# What does Trotter Product Formula mean?

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 en.wikipedia.org/wiki/Matrix_exponential en.wikipedia.org/wiki/Exponential_map#Lie_theory – yoyo Jun 23 '11 at 20:15

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).