Relationship between adjoint operators, trace-class operators, compact operators and density operators in Quantum-Mechanics I don't know much about Functional Analysis, but I was wondering about the following:
In Banach spaces it is possible to define for every continuous opertor $T:X \rightarrow Y$ an adjoint Operator $T':Y' \rightarrow X'$.
Now I heard that the set of trace-class Operators is the dual space of all compact Operators $K(X,Y)$. 
Question 1: Does this mean that if $T$ is compact then $T'$ is trace-class? 
I guess not, but Schauder's Theorem tells us that $T'$ is also compact. I guess that it is rather that for every $T$ compact, we have a dual map $T^*$ in the set of trace-class Operators such that $T^*(T) \in \mathbb{K}$.
Question 2: Is it generally true that if $T$ is trace-class, then $T$ is compact?
Question 3 ( if you cannot answer this one, no Problem ): In Quantum Mechanics we define density Operators such that they are positive semi-definite self adjoint Operators such that $\text{Tr(T)=1}$. My question is: Since these are obviously trace-class Operators, are they induced by some compact Operator $S$ such that a density Operator $T=S^*$?
 A: To answer your questions, we use the fact that an operator on a Hilbert space is compact if and only if it maps some orthonormal basis to a sequence converging to $0$.


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*Question 1: No. Consider the operator on $\ell^2$ given by
$$T(\{x_k\}) = \{k^{-1}x_k\}$$
and denote by $e_j$ the standard basis. $T$ is obviously self-adjoint an $Te_j = j^{-1}e_j \to 0$ so $T$ is compact. But $T' = T$ is not trace-class since
$$\sum_j |(e_j, Te_j)| = \sum_j j^{-1} = \infty.$$

*Question 2: Yes. If $T$ is trace-class, then for any orthonormal basis $\{e_j\}$ we have
$$\sum_j |(e_j, Te_j)| < \infty$$
and thus $T(e_j) \to 0$. Hence $T$ is compact.

*Question 3: Yes, but only in a trivial way. $S = T$ gives you what you want since $T$ is self-adjoint and trace-class, hence compact.
I apologize if this is wrong, but from the phrasing of your question it seems that you're a little bit confused about what is meant by the statement "the trace-class operators are the dual of the compact operators." This statement doesn't really have much to do with the adjoint of a compact or trace-class operator. It means that there exists an isometric isomorphism from the Banach space of trace-class operators on $H$ to the dual space of the Banach space of compact operators on $H$.
