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Commutator property question

  1. Nov 24, 2011 #1
    Hi,

    If we have two commuting operators A and B, is true that any function of A will commute with any function of B? I have a result which takes [itex][L_{z},r^{2}]=0[/itex] and claims that [itex][L_{z},r]=0[/itex]. How can this be proved? Thank you
     
  2. jcsd
  3. Nov 24, 2011 #2
    Note that A[itex]^{2}[/itex]B[itex]^{2}[/itex] = AABB = A(AB)B = A(BA)B since A and B commute.
    = (AB)AB = (BA)(BA) = B(AB)A = B(BA)A = B[itex]^{2}[/itex]A[itex]^{2}[/itex]

    i.e. A[itex]^{2}[/itex] commutes with B[itex]^{2}[/itex].

    In like manner one can prove that A[itex]^{n}[/itex] commutes with B[itex]^{m}[/itex].

    Note also that many functions of A can be expanded as power series in A. Similarly for functions of B.
     
  4. Nov 24, 2011 #3
    Thank you for replying grzz.

    I understand how to go from a lower power to a higher power. You showed that if A and B commute, then A^2 and B^2 also commute. But the reverse is what is being done in my question so how can I prove it for decreasing powers?

    Thank you.
     
  5. Nov 24, 2011 #4
    I did not read your qustion carefully!
     
  6. Nov 25, 2011 #5
    Can anyone else help? I have this idea but I am not sure if its correct.

    We have [itex]Lr^{2}-r^{2}L=0[/itex]

    So, [itex]Lr^{2}-rLr+rLr-r^{2}L=0[/itex] where I just added and subtracted the middle terms.

    Then [itex][L,r]r+r[L,r]=0[/itex].

    Would it be correct to say that since this is true for arbitrary r, [L,r] must be zero? I am not sure of this argument, especially the last step. Thank you

    EDIT: No, that's not right. So any help?
     
  7. Nov 25, 2011 #6
    It's not true in general.
    Counterexample: take the pauli matrices [itex][\sigma_x,\sigma_y^2]=[\sigma_x,1]=0[/itex], but [itex][\sigma_x,\sigma_y]=2i\sigma_z \neq0[/itex]
     
  8. Nov 25, 2011 #7
    Thank you aesir. I get it now.

    Looks like there aren't any shortcuts to prove [L,r]=0.
     
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