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Linearized momentum balance is ${\rho }_0{\partial }_t\tilde{u}=-\nabla \tilde{p}$. For a progressive acoustic plane wave $\tilde{p}=P(\hat{k}\cdot x-c_{0}t)$, integration givesThe integration constant is zero for the pure wave with no added uniform flow. The instantaneous acoustic intensity is $I=\tilde{p}\tilde{u}$, the pressure-work energy flux; time-averaged intensity is $⟨I⟩$. For complex harmonic pressure amplitude $P$, $⟨I⟩=|P{|}^2\hat{k}/(2{\rho }_0{c}_0)$.

Use phase convention $e^{i(k_{y}y-\omega t)}$, with $\omega =k{c}_0$, $k_y=k\sin \theta$, $k_x=k\cos \theta >0$. Let incident, reflected and transmitted pressure amplitudes be $1,R,T_p$. The two normal velocities at the membrane are $(1-R)\cos \theta /({\rho }_0{c}_0)$ and $T_p\cos \theta /({\rho }_0{c}_0)$, and both equal $-i\omega X_0$. Hence $T_p=1-R$ and $X_0=i{T}_p\cos \theta /({\rho }_0{c}_0\omega )$. Only normal velocity is matched for an inviscid fluid; tangential velocities need not agree.

Put $D_m=k^2(m{c}_0^2-T{\sin }^2\theta )$. The dynamic condition is $-D_m{X}_0+T_p-(1+R)=0$, or $-D_m{X}_0-2R=0$. Since $D_m\cos \theta /({\rho }_0{c}_0\omega )=1/\alpha$, elimination givesComplex conjugation gives the amplitudes for the opposite time-phase convention. At zero $D_m$, the formulas have the limiting values $R=0,T_p=1$ even though $\alpha$ itself is then infinite. A strongly inertial membrane has $\alpha \to 0$, giving almost total reflection.

For real $\alpha$, $|R{|}^2+|T_p{|}^2=1$. Thus the incoming net normal energy flux is $\cos \theta (1-|R{|}^2)/(2{\rho }_0{c}_0)$, equal to the transmitted flux $\cos \theta |T_p{|}^2/(2{\rho }_0{c}_0)$. Time-averaged energy flux is conserved: the lossless membrane stores and returns energy but has no mean dissipation.

The Rayleigh quotient is $R[\psi ]=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$, interpreted through the energy quadratic form for trial states in its form domain. Expanding in normalized energy eigenstates gives $R[\psi ]=\sum E_n|c_n{|}^2/\sum |c_n{|}^2\ge E_0$. If the ground state is unique, equality requires all coefficients outside its one-dimensional eigenspace to vanish. Thus equality means the same ground-state ray, allowing normalization and an overall phase.

Use the normalized trial function $\sqrt{2/a}\cos (\pi x/a)$ on $|x|<a/2$, zero outside. Its kinetic energy is ${\pi }^2{\hslash }^2/(2m{a}^2)$ and its mean square position is $c_0{a}^2$, where $c_0=1/12-1/(2{\pi }^2)$. The energy isMinimization gives $a^4={\pi }^2{\hslash }^2/(m^2{\omega }^2{c}_0)$ andFor the first excited well state use $\sqrt{2/a}\sin (2\pi x/a)$, zero outside. Now the kinetic energy is $2{\pi }^2{\hslash }^2/(m{a}^2)$ and the position factor is $c_1=1/12-1/(8{\pi }^2)$. Minimizing gives $a^4=4{\pi }^2{\hslash }^2/(m^2{\omega }^2{c}_1)$ andThis particular excited-state estimate is an upper bound. The trial function is odd and the oscillator ground state is even, so it is exactly orthogonal to the ground state. Its spectral expansion therefore contains only energies at least $E_1$, giving the same variational bound on that orthogonal subspace. An arbitrary trial state without this orthogonality would bound only $E_0$, not $E_1$. The compactly supported trial functions belong to the quadratic-form domain despite derivative jumps at the well edges; their kinetic energies are correctly computed as ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$.

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.