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Use Planck units and restrict to the physical Kähler metric domainPut $A=a{e}^{-\alpha S}$ and $w=A+b$, so that $W=C^3+w$. There are no gauged interactions specified, hence no additional D-term scalar potential. The supergravity F-term potential is $V=e^K(K^{i\bar{j}}{D}_{i}W\overline{D_{j}W}-3|W{|}^2)$, with $e^K=1/(x{y}^3)$. Its Kähler covariant derivatives of a superpotential areThe nonzero metric entries areThe determinant of this $T,C$ block is $9/y^3>0$, so it is positive definite. With the inverse-index convention $K_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, the raised components areKeeping these complex-conjugate off-diagonal entries in their correct index positions is essential.

To exhibit the no-scale supergravity cancellation, write the $T,C$ norm asHere $D_{C}W+\bar{C}{D}_{T}W=3C^2$. The $3|W{|}^2$ term cancels the negative term in the supergravity F-term potential. The result is the nonnegative scalar potentialThis is the matter-logarithm no-scale potential, specialized to the exponential superpotential.

Choose the supergravity auxiliary field convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. Substitution into the full inverse Kähler metric givesAn overall auxiliary-field phase is conventional. The $C$ dependence of $F^T$ cancels between the two metric terms. If all three auxiliary fields vanished at a finite physical point, $F^C=0$ would force $C=0$, $F^T=0$ would force $A+b=0$, and $F^S=0$ would then force $\alpha xA=0$. Thus $a=b=0$ is necessary. For any nontrivial choice of $a,b$, supersymmetry is broken at every finite physical point. The allowed exception $a=b=0$, $C=0$ is supersymmetric; the statement cannot hold for all arbitrary parameters without this qualification.

The no-scale volume runaway criterion also determines whether the minimum is attained. For fixed $S,C$, differentiation with respect to $y$ givesIt is negative unless both numerators vanish. Therefore a finite stationary vacuum must satisfyEvery such point is a global minimum with $V_{\min }=0$. For $a,b\ne 0$, setting $r=|b/a|$ and $z=\alpha x>0$ reduces existence toThe derivative of the magnitude function is $(1-z)e^{-z/2}/2$. It rises from $1$ at the excluded boundary $z=0$ to $2e^{-1/2}$ at $z=1$, then falls to zero. Consequently a finite zero-energy minimum exists exactly whenor in the separate trivial family $a=b=0$. There is one positive root for $0<r\le 1$, two for $1<r<2e^{-1/2}$, and one double root at equality. If only one of $a,b$ vanishes, or $r$ exceeds this maximum, there is no finite minimum: the infimum is zero as $\mathrm{Re}T\to \infty$. Calling this runaway a finite zero-energy vacuum would be incorrect.

At a nontrivial zero-energy minimum, $W=A+b=-\alpha xA\ne 0$, and the auxiliary expectation values reduce toThis explicitly verifies supersymmetry breaking with zero vacuum energy in no-scale supergravity. Both real components of $T$ are exact flat directions of a scalar potential on the minimum locus. Away from that locus, $\mathrm{Im}T$ is still absent from the potential, but $\mathrm{Re}T$ is not flat. At a minimum, $C$ is lifted at quartic order: at fixed $S,T$, $V=3|C{|}^4/[x(T+\bar{T}{)}^2]+O(|C{|}^6)$. A zero quadratic mass does not make $C$ an exact flat direction. The quartic stabilization at a double no-scale root likewise gives a vanishing quadratic mass, but no exact flat direction, for $\mathrm{Re}S$ at $z=1$. For $a=b=0$, the whole $C=0$ family has both $S$ and $T$ flat and supersymmetry unbroken.

Finally, the gravitino mass from a superpotential isAt a nontrivial zero-energy minimum this becomesIt depends on the unfixed volume modulus, so the model does not predict one numerical mass. For $a=b=0$, $C=0$, the gravitino is massless.

Set $x=T+\bar{T}>0$, the domain in which the Kähler potential is defined, and use Planck units. The nonzero entries of the Kähler metric and its inverse areThe Kähler covariant derivatives of a superpotential areDenote $W=C^3+B$ and $S=3C^2+\bar{C}W$. The supergravity F-term potential isHere the $T$ contribution is exactly $3|W{|}^2$, cancelling the negative term. This is the no-scale supergravity identity, givingThe cancellation is important: replacing the covariant derivatives by ordinary derivatives would miss both the matter dependence and the possibility of zero vacuum energy with supersymmetry breaking.

