelativistic QM, since spin is not derived from x and p.
     
  10. May 19, 2004 #9
    Yes, but you can derive the relations for orbital angular momentum and they are essentially the same.
     
  11. May 19, 2004 #10

    Tom Mattson

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    They're exactly the same, but that's not the point. The point is that spin isn't a function of x and p, and so the commutators for the spin operators must be added to the set of "fundamental" commutators.
     
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