My notes state that the spectral decomposition formula is of the form:
$$ \hat{A} = \hat{A}\hat{1} = \sum{\hat{A} } |A_i\rangle\langle A_i | = \sum{A_i } |A_i\rangle\langle A_i | $$
Now consider the Hamiltonian as $$ \hat{H} = \begin{bmatrix}E&K\\K*&E\end{bmatrix}$$ It would be easy to show that the eigenvalues are $$E_+ = E+|K|$$ and$$E_- = E-|K| $$and the eigenstates are as follows:
$$ |+\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} \frac{k}{|k|}\\ 1\\ \end{pmatrix} $$
$$ |-\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} \frac{-k}{|k|}\\ 1\\ \end{pmatrix} $$
I was expected then to show that the unitary operator $$\hat{U}(t) = e ^{it\hat{H}/\hbar}$$ can be defined as follows:
$$\hat{U}(t)=e^{-i\hat{H}t/\hbar}= e^{-iEt/\hbar}\begin{pmatrix} \cos(|K|t/\hbar) & -i\frac{K}{|K|}\sin(|K|t/\hbar) \\ -i\frac{K^*}{|K|}\sin(|K|t/\hbar) & \cos(|K|t/\hbar) \end{pmatrix}$$
According to my notes the appropriate application of the spectral decomposition formula to do so is
$$ \hat{U}(t) = e ^{it{\hat{H}}/\hbar} = e ^{it{E_+}/\hbar} |+\rangle\langle +| + e ^{it{E_-}/\hbar} |-\rangle\langle -| $$
However if we recall the spectral decomposition formula as: $$ \hat{U} = \hat{U}\hat{1} = \sum{\hat{U} } |U_i\rangle\langle U_i | = \sum{U_i } |U_i\rangle\langle U_i | $$
My question therefore is why is it appropriate to use the $|+\rangle$ and $|-\rangle$ states when they are the eigenstates of the $\hat{H}$ operator and not the $\hat{U}(t)$ one?