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Put $C=[A,B]$, which commutes with both operators by hypothesis. Differentiating $e^{\lambda A}B{e}^{-\lambda A}$ gives $C$, so this conjugate is $B+\lambda C$. Similarly the product $P=e^{\lambda A}{e}^{\lambda B}$ conjugates $A+B$ to itself: the two central commutator shifts cancel. HenceWith $F(0)=I$, the operator differential equation gives $F(\lambda )=e^{{\lambda }^{2}C/2}$. At $\lambda =1$ this proves the central-commutator Baker--Campbell--Hausdorff formulaThese formal manipulations apply on a suitable common domain, or to the oscillator exponentials below where the corresponding unitary identities are well defined.

The oscillator has $[a,a^{†}]=1$ and $H=\hslash \omega (a^{†}a+1/2)$. Its normalized states and energies areSince $\hat{x}=\sqrt{\hslash /(2m\omega )}(a+a^{†})$, put $\gamma =\mu \sqrt{\hslash /(2m\omega )}$. Taking $A=\pm i\gamma a^{†}$, $B=\pm i\gamma a$ gives $[A,B]={\gamma }^2$. The formula implies $e^{\pm i\mu \hat{x}}=e^{-{\gamma }^2/2}{e}^{\pm i\gamma a^{†}}{e}^{\pm i\gamma a}$. Averaging the two signs proves the requested normal ordering of the cosine.

Because $a|0⟩=0$, expansion of the remaining creation exponential givesOdd matrix elements vanish, the ground expectation is $e^{-{\gamma }^2/2}$, and the magnitude squared for $n=2p$ is $e^{-{\gamma }^2}{\gamma }^{4p}/(2p)!$. The nondegenerate second-order nondegenerate perturbation theory formula is $\Delta E_0=ϵV_{00}+{ϵ}^2{\sum }_{n>0}|V_{n0}{|}^2/(E_0-E_n)+O({ϵ}^3)$. ThereforeThe negative second-order sign comes from the higher unperturbed energies in every denominator.

For a normalized eigenstate $|n⟩$ with a nondegenerate energy eigenvalue, first-order nondegenerate perturbation theory and second-order nondegenerate perturbation theory give

Choose the $z$-axis along $\mathbf{E}$, so the perturbation isup to an irrelevant sign convention for the electron charge. By second-order nondegenerate perturbation theory,Parts (a) and (b) restrict the contributing bound states toFor hydrogen,and henceThereforeEvery denominator is negative, so the discrete-state contribution lowers the ground-state energy.