Quantum Mechanical Operators in the argument of an exponential In Quantum Optics and Quantum Mechanics, the time evolution operator 
$$U(t,t_i) = \exp\left[\frac{-i}{\hbar}H(t-t_i)\right]$$ 
is used quite a lot. 
Suppose $t_i =0$ for simplicity, and say the eigenvalue and eigenvectors of the hamiltionian are $\lambda_i, \left|\lambda_i\right>$.
Now, nearly every book i have read and in my lecture courses the following result is given with very little or no explanation:
$$U(t,0) = \sum\limits_i \exp\left[-\frac{i}{\hbar}\lambda_it\right]\left|\lambda_i\right>\left<\lambda_i\right|$$
This is quite a logical jump and I can't see where it comes from, could anyone enlighten me?
 A: Starting with:
$$U(t,t_i) = e^{\frac{-i}{\hbar }H(t-t_i)}$$
If $t_i=0$:
$$U(t,0) = e^{\frac{-i}{\hbar }Ht}$$
Using the identity: $\sum\limits_i \left|\lambda_i\right>\left<\lambda_i\right|=\mathbb{I}$
$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }Ht}\left|\lambda_i\right>\left<\lambda_i\right|$$  
Since the exponential of an operator is (by Taylor expanding): $e^H=\mathbb{I}+H+\frac{1}{2}H^2+\dots$
And: $H\left| \lambda_i \right> =\lambda_i \left| \lambda_i \right>$
You should be able to see that:
$$U(t,0) = \sum\limits_i e^{\frac{-i}{\hbar }\lambda_it}\left|\lambda_i\right>\left<\lambda_i\right|$$ 
A: Without loss of generality, let's take the $|\lambda_i\rangle$ to be orthonormal.  Notice that, by the spectral theorem, the hamiltonian can be written as follows:
$$
  H = \sum_i \lambda_i P_i, \qquad P_i = |\lambda_i\rangle\langle \lambda_i|
$$
Each operator $P_i$ is a projectors onto the subspace spanned by $|\lambda_i\rangle$.  Notice, in particular, that
$$
  P_i^2 = P_i, \qquad P_iP_j = P_jP_i = 0
$$
and a mathematical induction argument gives
$$
  P_i^n = P_i
$$
for all $n\geq 1$.  Now, for notational simplicity let
$$
  \mu = -\frac{i}{\hbar}t
$$
Then we have
$$
  U(t,0) = e^{\mu H} = \sum_{n=0}^\infty \frac{1}{n!}\mu^nH^n
$$
but notice that using the properties of projection operators written above, we have
$$
  H^n = \sum_{i_1, \dots, i_n}\lambda_{i_1}\cdots\lambda_{i_n}P_{i_1}\cdots P_{i_n} = \sum_i\lambda_i^nP_i
$$
and therefore
$$
  U(t,0) = \sum_i\sum_n\frac{1}{n!}(\mu\lambda_i)^nP_i = \sum_ie^{\mu\lambda_i}P_i
$$
as desired.
