Representation of time evolution operator and spectral decomposition I was preparing for my quantum mechanics exam, and I came to think about this question regarding the spectral representation of time evolution operator. Let's say we are given the Hamiltonian:
$\hat{H} = S_{-} S_{+} $.
What will be the time evolution operator in the spectral representation? Here is my shot at it. What worries me is that, shouldn't this Hamiltonian have two eigenstates? I am only getting one. Anyways, here is my approach:
The raising and lowering spin operators can be written, in terms of the z-basis, as:
$S_{+} = \hbar |+ z \rangle \langle -z|$, $S_{-} = \hbar |- z \rangle \langle +z | $.
Then, the Hamiltonian becomes:
$\hat{H} = \hbar^2 |- z \rangle \langle +z |+ z \rangle \langle -z| = \hbar^2 |-z\rangle \langle -z|. $
Now, the only eigenvector of this equation is $|-z\rangle$, with the corresponding eigenvalue, $\hbar^2$. So, we can write the spectral representation of the time evolution operator, given by:
$e^{i\hat{H}t/\hbar} = e^{i\hbar t} |-z\rangle \langle -z|$. 
Is this it? Or am I missing something here? Any help would be greatly appreciated!
 A: $\newcommand{\bra}[1] {\left< #1 \right|}
 \newcommand{\ket}[1] {\left| #1 \right>}
 \newcommand{\bracket}[2] {\left< #1 \vert #2 \right>}
$
Your calculation of the Hamiltonian
$$\hat{H} = \hbar^2 \ket{-z}\bra{-z}$$
is correct so far.
One eigenvector is $\ket{-z}$, with eigenvalue $\hbar^2$.
But you missed (as Dani already wrote in his comment):
another eigenvector is $\ket{+z}$, with eigenvalue $0$.
So you could also write (in a rather pedantic way):
$$\hat{H} = 0\ket{+z}\bra{+z} + \hbar^2 \ket{-z}\bra{-z}$$
Then it is more obvious, that the time evolution operator becomes:
$$e^{i\hat{H}t/\hbar} = 1 \ket{+z}\bra{+z} + e^{i\hbar t}\ket{-z}\bra{-z}.$$

Yet another equivalent approach is by using the matrix representation
(with $\ket{+z}$ and $\ket{-z}$ as base vectors)
and calculate the matrix exponential of a diagonal matrix.
We have
$$\hat{H}=\begin{pmatrix}0 & 0 \\ 0 & \hbar^2\end{pmatrix}.$$
Then we get
$$e^{i\hat{H}t/\hbar}
= \exp\left[i\begin{pmatrix}0 & 0 \\ 0 & \hbar^2\end{pmatrix}t/\hbar\right]
= \exp \begin{pmatrix}0 & 0 \\ 0 & i\hbar t\end{pmatrix}
= \begin{pmatrix}1 & 0 \\ 0 & e^{i\hbar t}\end{pmatrix}
$$
A: My quantum mechanics is a little rusty but I will attempt an answer. The problem is that the Hamiltonian is not Hermitean. A Hermitean operator in this case would have two eigenvectors. You need to add the hermitean conjugate of the S_{-}S_{+} term to the hamiltonian. From what I remember this is common practice in condensed matter physics. 
