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--- von-neumann-equation [June 09, 2026 at 14:43] – Ivan Janevski
+++ von-neumann-equation [June 10, 2026 at 23:55] (current) – external edit 127.0.0.1
@@ Line 32: / Line 32: @@
 $$ i\hbar\frac{\mathrm d}{\mathrm dt}\lvert\psi\rangle = H\lvert\psi\rangle $$
-Move $i\hbar$ to the other side (will be convenient later). Recall the reciprocal value of the imaginary unit flips the sign $1 / i = -i$.
+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" $\langle\psi\lvert$.
+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.
-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 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
@@ Line 67: / Line 60: @@
 That's the von Neumann equation!
+## List of code implementations
+ - [[von-neumann-equation-qutip|von Neumann equation (QuTiP)]]