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I would like to calculate something like $$K\left(z\right)=\dfrac{dF\left(A\left(z\right)+B\left(z\right)\right)}{dz}$$ i.e. the derivative of a function $F$ of a sum of two operators $A$ and $B$ with respect to the variable $z$, when $A$ and $B$ do not commute with each other: $\left[A,B\right]\neq0$ and when $$\dfrac{dA}{dz}=A^{\prime}\;;\;\dfrac{dB}{dz}=B^{\prime}$$ are known. $F$, $A$ and $B$ have all good properties for physics (expandable in series around every point of interest).

I thought it exists an integral representation for such a derivative (which I may have read once in a paper by Feynman or Glauber) but I'm totally unable to found it since a few days, so perhaps I was just dreaming one more time :-(

Actually, my problem is simpler, since $F$ is an exponential $F=e^{iz}$, and $A$ and $B$ are simple functions of $z$: $dA/dz=0$ and $B=B_{0}z$, with $dB_{0}/dz=0$ and so I have $\left[A,B_{0}\right]\neq0$. I can obviously expand for small $A$ and $B_{0}$ which gives me a hint of what I'm looking for, but I would like to know if I can do something more general. Thanks in advance.

PS: I ask this question here instead of Mathematic-SE since I believe it is not of great interest for mathematicians. Feel free to transfer this question to Math-SE.

FraSchelle
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    I guess by assumption also their derivatives do not commute. I wonder: does $[A,B]\neq 0$ imply $[A',B]\neq 0$? – levitopher Feb 23 '15 at 12:52
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    Possibly related: http://physics.stackexchange.com/q/15742/2451 – Qmechanic Feb 23 '15 at 12:56
  • @levitopher Yes, all derivative do not commute. Actually, $A$ and $B$ are expressible in term of Pauli matrices in my explicit case – FraSchelle Feb 23 '15 at 13:22
  • @Qmechanic Thanks a lot, the formula I was looking for is exactly the one you demonstrate in http://physics.stackexchange.com/a/41671/16689 Sorry to have missed this answer. – FraSchelle Feb 23 '15 at 13:23

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