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.
Best Answer
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.