Strange question about Hechler Recently during a lecture, my professor mentioned that forcing over any poset which is countable, separative, and atomless, is essentially the same as forcing over the Cohen poset, that is to say results in adding a Cohen real. 
My question is: Are there any other similar characterizations of "commonly used" forcing posets? Specifically, is there one for the Hechler poset?
The Hechler Poset/forcing notion $(H,\le)$ is given by setting $H=\omega^{\lt\omega} \times \omega^{\omega}  $, and defining the relation $(t,v) \le (s,u)$ iff $ ( t \supset s  \wedge (\forall n\in\omega) (u(n) \le v(n)) \wedge (\forall m \in dom(t \backslash s))(t(m) \gt u(m))$. When forcing with this poset, you end up adding an unbounded real to the ground model.
I understand that you cannot produce a model in which $\mathfrak{b}=\omega_2$ using product forcing, and that you need iterated forcing to do so. Moreover, the iterated forcing construction I've seen that produces $\mathfrak{b}=\omega_2$ in the forcing extension, used the finite support iteration of $\omega_2$ many copies of the Hechler poset. Is this evidence for the lack of such a characterization?
(I apologize in advance if this is an ill-stated question, I will change it accordingly if it is.)
 A: Here are a few additional examples of the kind you seek,
and in fact each of them directly generalizes the
characterization you mention of the forcing to add a Cohen
real. However, I know of no such characterization of Hechler forcing.


*

*The collapse forcing
 $\text{Coll}(\omega,\theta)$ is, up to forcing equivalence, the unique forcing
 notion of size $|\theta|$ necessarily collapsing
 $\theta$ to $\omega$. (Note, this includes the case of
 adding a Cohen real, since $\text{Add}(\omega,1)\equiv\text{Coll}(\omega,\omega)$.)
To see this, suppose that $\mathbb{Q}$ is a forcing notion
of size $|\theta|$ that necessarily collapses $\theta$ to
$\omega$. We may assume without loss of generality that
$\mathbb{Q}$ is separative, since the separative quotient
of $\mathbb{Q}$ is forcing equivalent to it and no larger
in size. Observe that below every condition in
$\mathbb{Q}$, there is an antichain of size $\theta$. Since
forcing with $\mathbb{Q}$ adds a function from $\omega$
onto $\theta$ and $\mathbb{Q}$ has size $\theta$, there is
a name $\dot g$ forced to be a function from $\omega$ onto
the generic filter $\dot G$. We build a refining sequence
of maximal antichains $A_n\subset\mathbb{Q}$ as follows. Begin
with $A_0=\{ 1\}$. If $A_n$ is defined, then let
$A_{n+1}$ be a maximal antichain of conditions such that
every condition in $A_n$ splits into $\theta$ many elements
of $A_{n+1}$, and such that every element of $A_{n+1}$
decides the value $\dot g(\check n)$. The union
$\mathbb{R}=\bigcup_n A_n$ is clearly isomorphic as a
subposet of $\mathbb{Q}$ to the tree $\theta^{\lt\omega}$, and
so it is forcing equivalent to
$\text{Coll}(\omega,\theta)$. Furthermore, $\mathbb{R}$ is
dense in $\mathbb{Q}$. To see this, fix any condition
$q\in\mathbb{Q}$. Since $q$ forces that $q$ is in $\dot G$,
there is some $p\leq q$ and natural number $n$ such that
$p$ forces via $\mathbb{Q}$ that $\dot g(\check n)=\check
q$. Since $A_{n+1}$ is a maximal antichain, there is some
condition $r\in A_{n+1}$ that is compatible with $p$. Since
$r$ also decides the value of $\dot g(\check n)$ and is
compatible with $p$, it must be that $r$ forces $\dot
g(\check n)=\check q$ also. In particular, $r$ forces
$\check q\in\dot G$, and so by separativity it must be that
$r\leq q$. So $\mathbb{R}$ is dense in $\mathbb{Q}$, as
desired. Thus, $\mathbb{Q}$ is forcing equivalent to
$\mathbb{R}$, which is forcing equivalent to
$\text{Coll}(\omega,\theta)$, as desired.

*If $\kappa^{\lt\kappa}=\kappa$, then the forcing $\text{Add}(\kappa,1)$ to add a Cohen subset
to $\kappa$ with conditions of size less than $\kappa$ is,
up to forcing equivalence, the unique ${\lt}\kappa$-closed
necessarily nontrivial forcing notion of size $\kappa$.
For this, one similarly builds up a dense tree of conditions
inside the poset that is isomorphic to $\kappa^{\lt\kappa}$, which
is forcing equivalent to $\text{Add}(\kappa,1)$.

*More generally, if $\theta^{\lt\kappa}=\theta$, then the forcing $\text{Coll}(\kappa,\theta)$
is the unique ${\lt}\kappa$-closed forcing notion
necessarily collapsing $\theta$ to $\kappa$ and having size $\theta$. (The
straightforward generalization uses separativity, but the
separative quotient of a partial order may no longer be $\lt\kappa$-closed, so there is an
issue about it; but my
student Norman Perlmutter found a solution avoiding the issue.)
