Schrödinger Equation as a limit of von Neumann equation How would I derive the Schrödinger Equation as a limit of the von Neumann equation?
The quantum Liouville equation (von Neumann equation) is given by
$$i \hbar \: \partial_t \rho = [ H, \rho ] \quad .$$
Similarly, the Schrödinger equation reads
$$ i \hbar \: \partial_t |\psi\rangle = H |\psi \rangle \quad .$$
I'm assuming that the Schrödinger equation is a limiting case of the Liouville (mixed states to pure state), but how would I write that mathematically?

Here is my attempt...
My guess is that it goes something like
$$\rho = \sum_i p_i | \psi_i \rangle \langle \psi_i | \quad .$$
Insert that into the Quantum Liouville equation:
\begin{align}
i \hbar \: \partial_t \rho &= \sum_i p_i [ H, | \psi_i \rangle \langle \psi_i |]\\
i \hbar \: \partial_t \rho &= \sum_i p_i (H  | \psi_i \rangle \langle \psi_i | - | \psi_i \rangle \langle \psi_i | H )\\
i \hbar \: \partial_t \rho &= \sum_i p_i E_i ( | \psi_i \rangle \langle \psi_i | - | \psi_i \rangle \langle \psi_i |  )  = 0 .
\end{align}
Which obviously doesn't work. Can anyone help?
For reference, I'm looking notation from here on Wikipedia.
 A: 
The quantum Liouville equation (von Neumann equation) is given by
$$i \hbar \: \partial_t \rho = [ H, \rho ] \quad .$$
Similarly, the Schrödinger equation reads
$$ i \hbar \: \partial_t |\psi\rangle = H |\psi \rangle \quad .$$


I'm assuming that the Schrödinger equation is a limiting case of the Liouville (mixed states to pure state), but how would I write that mathematically?

I would not describe the situation this way, but I would say they are certainly related. If you wanted my opinion on which is primal, I would say the Schrodinger equation.
Anyways, take your definition of
$$
\rho = \sum_i p_i |\Psi_i\rangle\langle\Psi_i|
$$
and take the "limit" where all $p_i\to 0$ except for one of them $p_0 \to 1$. This is called a "pure state," and pure states are the main purview of the Schrodinger equation, so we will be able to derive a relationship.
For the pure state we have:
$$
\rho_{\text{pure}} =  |\Psi_0\rangle\langle\Psi_0|\;.
$$
Setting $\hbar=1$, the Schrodinger equation for our state $|\Psi_0\rangle$ is:
$$
i\partial_i|\Psi_0\rangle = H|\Psi_0\rangle\;,\tag{A}
$$
and its Hermitian conjugate is:
$$
-i\partial_i\langle\Psi_0| = \langle\Psi_0|H.\tag{B}
$$
Therefore, the Liouville-von-Neumann equation follows from the Schrodinger equation since:
$$
i\partial_t\rho_{\text{pure}}=i\partial_t(|\Psi_0\rangle\langle\Psi_0|) 
$$
$$
= i\left(H|\Psi_0\rangle\langle\Psi_0| - |\Psi_0\rangle\langle\Psi_0|H\right)\tag{C}
$$
$$
=i[H,\rho_{\text{pure}}]\;,
$$
where the third line (tagged "C") follows by plugging in the the two forms of the Schrodinger equation in Eq. (A) and Eq. (B).

Update (writing up TF's comments as a answer to the converse question):
Assuming the Hamiltonian is time-independent, I can formally solve the LvN equation by writing:
$$
\rho(t) = U(t)\rho(0)U^\dagger(t)\;,
$$
where $U(t) = e^{-iHt}$.
Therefore, for a pure state:
$$
\rho(t) = |\Psi(t)\rangle\langle\Psi(t)| = U(t)|\Psi(0)\rangle\langle\Psi(0)|U^\dagger(t)\;,
$$
where we now identify
$$
|\Psi(t)\rangle = U(t)|\Psi(0)\rangle\;.
$$
Now, we see that
$$
\partial_t|\Psi(t)\rangle = (\partial_t U(t))|\Psi(0)\rangle
=-iH|\Psi(t)\rangle\;,
$$
which is the Schrodinger equation.
A: Here was an answer provided to me by a friend that I am also happy with.
It resembles the one from this textbook.
$$i \partial_t \rho = H \rho - \rho H $$
Assume $\rho = | \psi \rangle \langle \psi | $.
$$i \partial_t | \psi \rangle \langle \psi |= H | \psi \rangle \langle \psi | - | \psi \rangle \langle \psi | H $$
product rule
$$i (\partial_t | \psi \rangle ) \langle \psi | + i | \psi \rangle (\partial_t \langle \psi |) = H | \psi \rangle \langle \psi | -  | \psi \rangle \langle \psi | H $$
dot with $| \psi \rangle$ and assume normalized $\langle \psi |  \psi \rangle =1 $
$$i (\partial_t | \psi \rangle ) + i | \psi \rangle (\partial_t \langle \psi |)| \psi \rangle  = H | \psi \rangle - | \psi \rangle \langle \psi | H | \psi \rangle$$
use Schrodinger eqn $ - i \partial_t \langle \psi | = H \langle \psi |$ to cancel terms....
$$i \partial_t | \psi \rangle = H | \psi \rangle $$
