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For two linear operators $A$ and $B$, I am trying to find

$$ \lim_{n \to \infty} ||e^\frac{A+B}{n}-e^\frac{A}{n}e^\frac{B}{n}|| $$

Where $||.||$ is the operator norm. I have approached the problem using a Taylor expansion, and this is what I have so far:

$$ e^\frac{A+B}{n}-e^\frac{A}{n}e^\frac{B}{n}$$ $$=\sum_{i=0}^{\infty} \frac{(A+B)^i}{n^ii!}-(\sum_{i=0}^{\infty} \frac{A^i}{n^ii!})(\sum_{i=0}^{\infty} \frac{B^i}{n^ii!}) $$ $$=\sum_{i=0}^{\infty} \frac{(A+B)^i}{n^ii!}-(\sum_{i,j=0}^{\infty} \frac{A^iB^j}{n^{i+j}i!j!}) $$

At this point, I know that I need to get the first term into a double-sum format in order to make any progress. But I can't use the binomial expansion for $(A+B)^i$, because $A$ and $B$ don't commute.

I have considered trying to write the left term in terms of a non-commutative expansion over all $2^i$ permutations of $A$ and $B$, but I don't think that gets me anywhere.

For the proof I am working on, all I really need is that this limit be bounded above by a term that is $o(\frac{1}{n})$. Could anyone help me with this proof?

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  • $\begingroup$ The Lie Product formula may be useful here. (Unless you're trying to prove it, of course!) $\endgroup$ – Semiclassical Apr 25 '17 at 17:53
  • $\begingroup$ @Semiclassical, thank you, I have studed basic algebra such as rings and groups, but have not studied lie group theory. So this is probably a good stretch $\endgroup$ – Paul Apr 25 '17 at 18:29
  • $\begingroup$ ok nvm, that was actually what I'm trying to prove xD $\endgroup$ – Paul Apr 25 '17 at 18:33
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Short answer: Just keep the first few terms in the Taylor expansion.

For any operator $M$ we have that $e^{M}=I+M+O(\|M\|^2)$. Therefore $$ \exp\left(\frac{A+B}{n}\right)=1+\frac{A+B}{n}+O\left(\frac{1}{n^2}\right) $$ whereas $$ \exp\left(\frac{A}{n}\right)\exp\left(\frac{B}{n}\right)=\left[1+\frac{A}{n}+O\left(\frac{1}{n^2}\right)\right]\left[1+\frac{B}{n}+O\left(\frac{1}{n^2}\right)\right]=1+\frac{A+B}{n}+O\left(\frac{1}{n^2}\right). $$ It follows by the triangle inequality that $$ \left\|\exp\left(\frac{A+B}{n}\right)-\exp\left(\frac{A}{n}\right)\exp\left(\frac{B}{n}\right)\right\|=O\left(\frac{1}{n^2}\right). $$ Taking the limit as $n\to\infty$ gives the desired result.

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For convenience, denote $D = e^\frac{A+B}{n}-e^\frac{A}{n}e^\frac{B}{n}$. Organize the coefficients by powers of $n$. Write $$ \|D\| = \left\| \sum_{i=0}^{\infty} \frac{(A+B)^i}{n^ii!}-(\sum_{i=0}^{\infty} \frac{A^i}{n^ii!})(\sum_{i=0}^{\infty} \frac{B^i}{n^ii!}) \right\| = \\ \left\| (I - I) + \frac 1n\left[A + B - (A + B)\right] + \frac 1{n^2}\left[\frac{(A + B)^2}{2} - \frac{A^2 + 2A B + B^2}{2} \right] + \cdots \right\| = \\ \left\|\sum_{n=2}^\infty \frac {1}{n^2}C_n(A,B) \right\| \leq \sum_{n=2}^\infty \frac {1}{n^2}\left\|C_n(A,B) \right\| $$ Perhaps you could take it from there. Note that it is not necessary to explicitly compute $C_n(A,B)$, only to note that it can be written as a function of $A,B$.

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