In the convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$, the chiral supergravity auxiliary fields areAn overall auxiliary phase or sign convention does not change their vanishing conditions. These are upper-index fields, including the inverse Kähler metric, rather than just the covariant derivatives.

Since $V$ is nonnegative and vanishes at $C=0$, its minimum energy isThere are additional minima that should not be discarded by examining only real $C$. For $C=\rho e^{i\alpha }\ne 0$, the condition $S=0$ becomesIf $B\ne 0$, the magnitude equationhas exactly one positive root: its left side is strictly increasing from zero to infinity. There are three phases, differing by $2\pi /3$, determined by $e^{3i\alpha }=-B/|B|$. Thus for nonzero $B$ the finite zero-energy matter vacua consist of $C=0$ and these three nonzero phase-related values, each with arbitrary $T$ in the allowed half-plane. At a nonzero matter vacuum,At $C=0$, $W=B$.

For $B\ne 0$, every such minimum has $F^C=0$ but $F^T\ne 0$, so supersymmetry is spontaneously broken despite zero vacuum energy. Indeed no finite point anywhere is fully supersymmetric: $D_{T}W=0$ would force $W=0$, after which $D_{C}W=3C^2=0$ forces $C=0$ and then $B=0$. This proves breaking for the specified nonzero-constant case without relying just on the positivity of the potential.

For the allowed exceptional value (B=0), the requested breaking statement is false. Then $S=C^2(3+|C{|}^2)$ vanishes only at $C=0$, where $W=F^T=F^C=0$. Every finite allowed $T$ gives an unbroken supersymmetric vacuum with zero energy. This is an explicit counterexample to unqualified breaking for an arbitrary complex $B$.

Both real components of $T$ are flat directions of a scalar potential along every minimum: the imaginary component never appears in $V$, and when $S=0$ the entire $x$ dependence is multiplied by zero. The matter vacua are isolated in the $C$ plane; there is no continuous matter vacuum direction. At $B=0$ the potential begins quartically in $C$, so zero quadratic mass is not a flat vacuum manifold. Away from $S=0$, the real $T$ direction instead has $V\propto x^{-3}$ and a decompactification-type runaway towards zero as $x\to \infty$, not a further finite positive-energy minimum.

The gravitino mass from a superpotential isAt the $C=0$ branch it is $|B|/x^{3/2}$; at each nonzero branch it is $3\rho e^{{\rho }^2/2}/x^{3/2}$. Its value is not fixed because the $T$ modulus is flat. For $B=0$ it vanishes at the supersymmetric minimum. For nonzero $B$, the super-Higgs mechanism absorbs the Goldstino associated with $F^T$ into the massive gravitino. The zero-energy relation is explicitly satisfied:which balances the auxiliary contribution against the gravitational negative term.

The numerical fractions describe physically very different components. Use $c=1$ so that pressure and energy density have the same units. The stress-energy conservation equation for a homogeneous component, together with its equation of state $p_i=w_i{\rho }_i$, givesNonrelativistic baryons and cold dark matter have $w\simeq 0$ and dilute as $a^{-3}$; radiation has $w=1/3$ and dilutes as $a^{-4}$; vacuum energy has $w=-1$ and remains constant. The curvature term scales as $a^{-2}$ in the Friedmann equation but is a geometric contribution, not an additional material fluid.

There is consequently a coincidence in the present mixture. Normalize $a_0=1$. ThenAt early times vacuum is negligible, while at late times it dominates overwhelmingly. Observing order-unity matter and vacuum fractions during their comparatively recent crossover motivates the cosmic coincidence problem: why should this crossover occur near our observing epoch?

The absolute scale gives a second puzzle. Restoring SI units, $H_0\simeq 2.27\times 10^{-18}{\mathrm{s}}^{-1}$ yieldsIts energy density is ${\rho }_{\Lambda ,0}{c}^2\simeq 5.8\times 10^{-10}\mathrm{J}{\mathrm{m}}^{-3}$. Quantum vacuum contributions associated with ordinary particle-physics scales can be far larger than this observed value. Explaining the tiny net value, including cancellations among contributions, is the cosmological constant problem. The required negative pressure also differs qualitatively from ordinary dilute matter.

The matter budget supplies another explanatory gap: only $0.045$ is baryonic, while $0.255$ must be nonbaryonic matter in this benchmark. Together with $0.7$ in vacuum energy, most of the cosmic density is in components whose microscopic explanation is not supplied by fitting their fractions. Finally, near-zero curvature raises the flatness problem. For nonzero $k$,During radiation domination $H\propto a^{-2}$, so $|{\Omega }_k|\propto a^2$; during matter domination $H\propto a^{-3/2}$, so $|{\Omega }_k|\propto a$. Near-flatness today therefore requires extremely small early curvature in an unmodified decelerating history. Cosmic inflation offers a mechanism that reduces this curvature fraction through accelerated expansion.

