Hilbert Space multiplication Operator, shift operator I have this problem and am not sure how to even approach it..
Hilbert space $l^2(\mathbb{Z})$ with orthonormal basis$ $$(e_n)$ and Hamiltonian operator $He_n=i(e_{n+1}-e_{n-1})$
a)I need to use Fourier series to find a multiplication operator which is unitary equivalent to H
b)Use a) to show that there is a no vector $h \in l^2(\mathbb{Z})$ , $h\neq 0$ such that $exp(-itH)h=h$ for all $t \in R$
 A: As pointed out by MaoWao, consider the following Fourier transform operator
${\cal F}: \ell^2(\mathbb{Z}) \rightarrow L^2(S^1)$
$$
{\cal F}[f(n)](k) = \hat{f}(k) = \frac{1}{\sqrt{2 \pi}} \sum_n e^{i kn} f(n)
$$ 
with inverse
$$ 
{\cal F}^{-1}[g(k)](n) = \frac{1}{\sqrt{2 \pi}} \int_0^{2 \pi} dk e^{-i kn} g(k)
$$
For point a) 
First compute the Fourier transform of the standard basis. Note that $e_n(m) = \delta_{n,m}$ (Kronecker delta). So
\begin{align}
\hat{e_n}(k) &= \frac{1}{\sqrt{2 \pi}} \sum_m e^{i km} \delta_{n,m}\\
 & = \frac{e^{ikn}}{\sqrt{2\pi}} 
\end{align}
Now use
$$
{\cal F} H{\cal F}^{-1} {\cal F} e_n = i ({\cal F}e_{n+1} - {\cal F}e_{n-1})
$$
which means 
\begin{align}
{\cal F} H{\cal F}^{-1} e^{ikn} &= i (e^{ik(n+1)}- e^{ik(n-1)} ) \\
&= -2\sin(k) e^{ikn}
\end{align}
Hence $H$ is unitarily equivalent to multiplication by $-2\sin(k)$. Note that the sign is conventional: a different choice of signs in the Fourier operator would have led to $2\sin(k)$. 
For point b)
Now Fourier transform the equality $ \exp{itH}h = h$ to get
$$
\exp{ ( it {\cal F} H{\cal F}^{-1} )} \hat{h} = \hat{h} \tag{1}
$$
Hence you need to find a non-zero function $\hat{h}(k)$ in $L^2(S^1)$ such that
$$
\exp{[-2it \sin(k)]} \, \hat{h}(k) = \hat{h}(k)
$$
for all $k\in [0,2\pi)$ (and all t) which is impossible.
Note added:
In fact $\sin(k)$ is non-zero for all $k$ (in the interval) except $k=0,\pi$. Hence, in order to satisfy (1), the function $\hat{h}(k)$ should be zero for all $k$ except possibly $k=0,\pi$. This is clearly the zero function almost everywhere. To add some intuition you can reason as follows. If such a function existed, it would try to be "infinite" at $k=0$ (or $k=\pi$ or both), in order to have a finite (non-zero) norm. This is achieved by a (Dirac) delta function --which of course is not a function--. So, "in a sense", $\hat{h}(k) = \delta(k)$, is a solution of $\exp{itH}h=h$. If you Fourier transform back you get the function (vector) with components
$$
h(n) = \frac{1}{\sqrt{2 \pi}}
$$
You can plug this in your equation defining $H$ and you will notice that it leads to $H h = 0$. So it would seem that $h$ is an eigenvector of $H$ with eigenvalue zero. However the vector with components $h(n) \propto 1$ is not normalizable, hence such $h$ does not belong to $\ell^2(\mathbb{Z})$. 
