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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.

Define the signed dimensionless displacementThe spatial translation operator is $T(x_0)=e^{-ip{x}_0/\hslash }$, and the phases of the new energy eigenstates are fixed by $|n^{\prime }⟩=T(x_0)|n⟩$. Their expansion coefficients are thereforeSince $p=i\sqrt{m\hslash \omega /2}(a^{†}-a)$,Set $A=-\lambda a^{†}$ and $B=\lambda a$. Their commutator is $[A,B]={\lambda }^{2}I$, which commutes with both operators. Thus the stated special case of the Baker--Campbell--Hausdorff formula is applicable and givesBecause $a|0⟩=0$,Taking the overlaps with $⟨n|$ therefore provesThis is the coherent state after shifting an oscillator equilibrium. Its normalization follows from $e^{-{\lambda }^2}{\sum }_n{\lambda }^{2n}/n!=1$.

There is a sign qualification in the printed expansion: $(m\omega x_0^2/(2\hslash ){)}^{n/2}=|\lambda {|}^n$ agrees with $(-\lambda {)}^n$ when $x_0\le 0$, equivalently $\gamma \ge 0$. For a general signed displacement $x_0>0$, the odd coefficients have the opposite sign in the specified translated basis. The squared displacement alone fixes the occupation probabilities, not these coherent relative phases. The two-exponential identity also requires the central-commutator condition used above; it is not valid in that simple form for arbitrary operators.