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--- von-neumann-equation [June 09, 2026 at 14:05] – Ivan Janevski
+++ von-neumann-equation [June 10, 2026 at 23:55] (current) – external edit 127.0.0.1
@@ Line 22: / Line 22: @@
 $$\rho = \lvert\psi\rangle\langle\psi\lvert$$
-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 $\langle\psi\lvert$
+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 $\langle\psi\lvert$
-$$\frac{\mathrm d\rho(t)}{\mathrm dt}
+$$\frac{\mathrm d\rho}{\mathrm dt}
-= \frac{\mathrm d}{\mathrm dt}\left(\lvert\psi(t)\rangle\langle\psi(t)\lvert\right)
+= \frac{\mathrm d}{\mathrm dt}\left(\lvert\psi\rangle\langle\psi\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)$$
+= \left(\frac{\mathrm d}{dt}\lvert\psi\rangle\right)\langle\psi\lvert + \lvert\psi\rangle\left(\frac{\mathrm d}{\mathrm dt}\langle\psi\lvert\right)$$
-Now, start from the Schrodinger equation:
+Now, start from the Schrodinger equation for "ket" $\lvert\psi\rangle$.
-$$i\hbar\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle = H\lvert\psi\rangle$$
+$$ i\hbar\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle = H\lvert\psi\rangle $$
+Move $i\hbar$ to the other side (this will be convenient later). Recall the reciprocal value of the imaginary unit flips the sign $1 / i = -i$.
+$$ \frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle = -\frac{i}{\hbar}H\lvert\psi\rangle $$
+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.
+$$\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle = -\frac{i}{\hbar}H\lvert\psi\rangle
+\qquad \frac{\mathrm d}{\mathrm dt}\langle\psi\lvert = \frac{i}{\hbar}\langle\psi\lvert H$$
+Multiply the "ket" equation with $\langle\psi\lvert$ from the right and multiply the "bra" equation with $\lvert\psi\rangle$ from the left:
+$$\left(\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle\right)\langle\psi\lvert = -\frac{i}{\hbar}H\lvert\psi\rangle \langle\psi\lvert
+\qquad \lvert\psi\rangle\left(\frac{\mathrm d}{\mathrm dt}\langle\psi\lvert\right) = \frac{i}{\hbar}\lvert\psi\rangle\langle\psi\lvert H$$
+Add both "ket" and "bra" equations up:
+$$\left(\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle\right)\langle\psi\lvert + \lvert\psi\rangle\left(\frac{\mathrm d}{\mathrm dt}\langle\psi\lvert\right) = -\frac{i}{\hbar}(H\lvert\psi\rangle \langle\psi\lvert + \lvert\psi\rangle\langle\psi\lvert H)$$
+The left hand side is the time derivative of $\lvert\psi\rangle\langle\psi\lvert$ after applying the chain rule seen in the beginning. First, undo the chain rule. Then replace $ \lvert\psi\rangle\langle\psi\lvert$ with density matrix $\rho$.
+$$\frac{\mathrm d\rho}{dt} = -\frac{i}{\hbar}(H\rho - \rho H)$$
+Define the commutator $[X, Y] = XY - YX$. Then use the commutator to replace $(H\rho - \rho H)$ with $[H, \rho]$.
+$$\frac{\mathrm d\rho}{dt} = -\frac{i}{\hbar}[H, \rho]$$
+That's the von Neumann equation!
+## List of code implementations
+ - [[von-neumann-equation-qutip|von Neumann equation (QuTiP)]]