A question on two self-adjoint operators with same spectrum What could be an example of two self-adjoint operators $T_1,T_2\in \mathcal B(E)$,where $E$ is a hilbert space,such that both of them have cyclic vectors and have same spectrum but they are not unitarily equivalent?
What if one of them is compact and injective? Are two operators unitarily equivalent then?
I know that, $T_1$ and $T_2$ both are unitarily equivalent to some multiplication operator on $L^2(\sigma(T_i),\mu)$. So,I was thinking in terms of multiplication operators but could not come up with an example.
Any help or hint would be appreciated. Thanks in advance.
 A: A  self-adjoint operator $T$ admitting a cyclic vector is, as you said,  necessarily the multiplication operator on $L^2(\sigma(T),\mu)$ given by
$$
  T(f)x=xf(x).
   $$
So, once $\sigma(T)$ is fixed, all you have to determine is the measure $\mu$ which necessarily has full support.
It is also not hard to see that two equivalent, that is, mutually absolutely continuous measures lead to unitarily equivalent operators.
In the special case of compact operators, the spectrum always contains zero and is discrete, except for the fact that zero may be an accumulation point.  In this case two measures $\mu_1$ and $\mu_2$ with full support are equivalent iff they both assign zero for $\{0\}$ (in which case the associated operators are both injective) or they both do not (in which case both operators have a one-dimensional kernel).
As a consequence, if two injective, self-adjoint, compact operators admit cyclic vectors and share spectrum then, yes, they are unitarily equivalent.
On the other hand suppose you take the counting measure $\mu_1$ on the set
$$X=\{0\}\cup\{1/n:n\in\mathbb N\}$$
and let $\mu_2$ be the same measure, except that $\mu_2$ assigns zero to $\{0\}$.   Then the corresponding operators will provide an example to the first paragraph in your question in which $T_2$ is injective and compact.
