Intuitive meaning of the exponential form of an unitary operator I'm an undergraduate student in Chemistry currently studying quantum mechanics and I have a problem with unitary transformations. 
Here in my book, it is stated that 

Every unitary operator $\hat{\mathcal{U}}$ can be written in an exponential form as follows:
  $$\mathrm{e}^{-i\alpha\hat{\mathcal{T}}}=\sum_{k=0}^{\infty}\dfrac{1}{k!}\left(-i\alpha\right)^{k}\hat{\mathcal{T}}^{k}
 $$

Provided that I have no knowledge of Lie Group/Algebra, my questions are:


*

*Why a unitary operator can be always represented by an exponential form?

*What is the intuitive mathematical meaning of the exponential form/matrix?

*What is the relation between the operator $\hat{\mathcal{U}}$ and the operator $\hat{\mathcal{T}}$?


Thanks in advance!
 A: The spectral theorem for normal operators on a Hilbert space gives you the result, but you don't get a unique selfadjoint $T$. Any unitary $U$ can be written as
$$
                            U = \int_{|\lambda|=1}\lambda dE(\lambda),
$$
where $E$ is the unique spectral measure associated with $U$. Choose a branch of $\log$ and define
$$
                         T = -i\int_{|\lambda|=1}\log(\lambda)dE(\lambda).
$$
Then $\log(\lambda)$ is purely imaginary and $T=T^{\star}$. By the functional calculus,
$$
      e^{iT} = \exp\left\{\int_{|\lambda|=1}\log(\lambda)dE(\lambda)\right\}
             = \int_{|\lambda|=1}e^{\log(\lambda)}dE(\lambda) = U.
$$
This $T$ is rarely useful to you as an observable because $T$ is a bounded linear operator with $\|T\| \le 2\pi$. Exponentiation does this. If $H$ is a selfadjoint linear operator (possibly unbounded),
$$
                           H = \int_{\mathbb{R}} \lambda dF(\lambda)
$$
and
$$
                e^{iH} = \int_{\mathbb{R}}e^{i\lambda}dF(\lambda),
$$
but there's no general way to find $H$ from $e^{iH}$. If you know $e^{itH}$ for $t$ in a range of values $[0,\epsilon]$ for some $\epsilon > 0$, then you can recover $H$ using differentiation in $t$, but not just from knowing $e^{iH}$.
