States are time-independent: $|\psi_H\rangle = |\psi_S(t_0)\rangle$.

Operators evolve in time: $A_H(t) = U^\dagger(t, t_0)A_S U(t, t_0)$.
Derivation: $\frac{dA_H}{dt} = \frac{dU^\dagger}{dt}A_S U + U^\dagger A_S \frac{dU}{dt} + U^\dagger \frac{\partial A_S}{\partial t} U$. Using $\frac{dU}{dt} = -\frac{i}{\hbar}HU$ and $\frac{dU^\dagger}{dt} = \frac{i}{\hbar}U^\dagger H$: $\frac{dA_H}{dt} = \frac{i}{\hbar}U^\dagger H A_S U - \frac{i}{\hbar}U^\dagger A_S H U + U^\dagger \frac{\partial A_S}{\partial t} U$ $\frac{dA_H}{dt} = \frac{i}{\hbar}(U^\dagger H A_S U - U^\dagger A_S H U) + \left(\frac{\partial A_S}{\partial t}\right)_H$. Since $H_H = U^\dagger H U$, and $H_S = H$, we can write $H = U H_H U^\dagger$. $\frac{dA_H}{dt} = \frac{i}{\hbar}(H_H A_H - A_H H_H) + \left(\frac{\partial A_S}{\partial t}\right)_H$. This gives the Heisenberg Equation of Motion: $$\frac{dA_H}{dt} = \frac{1}{i\hbar}[A_H, H_H] + \left(\frac{\partial A_S}{\partial t}\right)_H$$ If $A_S$ has no explicit time dependence, the last term is zero.

Momentum as Generator of Translation in Space: The operator for spatial translation by $a$ is $T_a = e^{-iaP/\hbar}$. $$T_a|\psi(x)\rangle = |\psi(x-a)\rangle$$ Derivation: For infinitesimal translation $\delta a$: $T_{\delta a} = I - \frac{i}{\hbar}P\delta a$. $T_{\delta a}\psi(x) = (I - \frac{i}{\hbar}(-i\hbar\frac{\partial}{\partial x})\delta a)\psi(x) = (I - \delta a\frac{\partial}{\partial x})\psi(x) = \psi(x) - \delta a\frac{\partial\psi}{\partial x}$. This is the Taylor expansion of $\psi(x-\delta a)$. Repeated application or exponentiation leads to $T_a|\psi(x)\rangle = |\psi(x-a)\rangle$.

Hamiltonian as Generator of Translation in Time: The time-evolution operator $U(t, t_0) = e^{-iH(t-t_0)/\hbar}$ generates translation in time. $$|\psi(t)\rangle = e^{-iH(t-t_0)/\hbar}|\psi(t_0)\rangle$$

Example: Simple Harmonic Oscillator in Heisenberg Picture

Hamiltonian: $H = \frac{P^2}{2m} + \frac{1}{2}m\omega^2 X^2$. (No explicit time dependence)
Heisenberg equations for $X_H(t)$ and $P_H(t)$: $\frac{dX_H}{dt} = \frac{1}{i\hbar}[X_H, H] = \frac{1}{i\hbar}[X_H, \frac{P_H^2}{2m}] = \frac{1}{i\hbar}\frac{1}{2m}(X_H P_H^2 - P_H^2 X_H)$. Using $[X, P^2] = XP^2 - P^2X = XPP - PPX = XPP - PXP + PXP - PPX = (XP-PX)P + P(XP-PX) = i\hbar P + P(i\hbar) = 2i\hbar P$. So, $\frac{dX_H}{dt} = \frac{1}{i\hbar}\frac{1}{2m}(2i\hbar P_H) = \frac{P_H}{m}$. $\frac{dP_H}{dt} = \frac{1}{i\hbar}[P_H, H] = \frac{1}{i\hbar}[P_H, \frac{1}{2}m\omega^2 X_H^2] = \frac{1}{i\hbar}\frac{1}{2}m\omega^2(P_H X_H^2 - X_H^2 P_H)$. Using $[P, X^2] = PX^2 - X^2P = PX^2 - XPX + XPX - X^2P = P(X^2-PX) + (PX-XP)X = -i\hbar X - i\hbar X = -2i\hbar X$. So, $\frac{dP_H}{dt} = \frac{1}{i\hbar}\frac{1}{2}m\omega^2(-2i\hbar X_H) = -m\omega^2 X_H$. These are the classical equations of motion: $$\frac{dX_H}{dt} = \frac{P_H}{m}$$ $$\frac{dP_H}{dt} = -m\omega^2 X_H$$ Differentiating the first equation: $\frac{d^2X_H}{dt^2} = \frac{1}{m}\frac{dP_H}{dt} = \frac{1}{m}(-m\omega^2 X_H) = -\omega^2 X_H$. The solutions are $X_H(t) = X_H(0)\cos(\omega t) + \frac{P_H(0)}{m\omega}\sin(\omega t)$ and $P_H(t) = P_H(0)\cos(\omega t) - m\omega X_H(0)\sin(\omega t)$.

Classical Limit (Correspondence Principle): Quantum mechanics must reproduce classical mechanics in the limit of large qu