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The initial quantum state is a coherent state. Since $a|\alpha ⟩=\alpha |\alpha ⟩$ and the state is normalized, its mean occupation number is $N=⟨a^{†}a⟩=|\alpha {|}^2$. Each quantum harmonic oscillator energy eigenstate acquires the phase $e^{-i\omega (n+1/2)t}$, soThe prefactor is a global phase and does not change an expectation value. Inverting the definitions of the annihilation operator and creation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. If $\alpha =\sqrt{N}{e}^{i\phi }$, thenFor a classical harmonic oscillator with amplitude $A$, the conserved energy is $E=m{\omega }^2{A}^2/2$. Setting $E=\hslash \omega N$ gives exactly the same amplitude, $A=\sqrt{2\hslash N/(m\omega )}$. Thus these coherent-state oscillator expectation values follow the classical orbit.

The full quantum mean energy is nevertheless $⟨H⟩=\hslash \omega (N+1/2)$: the extra zero-point energy belongs to the nonzero position and momentum uncertainties required by the Heisenberg uncertainty relation, rather than to the motion of the packet centre. In particular, $N=0$ gives zero mean displacement but a finite ground-state width. A coherent state follows a classical mean trajectory while retaining finite position and momentum uncertainties.

The quantum state is continuous through a sudden finite-force quantum quench, so it is still $|0⟩$ at $t=0^{+}$. DefineThe new Hamiltonian operator is $H=\hslash \omega (b^{†}b+1/2)-\hslash \omega {\lambda }^2$. The old Fock vacuum obeys $b|0⟩=-\lambda |0⟩$, so the coherent state after shifting an oscillator equilibrium rotates with amplitude $-\lambda e^{-i\theta }$ relative to the new equilibrium. In the original oscillator basis, the full state isOne way to obtain the phase is to writeand apply the displacement operator composition rule. The coherent-state oscillator expectation values giveThe position uncertainty stays $\sqrt{\hslash /(2M\omega )}$. These mean values obey the forced classical equation, while the nonzero variance retains the quantum width.