Simple question about Schrodinger equation (time independent) For a quantum mechanical description of a system (like a small molecule) we can write:
$$\langle\psi|\hat {H}|\psi\rangle = \overline E$$
Question: 
Is that energy the same as zero Kelvin energy obtained by statistical mechanics (using $E_n$ energies and partition functions)?
 A: The average $\langle E\rangle$ you have on the right is not to be understood in the sense of statistical mechanics (and thus as a temperature).
The wavefunction $\psi$ can be a linear combination of states $\psi_n$ of definite energy of your molecule.  The $\psi_n$’s are solutions of the Schrodinger equation.  Thus, 
\begin{align}
\psi(x,t)&=\sum_i a_i \psi_i(x,t)\,  ,\qquad \sum_i \vert a_i\vert^2=1\, , \tag{1} \\
\hat H\psi_i(x,t)&=E_i\psi_i(x,t)\, . 
\end{align}
With this wavefunction the average energy
$$
\langle E\rangle = \sum_i \vert a_i\vert^2 E_i \tag{2}
$$
is a weighted average of the possible energies of your molecule, with the modulus square $\vert a_i\vert^2$ of the complex amplitude $a_i$ functioning as statistical weight.  $\psi(x,t)$ given in (1) is an example of a pure state.
For instance, if you have a hydrogen wavefunction of the form
$$
\psi(r)=a_1\psi_{100}(r)+a_2\psi_{200}(r)
$$
then the average energy of this system is 
$$
\vert a_1\vert^2\times (-13.6) + \vert a_2\vert^2 \times (-13.6/4).  \tag{3}
$$
Measuring the energy, you will sometimes get the value $-13.6$ (this outcome will occur $\vert a_1\vert^2$ of the time) and sometimes get the value $-13.6/4$ (this outcome will occur $\vert a_2\vert^2$ of the time).  The average energy is exactly given by (3).
In quantum statistical mechanics one introduces the concept of a mixted state.  For mixed states one cannot define a wavefunction for the system as in (1).  The system is described using a matrix that represents a statistical mixture of wavefunction that is not of the form given in (1); the average energy for this statistical mixture is also not of the form (2) since the latter comes from (1), which does not exist for mixted states.
A: Statistical mechanics does not really apply to the kind of system your equation refers to: your equation is good for a pure quantum state $\psi$. In statistical mechanical terms, the system's microstate is exactly and fully specified by $\psi$, and you can't really therefore talk about a system's temperature when it is in a pure state.
Recall that pure states can be non ground states, or in a superposition containing non ground states. So there is, in general, no relationship definable with a "$0{\rm K}$ energy".
A classical mixture of pure quantum states can describe a statistical mechanical system, can therefore be assigned a temperature and is described by a density matrix $\rho$. In such a case, your equation becomes replaced by:
$$\langle E\rangle = \mathrm{tr}(\rho\,\hat{H})$$
Perhaps a kind of converse to your question makes more sense and yields a more concrete answer: as we cool a statistical mechanical system down, and if the ground state is nondegenerate, the system becomes "forced" into a state where all particles / ensemble members are in the ground state. If, further, the ground state is nondegenerate (i.e. there is only one ground quantum energy eigenstate), then the $0{\rm K}$ state is a multiparticle pure quantum state and this state is the ground energy eigenstate. User Joshphysics analyzes this behavior in more detail in his answer here. 