The apparent contrivances are the recent matter-vacuum coincidence, the tiny vacuum-energy scale, the near-flat initial geometry and the unexplained microscopic dark components. Their observationally inferred values impose concrete requirements on a deeper physical account.

The hierarchy problem concerns the enormous separation between the electroweak scale and a possible fundamental ultraviolet scale such as the reduced Planck mass: $v\simeq 246\mathrm{GeV}$ whereas $M_{\mathrm{Pl}}\simeq 2.4\times 10^{18}\mathrm{GeV}$. For an elementary Higgs field, heavy states or a physical high-energy ultraviolet cutoff generically generateHere $\kappa$ represents the appropriate dimensionless couplings. Maintaining a small physical Higgs mass then requires a very accurate cancellation against its bare mass parameter. The problem is sensitivity to actual high-energy physics, rather than merely the presence of a quadratic divergence in one choice of regulator.

The cosmological constant problem concerns the much smaller gravitationally observable vacuum energy. In the Einstein equation, a constant term in the effective action contributes to the cosmological constant, with ${\Lambda }_{\mathrm{eff}}={\Lambda }_{\mathrm{bare}}+{\rho }_{\mathrm{vac}}/M_{\mathrm{Pl}}^2$ in the usual normalization. Generic zero-point contributions scale as ${\Lambda }_{\mathrm{UV}}^4$, and vacuum transitions also change ${\rho }_{\mathrm{vac}}$. The dark-energy density is only of order $10^{-47}{\mathrm{GeV}}^4\sim (\text{a few meV}{)}^4$. The puzzle is why the total remains so small despite much larger individual contributions. A constant shift of the nongravitational Hamiltonian cannot remove its gravitational effect.

Supersymmetry relates bosonic and fermionic degrees of freedom, including their interactions. Their opposite loop signs cancel the quadratic contributions to a scalar field mass in the unbroken theory. With soft supersymmetry breaking, the surviving correction is controlled by the mass splitting, schematicallyThus a sufficiently low soft-breaking scale protects a small electroweak scale against a much higher ultraviolet scale. This is the useful supersymmetric response to the hierarchy problem.

Exact global supersymmetry also cancels vacuum energy: the supersymmetry algebra expresses the Hamiltonian as a positive sum of anticommutators, so a vacuum annihilated by every supercharge has zero energy. Equivalently, with canonical kinetic terms,vanishes in a supersymmetric vacuum. However, ordinary particles and their superpartners cannot have the identical mass spectrum of unbroken supersymmetry, so a useful phenomenological theory needs supersymmetry breaking. Its vacuum energy is generically of order the fourth power of the breaking scale; even a scale of order $10^2\mathrm{GeV}$ gives about $10^8{\mathrm{GeV}}^4$, enormously above the required value. In supergravity, a negative term can cancel positive breaking contributions:but arranging a tiny residual is a new tuning. Consequently supersymmetry can protect the hierarchy after soft breaking, while its exact vacuum-energy cancellation does not by itself survive realistic breaking.

One higher-dimensional proposal uses large extra dimensions. Take $n$ flat compact dimensions of common radius $R$, with gravity propagating in the bulk and Standard Model fields confined to a four-dimensional brane. Reducing the gravitational Einstein-Hilbert action givessoA fundamental gravitational scale $M_{\ast }$ near a TeV then produces apparently weak four-dimensional gravity through dilution into the large compact volume. For $n=2$ and $M_{\ast }=1\mathrm{TeV}$, $R=M_{\mathrm{Pl}}/(2\pi M_{\ast }^2)$ is about $0.075\mathrm{mm}$, using $1{\mathrm{GeV}}^{-1}\simeq 1.97\times 10^{-16}\mathrm{m}$. At distances much smaller than $R$, the gravitational potential scales as $r^{-(n+1)}$; at distances much larger than $R$ it scales as $r^{-1}$. Confining charged Standard Model fields to the brane avoids a light tower of charged Kaluza-Klein modes associated with the large radius. The primary explanatory challenge is radius stabilization: one must generate and stabilize $R{M}_{\ast }\gg 1$, rather than simply postulate a huge volume and transfer the original hierarchy into a modulus.

