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Thread: More on commutators

  1. #1
    Senior Member oldguynewstudent's Avatar
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    More on commutators

    First, thanks for your previous help. It was greatly needed and appreciated.

    So we have proved [A,Bn]=nBn-1[A,B]

    Next we have: If F(x)= \sum\limits_n f_{n} x^{n} is any Taylor Series expandable function with an appropriate radius of convergence, and A and B satisfy the conditions already stated, show that the operator function F(B) obeys [A,F(B)]=F^{'}(B)[A,B]=[A,B]F^{'}(B), where F^{'}(x)=dF/dx.

    F(x)=\sum\limits_n f_{n} x^{n} so F^{'}(x)=n f_{n} x^{n-1}

    Using earlier proof

    [A,F(B)]=[A,\sum\limits_n f_{n} B^{n}]=\sum\limits_n f_{n} B^{n-1}[A,B]=[A,B]\sum\limits_n f_{n} B^{n-1}=F^{'}(B)[A,B]=[A,B]F^{'}(B)

    Now the last part: Use prior result to evaluate [X,e^{i K_{x} L}] where L is a constant having units of length.

    I got [X,e^{i K_{x} L}]=[X,K_{x}] \sum\limits_x (iLe^{i K_{x} L})

    Would someone be able to verify the second part and the last part?

    Thank you
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  2. #2
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    Re: More on commutators

    The idea behind this, is to show your last result extends to polynomials, and we can extend this to convergent power series.

    Note that we ought to prove $[A,B+C] = [A,B] + [A,C]$, and that $[A,kB] = k[A,B]$ (i.e. that $\delta_A$ is linear).

    This means that for any POLYNOMIAL $F(x)$, we have:

    $[A,F(B)] = F'(B)[A,B] = [A.B]F'(B)$ (the last equality holds because any polynomial in $B$ commutes with $[A,B]$ since $B$ does).

    Let's denote $F_k(x)$ to be the $k$-th degree partial sum of the Taylor series expansion for $F$.

    Then $\displaystyle [A,F(B)] = [A,\lim_{k \to \infty} F_k(B)] = \lim_{k \to \infty} [A,F_k(B)] = \lim_{k \to \infty} F_k'(B)[A,B] = \lim_{k \to \infty} [A,B]F_k'(B) = [A,B](\lim_{x \to \infty} F_k'(B))$

    (note: the last step assume that operator composition is continuous, and the second step assumes so for the commutator bracket!)

    $= [A,B]F'(B)$ (within our radius of convergence-that is, we can differentiate "term-by-term").

    It is not quite clear to me what $K_x$, so I cannot evaluate the second part. If it is an operator that is parameterized by some $x$ ( vector? scalar?), it seems to me that:

    $D(e^{iK_xL}) = iLe^{iK_xL}$, so that:

    $[X,e^{iK_xL}] = [X,K_x]iLe^{iK_xL}$ (if $L$ is a (real) constant, then $iL$ is just a complex constant, and the complex exponential Taylor series converges everywhere in $\Bbb C$).

    For this to hold, you would (of course) have to verify that $[X,K_x]$ and $K_x$ commute.
    Thanks from oldguynewstudent
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  3. #3
    Senior Member oldguynewstudent's Avatar
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    Re: More on commutators

    Thank you so much for the help. I was of course in my normal state of confusion and bewilderment. Thank you for making the solution clear.
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