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Operator equation

  1. Dec 18, 2006 #1
    let be A and B 2 operator so their commutator is:

    [tex] [A,B]=1 [/tex]

    then my question is if A and B are dependant of x,y,z,t then what would be the value of commutator:

    [tex] [\partial_{c} A , \partial_{b} B]=? [/tex]

    where c and b can be x,y,z or t..thanks...:redface: :redface:
     
  2. jcsd
  3. Dec 18, 2006 #2

    cristo

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    Well, if A is independent of x^b, then [tex] \partial_b A = \frac{\partial A}{\partial x^b}=0 \forall b [/tex]. Similarly for B, and so the commutator is zero.
     
  4. Dec 18, 2006 #3

    dextercioby

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    One doesn't know. Iff one specifies the domain and ranges of the partially differentiated operators, can one compute the value of the commutator.

    Daniel.
     
  5. Dec 18, 2006 #4
    I'm not so sure, but after a fast calculus i found out that it should be zero the answer.
    Using the propertie:

    [tex] [AB,C]=A[B,C]+[A,C]B [/tex]

    and remebering that:

    [tex] [\partial_{c}, \partial_{b}]=0 [/tex]
    since b and c are indip. variables.
    you reach this condition:

    [tex] [\partial_{c}A, \partial_{b}B]=[\partial_{b}B, \partial_{c}A]= [/tex]

    wich means that:

    [tex] [A,B]=[B,A]=AB-BA=BA-AB[/tex]

    so AB=-AB ==> i think zero is the only solution.

    Bye MARCO:smile:
     
  6. Dec 18, 2006 #5
    The commutator product is antisymmetric under interchange, no symmetric, so everything following this point is incorrect. I worked out the algebra (took about a page) and found that

    [tex] \left [ \partial_c A, \partial_b A \right ] = \partial_c \left ( \left [ A, \partial_b \right ] B \right ) + \partial_b \left ( \left [ B, \partial_c \right ] A \right ) [/tex]

    although I might have made a computational error.
     
    Last edited: Dec 18, 2006
  7. Dec 18, 2006 #6
    i know is antysemmetric but if you keep expanding your calculus you find
    a symmetric relation wich tells you that the commutator must be zero since:
    [A,B]=-[B,A] and also [A,B]=[B,A]
     
  8. Dec 19, 2006 #7

    dextercioby

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    I see you imply that

    [tex] \partial_{\mu}[A,B]=\left[\partial_{\mu}A, B\right] +\left[A,\partial_{\mu}B\right] [/tex]

    ,which is completely incorrect.

    Daniel.
     
  9. Dec 19, 2006 #8
    I'm not sure how I imply that at all. Lemme go through my steps, maybe you can find my error.

    [tex]
    [ \partial_c A, \partial_b B ] & = & \partial_c [A, \partial_b B] + [\partial_c, \partial_b B] A
    [/tex]
    [tex]
    & = & - \left \{ \partial_c[\partial_b B, A] + [\partial_b B, \partial_c] A \right \} + \left \{ \partial_b [B, \partial_c] + [\partial_b, \partial_c] B \right \} A
    [/tex]
    [tex]
    & = & - \partial_c \partial_b [B, A] - \partial_c([\partial_b, A] B) + \partial_b ([B, \partial_c ] A)
    [/tex]
    [tex]
    & = & - \partial_c ([\partial_b, A] B) + \partial_b([B, \partial_c] A) [/tex]
     
    Last edited: Dec 19, 2006
  10. Dec 20, 2006 #9

    dextercioby

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    Let's see whether i got the problem right. It happened to me a few days to get a problem wrong and therefore make erroneous statements.

    1.What is [itex] [,] [/itex] supposed to mean ?
    2.Who is A/B ?
    3. Why can you take out the (possibly) space(-time) derivative from the bracket structure ?

    Daniel.
     
  11. Dec 20, 2006 #10
    Maybe ive made some mistake during the calculus.
    Well i bet that the problem is that we dont know in wich space A and B are defined. But if the space is L^2(RN) i hope u understand what i mean. We can use Von-Neumann theorem wich say:
    if two self-adjoint oerator A and B have the Heisenberg commutation rules, there exists u:
    q=uAù and p=uBù.

    so we can treat the problem using all we know about the position and the momentum operator.
    Sorry i forgot to tell you that u is a unitary trasformation so that
    uù=ùu=1.
     
  12. Dec 20, 2006 #11
    1. Commutator brackets
    2. I suppose two arbitrary operators
    3. I'm using commutator identities to "pull those things out".
     
  13. Dec 21, 2006 #12

    dextercioby

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    1. Thought so myself.
    2. Acting on what kinda space ?
    3. If you make an assumption about 2., then i can argue against 3.

    Daniel.
     
  14. Dec 21, 2006 #13

    dextercioby

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    In the way it's formulated, i'd say the problem is in the realms of quantum field theory, in which A and B are field operators depending on [itex] x^{\mu} [/itex] and the derivative [itex] \partial_{a} [/itex] is actually the more familiar [itex] \partial_{\mu} [/itex].

    The sad thing is that the problem is too abstractly (read "vaguely") formulated and all we can do is make convenient assumptions, as the OP never bothered to answer to our (possible) solutions.

    Daniel.
     
  15. Dec 21, 2006 #14
    Yes i know what you mean, but, you know for sure that the fock space still an Hilbert space (IT'S JUST A COUNTOUBLY ADDITION OF SUCH THOSE SPACES) so i think the theorem works good anyway.
    I'm pretty sure also the question was about the relativistic wiew. It doesn't change the algebraic properties of the operators a lot. P (the momentum) still a "illimitato" operator, not bounded it's right?
    what ever you say. You can pick a sequence of function (fn) in the Fock space that: (||Pfn||)/(||fn||)=infinity when n goes too infinity. where the norm/scalar product its the appropriate of the Hilbert space in exam. for example on L2(0,+infinity) fn=exp(-nx).

    Bye Marco
     
  16. Dec 21, 2006 #15

    dextercioby

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    Veramente giusto. Unfortunately though, using unitary transformations to bring the problem back to canonical (anti)commutation relations for the fundamental fields of the theory (i.e. analogues of the x and p in qm of systens with finite # of degrees of freedom) is not a great simplification, since the fock space is not unique and this reflects itself in different physical models for basically the same comm.relations.

    Either way, this is already philosophy and not mathematics.

    Daniel.
     
  17. Dec 21, 2006 #16
    Dexter, beware that you do not become too theoretical a physicist. :wink:
    People will get scared of you. :rofl:


    marlon
     
  18. Dec 21, 2006 #17
    What kind of space do you like? I viewed A and B as operators that were functions of x, p and t in your standard classical quantum mechanics. What else could they be?
     
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