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Commutator of function of operators

  1. Aug 11, 2014 #1
    According to my teacher, for any two operators A and B, the commutators
    [f(A),B]=[A,B]df(A)/dA
    and [A,f(B)]=[A,B]df(B)/dB
    He did not give any proof.
    I can easily prove this for the particular cases
    [f(x),p]=[x,p]df(x)/dx
    and [x,pn]=[x,p]npn-1
    But I don't see how the general formula is true. I also can't find it in any textbooks.
     
  2. jcsd
  3. Aug 12, 2014 #2

    bhobba

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    I am not sure that's true.

    Expand f(A) in terms of a power series. Take the derivative of that to give df(A)/dA and then check if equal. I cant see how it is without further assumptions.

    Thanks
    Bill
     
    Last edited: Aug 12, 2014
  4. Aug 12, 2014 #3
    But he used this formula to prove eAeB=eA+B+[A,B]/2.
    To be exact, he took [eAt,B]=[A,B]teAt in one step of the proof.
     
  5. Aug 12, 2014 #4

    bhobba

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    Expand them out and have a look at the first term - are they equal?

    To be exact the first term on LHS doesn't have t in front - yet the first term on the RHS does.

    Added Later:
    Wait a minute - the first term contains I which always commutes so is zero. That might be the assumption he is making - if the first term of f(A) commutes with B it may hold. I will check at get back.

    Yea - the assumption he is making is the power series expansion is in terms of scalars ie f(A) = a0 + a1A + a2A^2 ..... and a0, a1 etc are scalars.

    I think if you insert and equate then the formula are true.

    Thanks
    Bill
     
    Last edited: Aug 12, 2014
  6. Aug 12, 2014 #5

    Orodruin

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    This is only true if
    $$
    [A,[A,B]] = [B,[A,B]] = 0.
    $$
    That is, the commutator of A and B must commute with A and B for the formula to be true. In the case of x and p, their commutator is a scalar and it therefore holds. However, there is no guarantee for arbitrary A and B. The relation
    $$
    e^A e^B = e^{A+B + [A,B]/2}
    $$
    is a special case of the Baker-Campbell-Hausdorff formula (see http://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula) for when ##[A,B]## commutes with ##A## and ##B##.
     
  7. Aug 12, 2014 #6

    Orodruin

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    Just to expand a bit on the series expansion.

    Expanding ##f(a)## gives
    $$
    f(a) = \sum_{n=0}^\infty \frac{d^n f}{da^n} \frac{a^n}{n!} \equiv
    \sum_{n=0}^\infty k_n a^n
    $$
    and thus
    $$
    \frac{df(a)}{da} = \sum_{n=0}^\infty k_n n a^{n-1} = \sum_{n=1}^\infty n k_n a^{n-1}.
    $$
    Now, in the same way
    $$
    f(A) = \sum_{n=0}^\infty k_n A^n,
    $$
    which implies
    $$
    [f(A),B] = \sum_{n=0}^\infty k_n [A^n, B].
    $$
    We can here make use of the commutator relation
    $$
    [A^n, B] = \sum_{k = 1}^n A^{n-k} [A,B] A^{k-1}
    $$
    to rewrite this as
    $$
    [f(A),B] = \sum_{n=0}^\infty k_n \sum_{k = 1}^n A^{n-k} [A,B] A^{k-1}.
    $$
    Now, if ##[A,[A,B]] = 0##, then
    $$
    A^{n-k} [A,B] A^{k-1} = [A,B] A^{n-1}
    $$
    and
    $$
    \sum_{k = 1}^n A^{n-k} [A,B] A^{k-1} = [A,B] A^{n-1} \sum_{k=1} ^n 1 = n [A,B] A^{n-1}.
    $$
    The expression would then simplify to
    $$
    [f(A),B] = [A,B] \sum_{n=0}^\infty k_n n A^{n-1} = [A,B] f'(A).
    $$
    The last expression is of course also equal to ##f'(A) [A,B]##, since ##[A,B]## commutes with ##A##.
     
  8. Aug 12, 2014 #7
    Thanks a ton.
     
  9. Aug 12, 2014 #8

    bhobba

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    Yea - me too.

    It had me scratching my head a bit as well.

    Thanks
    Bill
     
  10. Aug 12, 2014 #9

    strangerep

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    Heh, now show that the formula extends to product functions, and to functions of the form 1/f(x). :tongue:
     
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