von-Neumann equation is basically the Schrodinger equation where the quantum state is represented by a density matrix $\rho$, rather than a state vector $\lvert\psi\rangle$. It takes the following form.

$$\frac{\mathrm d\rho}{\mathrm dt} = -\frac{i}{\hbar}[H, \rho]$$

Where $H$ is the Hamiltonian and $\rho$ is the density matrix and they both vary in time. Meanwhile $i$ is the imaginary unit and $\hbar$ is the reduced planck constant and they are both cosntants.

Alternative notation using Newton's dot notation:

$$\dot\rho = -\frac{i}{\hbar}[H, \rho]$$

The notation $[H, \rho]$ indicates the commutator. By definition $[H, \rho] = H\rho - \rho H$ (recall matrix multiplication is not necessarily commutative so this expression is not zero).

$$\dot\rho = -\frac{i}{\hbar}(H\rho - \rho H)$$

Recall that matrix multiplication is not necessarily

First, recall that a state vector $\lvert\psi\rangle$ has a density matrix representation $\rho$ inthe following way:

$$\rho = \lvert\psi\rangle\langle\psi\lvert$$

Desnity matrix $\rho$ varies in time, so we can write $\rho = \rho(t) = \lvert\psi(t)\rangle\langle\psi(t)\lvert$. But writing “$(t)$” everywhere would saturate the notation so we'll proceed without it but keep it mind! Apply the derivative with respect to time $t$ and use the chain rule against ket $\lvert\psi\rangle$ and bra $\langle\psi\lvert$

$$\frac{\mathrm d\rho}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt}\left(\lvert\psi\rangle\langle\psi\lvert\right) = \left(\frac{\mathrm d}{dt}\lvert\psi\rangle\right)\langle\psi\lvert + \lvert\psi\rangle\left(\frac{\mathrm d}{\mathrm dt}\langle\psi\lvert\right)$$

Now, start from the Schrodinger equation for “ket” $\lvert\psi\rangle$. Apply the Hermitian adjoint operation to both sides of the equation (also known as the “dagger” $\dagger$ operation) to obtain the dual Schrodinger equation equation for “bra” $\langle\psi\lvert$.

Recall that the Hamiltonian $H$ is Hermitian, so dagger has no effect on it $H^\dagger = H$. On the other hand, the dagger operaton turns a “ket” $\lvert\psi\rangle$ into “bra” $\langle\psi\rangle$ – $(\lvert\psi\rangle)^\dagger = \langle\psi\lvert$ – and flips the imaginary unit $(i)^\dagger = -i$. The dagger operation doesn't depend on time so it goes through the time derivative. The reduced planck constant $\hbar$ is unaffected.

$$i\hbar\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle = H\lvert\psi\rangle \qquad -i\hbar\frac{\mathrm d}{\mathrm dt}\langle\psi\lvert = \langle\psi\lvert H$$

Move $i\hbar$ to the right hand side. Recall $1 / i = -i$

$$\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle = -\frac{i}{\hbar}H\lvert\psi\rangle \qquad \frac{\mathrm d}{\mathrm dt}\langle\psi\lvert = \frac{i}{\hbar}\langle\psi\lvert H$$