A second proposal is the Randall–Sundrum model. Compactify on an interval, equivalently an orbifold with branes at $y=0$ and $y=L$, and use the warped metricA negative bulk cosmological constant and suitably related brane tensions support this geometry. Integrating over the full orbifold interval $[-L,L]$ givesFor a scalar field $h$ on the brane at $L$, its induced metric gives the kinetic coefficient $e^{-2kL}$ and mass coefficient $e^{-4kL}{m}_0^2$. The canonical field is $h_c=e^{-kL}h$, henceWith $k$ and $M_5$ of the same fundamental order, an exponent $kL\simeq 35$--$37$ converts a fundamental mass into a weak-scale mass, without requiring a large proper compactification volume.

Again the separation must be selected dynamically. The original warped solution leaves a massless radion, so merely choosing $L$ does not complete the explanation. For example, the Goldberger–Wise stabilization mechanism introduces a bulk scalar field with boundary values $v_0,v_L$ and a small mass $m_{\chi }$. Its static equation and solution areFor small backreaction and large $kL$, the leading radius-dependent potential is proportional to $e^{-4kL}(v_L-v_0{e}^{-ϵkL}{)}^2$. Its leading minimum obeys $kL\simeq {ϵ}^{-1}\log (v_0/v_L)$, so a modest boundary-value ratio and a small $ϵ$ can give the required exponent. Corrections and the residual four-dimensional cosmological constant must still be controlled. Both proposals need a stable dynamical origin for the compactification geometry; their volume or warp relations alone do not supply it.

Use natural units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Legendre transform in mechanics givesFor canonical quantization of a real scalar field, impose the equal-time canonical commutation relationsThe Heisenberg picture equation $\dot{O}=i[H,O]$ then gives $\dot{\varphi }=\pi$. For the other equation, the momentum term commutes with $\pi$; commuting the gradient term past $\pi$ and integrating the derivative of the Dirac delta function by parts gives $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$. ConsequentlyThe Fourier transform reduces this Klein-Gordon equation to oscillators of frequency $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$. Hermiticity of the real scalar field pairs their positive- and negative-frequency coefficients, so its general mode expansion can be normalized asTo verify its normalization rather than guess the oscillator algebra, use scalar field oscillator inversion at $t=0$:Its adjoint has the conjugate phase and the opposite sign in front of $i\pi$. The two mixed field-momentum commutators giveThe Fourier representation of the Dirac delta function sets the energies equal in the last step. Similarly, $[a(\mathbf{p}),a(\mathbf{q})]$ has a factor $E_{\mathbf{q}}-E_{\mathbf{p}}$ multiplying ${\delta }^3(\mathbf{p}+\mathbf{q})$ and vanishes; the two-creator commutator vanishes as well.

Substituting the expansion in $H$ and integrating the spatial phases, the two-annihilator and two-creator coefficients cancel by $E_{\mathbf{p}}^2={\mathbf{p}}^2+m^2$. The free real scalar Hamiltonian in oscillator variables isIt differs from $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}a/(2\pi {)}^3$ by its constant vacuum energy; normal ordering removes that constant. Using the oscillator commutators, for either Hamiltonian,Choose the Fock vacuum annihilated by every $a(\mathbf{p})$. Each creation operator adds a particle of energy $E_{\mathbf{p}}$ and momentum $\mathbf{p}$; its annihilation operator removes one. Their mutual commutativity gives bosonic Fock space and symmetric many-particle states. Thus the scalar mass-shell frequency has a particle interpretation, rather than merely describing a classical oscillation.

For $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$, vacuum contraction and time ordering give the Feynman propagatorIn the scalar Feynman propagator pole prescription, the integral of $i{e}^{-i{p}^{0}t}/((p^0{)}^2-E_{\mathbf{p}}^2+i0)$ has poles at $E_{\mathbf{p}}-i0$ and $-E_{\mathbf{p}}+i0$. Close clockwise below for $t>0$ and counterclockwise above for $t<0$. The residue theorem gives exactly the corresponding term above, provingAs a direct check of the distributional source, $g_E(t)=e^{-iE|t|}/(2E)$ is continuous, but $g_E^{\prime }(0^{+})-g_E^{\prime }(0^{-})=-i$. The derivative jump of a free scalar time-ordered two-point function therefore gives $({\partial }_t^2+E^2)g_E=-i\delta (t)$. Spatial Fourier inversion yields

Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.

