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For some reason, I have to work with Trotter product formula recently, but I do not have a strong background in functional analysis.

The following is the statement of the formula from MathWorld

When A and B are self-adjoint operators, $$ e^{t(A+B)} = \lim_{n \to +\infty}(e^{tA/n}e^{tB/n})^n $$

My questions are:

  • What does the exponential of an operator mean precisely?

  • How to interpret the convergence? In terms of some norm?

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As a first step, I suggest to try to understand it for $n \times n$-matrices. If $A$ and $B$ are bounded operators on a Banach space, the proof for matrices carries over quite painlessly. If $A$ and $B$ are unbounded, however, you need to assume much more (essentially you need to guarantee that all $A,B$ and $A+B$ generate contraction semigroups), the Hille-Yosida theorem is relevant here (the contraction semigroup generated by $A$ is traditionally denoted by $e^{At}$ for good reasons). Convergence is in the strong operator topology. – t.b. Jun 23 '11 at 19:49
$e^A=\sum_{k=1}^{\infty}A^k/k!$. if $A$ comes from some normed space (for instance the euclidean norm on $\mathbb{R}^{n^2}$ for an $n\times n$ matrix or $\sup_{|x|=1} |Ax|/|x|$ for a bounded operator on a banach space) then you can take convergence wrt that norm. see maybe – yoyo Jun 23 '11 at 20:15
up vote 10 down vote accepted

In this case $e^{tA}$ denotes the strongly continuous semigroup generated by the operator $A$ which does not have to be bounded (otherwise it is quite boring).

I suggest you take a look at the book "A Short Course In Operator Semigroups" by Engel and Nagel.

The book is in the yellow sale and available on springerlink.

This formula can be deduced from the Chernoff product formula so you could also look for that. Check out Chapter IV.2.

In the Seventh Internet Seminar 2003/04 semigroups were also treated and the book I'm recommending is based on those notes. Those notes can probably be found online.

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The only thing I can find about that internet seminar is this: It seems like the schedule covered great things. The Ornstein-Uhlbeck semigroup!! – Jonas Teuwen Jun 23 '11 at 20:05
Perfect reference. By the way, you mean Ornstein-UhlENbeck (see also here on MO :) – t.b. Jun 23 '11 at 20:10
@Theo: Oh, then I have always done it wrong. Thanks for the correction. – Jonas Teuwen Jun 23 '11 at 20:11
Thank you so much. I'm actually working on Trotter-Kato approximation. This is the exact book I want to see. – Hawii Jun 23 '11 at 22:42
@Hawii: No problem! There is also the bigger brother of this book by the same authors but I didn't read it. I had a seminar where we studied the book I have linked and my subject was the Chernoff product formula (which has the Trotter product formula as corollary). Why do I say this? Well, because I have read the book and to me it is very readable (and this also seemed to be the case for the other participants) so I can recommend it. – Jonas Teuwen Jun 23 '11 at 22:53

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