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von-neumann-equation

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von-Neumann equation

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.

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

Notation

Alternative notation using Newton's dot notation:

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

The notation $[H, \rho]$ indicates the commutator. By definition $[H, \rho] = H\rho - \rho H$. So

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

Recall that matrix multiplication is not necessarily

Derivation

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

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

Assume the density matrix depends on time $\rho = \rho(t)$. Apply the derivative with respect to time $t$ and use the chain rule against ket $\lvert\psi\rangle$ and bra $\langle\psi\lvert$

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

Now, start from the Schrodinger equation: $$i\hbar\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle = H\lvert\psi\rangle$$

von-neumann-equation.1781013760.txt.gz ยท Last modified: by Ivan Janevski