Power of approximate exponential of operator sum

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Your question is a bit open ended and dependent on the matrix norms involved, etc... Purely formally, if you introduce a parameter t, something tells me you are interested in the Zassenhaus formula, $$ e^{t(A+B)}= e^{tB}~ e^{tB} ~e^{-\frac{t^2}{2} [A,B]} \\ \times ~ e^{\frac{t^3}{6}(2[B,[A,B]]+ [A,[A,B]] )} ~ e^{\frac{-t^4}{24}([[[A,B],A],A] + 3[[[A,B],A],B] + 3[[[A,B],B],B]) } \cdots $$

Anyway, the expression you write is, by straightforward application of the CBH formula $$ \widetilde{C}=e^{tA/2}e^{tB/2}e^{tA/2}e^{tB/2}= (e^{tA/2}e^{tB/2})^2\\ = ( e^{\frac{t}{2}(A+B)+\frac{t^2}{8}[A,B]+ O(t^3) } )^2 \\ =e^{t(A+B)+ \frac{t^2}{4}[A,B]+ O(t^3) }, $$ which does not look like a good "approximation". The last step follows from the fact that the commutator of the exponent with itself upon squaring vanishes, so the squaring act has a degenerate, terminating CBH expansion!!

You could easily eliminate the $O(t^2)$ terms in the exponent by considering $$ e^{tA/2}e^{tB}e^{tA/2} =e^{t(A+B)+ O(t^3) }, $$ instead.

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Updated on August 01, 2022

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  • johnfrog
    johnfrog over 1 year

    Say $A$ and $B$ are two non-commuting operators in an Hilbert space (think matrices, for instance). Then, one way we can approximate $C=e^{A+B}$ is as $\widetilde{C}=e^{A/2}e^{B/2}e^{A/2}e^{B/2}\simeq C$.

    Question: Does it hold that $\widetilde{C}^{2}=e^{A}e^{B}e^{A}e^{B}$? It doesn't, right?

  • johnfrog
    johnfrog over 4 years
    Thank you, Prof. Cosmas. However, I don't understand two things in the second expression: 1) why you identify what I wrote with the square of the expansion up to third order in $t$, and 2) why, in the last step, we can simply multiply the exponent by $2$ even if $A$ and $B$ don't commute, while keeping the expansion up to third order in $t$ and saying nothing about the effect of the exponent $2$ (which could be higher).
  • Cosmas Zachos
    Cosmas Zachos over 4 years
    I edited the answer to answer your questions: you must use the CBH expansion. Squaring is CBH-multiplying an exponential with itself, so the commutator of the exponents vanishes, as they are the same! The effective exponent then is just the sum of the original exponent with itself, so twice that exponent. The expansion is in t, suitably defined when you normalize the matrices by their norm.
  • johnfrog
    johnfrog over 4 years
    Thank you for taking the time to make it clear!
  • johnfrog
    johnfrog over 4 years
    I didn't understand the full significance of the normalization. Can you quickly make some comment to clarify this point? Isn't it enough to make $t$ small enough as compared to the inverse of the smallest eigenvalue of $A+B$? Is the normalization just a convenience to be able to define simply an expansion for $t<<1$ or does it actually improve the approximation? Thank you!
  • Cosmas Zachos
    Cosmas Zachos over 4 years
    You got it: with a sufficiently small t as you guessed, all is fine. Normally, you set it equal to 1, but if the eigenvalues of the matrices were huge, the "corrections" would dominate the result....