Sum the diagonal spinor entries of the supercharge anticommutator. Since $\mathrm{tr}{\sigma }^i=0$ and $\mathrm{tr}{\sigma }^0=2$, the Hamiltonian isFor a normalized vacuum, energy positivity in global supersymmetry follows fromIn the unbroken case, all supercharges annihilate the vacuum, giving $E_{\mathrm{vac}}=0$. Conversely, zero energy forces each nonnegative norm to vanish, so the vacuum is supersymmetric. The algebra fixes the additive zero of energy here. For an infinite homogeneous vacuum, use a finite-volume regulator and interpret the result as its vacuum energy density.

Apply the Grassmann Gaussian integral to a regulated finite collection of field components. Up to a field-independent measure normalization and phase,In the continuum this is a formal functional determinant, including spinor and spacetime indices. Its meaningful definition requires a regulator and boundary conditions. A complex Dirac field supplies a determinant, not the inverse square root obtained for a real commuting field.

The comparison is clean after Wick rotation. A free real scalar field with positive Euclidean operator $P_E=-{\partial }_E^2+m^2$ hasTaking a logarithm gives $-\frac{1}{2}\mathrm{Tr}\log P_E$ for the scalar and $+\mathrm{Tr}\log K_E$ for the Dirac field. The opposite statistics sign is the determinant counterpart of the minus sign for a closed fermion loop. The magnitude also differs because a Dirac field has several independent spin and antiparticle degrees of freedom.

The vacuum energy sign of a fermionic oscillator makes the comparison explicit. A real scalar mode contributes $+E/2$; each independent fermionic oscillator contributes $-E/2$. There is one oscillator per scalar momentum, but a massive Dirac field has two particle and two antiparticle oscillators. Thus the vacuum energy densities are formallyEquivalently, the large Euclidean-time vacuum functional behaves as $\log Z_E\sim -\beta V{\rho }_{\mathrm{vac}}$. Both displayed vacuum energies are ultraviolet divergent; a common regularization in quantum field theory and the appropriate renormalization are needed before comparing them. The zero-point vacuum energies of bosons and fermions have opposite signs, but do not cancel without matching masses and degrees of freedom. Normal ordering removes an additive vacuum constant in nongravitating flat-space theory. Coupling to the metric in general relativity makes that constant contribute to the cosmological constant, so it cannot simply be discarded without a renormalization condition.

The quantum harmonic oscillator has $\hat{H}=({\dot{\hat{q}}}^2+{\omega }^2{\hat{q}}^2)/2$, with $\omega >0$. Using $\hat{a}|0⟩=0$ and the canonical commutation relation givesFor $q=r{e}^{is}$, the Wronskian normalization isConsequentlyEquality requires $\dot{r}=0$ and $r^2=1/(2\omega )$. The minimum-energy normalized oscillator mode is thereforeThe constant phase $\alpha$ is irrelevant. This mode also solves the oscillator equation of motion; an arbitrary squeezed mode would have larger vacuum energy.

For inflation, introduce the Mukhanov-Sasaki variable $v=z\mathcal{R}$. Since $z$ depends only on conformal time,Integrating the cross term by parts gives the canonical bulk actionup to the boundary term $-\frac{1}{2}\int d^{3}x[(z^{\prime }/z)v^2{]}_{\mathrm{boundary}}$. The resulting Euler-Lagrange field equation and its Fourier transform areAt leading order in the slow-roll approximation, retain a small positive, nearly constant $ϵ$ in $z=a\sqrt{2ϵ}$, while approximating $a=-1/(H\tau )$ and $H$ as constant. Then $z^{\prime \prime }/z\simeq 2/{\tau }^2$, soStrict exact de Sitter spacetime would have $ϵ=0$ and would not supply the nonzero curvature kinetic coefficient assumed here; the calculation is the leading quasi-de Sitter limit, not a substitution of zero into $z$.

For $-k\tau \gg 1$, the expansion correction is negligible and each canonical mode is a quantum harmonic oscillator of conformal frequency $k$. The Bunch-Davies vacuum selects its positive-frequency, minimum-energy mode $e^{-ik\tau }/\sqrt{2k}$ in that early subhorizon regime. It does not minimize an instantaneous Hamiltonian after the effective squared frequency has become negative outside the horizon.

A basis of exact solutions to the leading mode equation is $e^{-ik\tau }(1-i/(k\tau ))$ and its complex conjugate. Write the normalized Bogoliubov transformation combination asThe Bunch-Davies vacuum boundary condition sets ${\beta }_k=0$ and ${\alpha }_k=1$ up to phase. HenceSubstitution verifies the equation, and $v_k{v}_k^{\ast \prime }-v_k^{\prime }{v}_k^{\ast }=i$ verifies the Wronskian normalization.