What is the state complexity of the copy language? Let a number $n$ be given.  Consider the following language $L_n = \{ \; ww \; \vert \; w \in \{0,1\}^{n} \; \}$.

In words, $L_n$ is the set of copy strings of length $2n$.

Consider the following state complexity function $s$ such that $s(n)$ is the number of states in the smallest Pushdown Automata that recognizes $L_n$.

Question: Can you formally prove any meaningful lower bound for $s(n)$?

My Conjecture: $s(n) = 2^{\Theta(n)}$.

Known Upperbound: $s(n) \leq \mathrm{poly}(n) \cdot 2^{\frac{n}{2}}$.

Rules:

(1) The stack alphabet must be binary.

(2) The input tape is one-way and can't stop on any input character.
The technique described by Yuval: 

Do there exists polynomial size CFG that describe this finite language?

(
you may also read:
Lower bounds on the size of CFGs for specific finite languages
)

allows to show very easily an exponential lower bound for CFGs.
Let $G$ a grammar in Chomsky Normal Form for $L_n$.
For every word $w\in \{0,1\}^n$ there exists at least one non-terminal $A(w)$ accepting a subword $s(w)$ of $ww$ having length between $n/2$ and $n$. Let $p(w)$ be a position in $ww$ where this subword occurs. There are at least $n/2$ bits common to all words $w,w'$ such that $A(w)=A(w')$ and $p(w)=p(w')$. Consequently, there can be at most $2^{n/2}$ words that have the same $A(w)$ and $p(w)$. Hence there are at least $2^{\Theta(n)}$ non-terminals.

Furthermore, the PDA can be converted into a CFG in CNF, of polynomial size so this also gives the $2^{\Theta(n)}$ bound on the state complexity of $L_n$.