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Use the condensate wavefunction convention $\int |\psi {|}^2{d}^{3}r=N$, so $n(\mathbf{r})=|\psi (\mathbf{r}){|}^2$ is the number density and $\psi /\sqrt{N}$ is a normalized one-particle orbital. The contact interaction strength is $U_0=4\pi {\hslash }^{2}a/m$ for dilute atoms with s-wave scattering length $a$. The chemical potential $\mu =\partial E/\partial N$ fixes the stationary time dependence $e^{-i\mu t/\hslash }$ and enforces normalization when minimizing the energy. The Gross–Pitaevskii equation follows from an energy functional that contains the pair-counting factor $1/2$:Varying $E-\mu \int |\psi {|}^2{d}^{3}r$ gives the stationary Gross–Pitaevskii equation; the interaction contribution to that equation is $U_{0}n\psi$, without the factor $1/2$.

For the broadening discussed here assume repulsive interactions, $a>0$. The Thomas–Fermi approximation for a condensate neglects the gradient kinetic energy compared with the trapping and interaction energies. At fixed spatial scale, increasing $N{U}_0$ increases the mean-field potential. Expansion then reduces the kinetic energy further: for a cloud radius $R$, its scale per atom is ${\hslash }^2/(m{R}^2)$, whereas the trap energy scale is $m{\omega }^2{R}^2$. Their ratio decreases as $(a_0/R{)}^4$. In the weak-interaction limit the kinetic energy instead balances confinement, giving the Gaussian wave packet ground state of the quantum harmonic oscillator.

The spherical Thomas–Fermi condensate hasIntegrating this density gives $N=8\pi \mu R^3/(15U_0)$. ThusAt fixed $N$, stronger repulsion produces a wider, lower wavefunction: the Gaussian wave packet crosses over towards the square root of an inverted parabola, with an almost flat central region and a steeper outer drop. The density is the inverted parabola; the wavefunction is its square root. The exact wavefunction remains smooth and has a tail, since kinetic energy cannot be neglected near the Thomas–Fermi condensate edge rounding region. The healing length must be small compared with the cloud size in the bulk. Attractive interactions would not follow this large-$N$ expansion argument.

The original plot compares the noninteracting endpoint with two normalized repulsive spherical Thomas–Fermi condensate profiles. The latter are asymptotic shapes, rather than exact intermediate solutions of the Gross–Pitaevskii equation. Each has the same integrated atom number; $\chi =Na/a_0$ labels the illustrative interaction strengths.

Choose $A$ real and positive. The Gaussian integral givesThe three zero-point energies add for each atom. The ideal-gas chemical potential is a one-atom energy; the total energy includes all $N$ atoms.

For weak interactions, the first-order correction in perturbation theory is the energy functional of the contact interaction evaluated on the unperturbed Gaussian wave packet. The integral needed isConsequently the Gaussian interaction energy of a trapped condensate isHere $\hslash \omega ={\hslash }^2/(m{a}_0^2)$. These are the mean-field expressions consistent with the supplied equation and $\int |\psi {|}^2=N$. Exact first-order perturbation theory for precisely $N$ atoms in one orbital replaces $N^2$ by $N(N-1)$ through the number of distinct pairs; this distinction is negligible in the large-$N$ regime of the crossover estimate.

To judge the ideal-gas approximation, compare the interaction energy with the total ideal energy, or compare the interaction energy per atom with $\hslash \omega$:Thus the crossover from Gaussian to Thomas–Fermi condensate is governed by $\chi =Na/a_0$: perturbation theory is reliable for $\chi \ll 1$, while the regime of the Thomas–Fermi approximation for a condensate requires $\chi \gg 1$. The characteristic crossover number is $N_c\sim a_0/a$; there is no sharp critical atom number here. If one specifically defines crossover by $E_I=E_0$ using the weak-interaction expression, it gives $N_c=(3\sqrt{2\pi }/2)a_0/a$. That prefactor is a convention-dependent estimate extrapolated to the edge of perturbation theory, not a phase transition. Comparing $E_I$ itself to a single $\hslash \omega$ would incorrectly omit a factor of $N$.

For the numerical estimate use $m\simeq 87\times 1.66054\times 10^{-27}\mathrm{kg}$ and $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe particular equality $E_I=E_0$ would instead give about $1.15\times 10^3$ atoms. These quantify different choices within a smooth crossover; a well-developed spherical Thomas–Fermi condensate profile requires going substantially beyond the characteristic scale.

For this quantum harmonic oscillator, ${\hat{H}}^{(1)}=x^4$. The Gaussian integral givesThe distinction between the coefficient and the actual first-order shift matters: the energy shift is $3\lambda /4$, so $E_0=1/2+3\lambda /4+O({\lambda }^2)$ in the stated atomic units.

For any $\alpha >0$, the Gaussian function is the exact nodeless ground wavefunction of a quantum harmonic oscillator withDirect substitution cancels the $x^2$ terms. An additive constant in the potential would simply shift the eigenvalue.

A natural comparison uses the same optimized width as part (i), $\alpha =3^{1/3}$. The harmonic ground energy is then $0.72112479$, compared with the quartic Gaussian function expectation $0.54084359$. For this common normalized trial,while the kinetic energy expectation value is $\alpha /4$ in both cases. The quartic potential lies below this harmonic potential for $0<|x|<\alpha$ and above it for $|x|>\alpha$; the difference weighted by the Gaussian function is negative, giving the lower expectation.

If the standard unit-frequency oscillator is chosen instead, $\alpha =1$ gives $E_{0,\mathrm{ho}}=1/2$, below the optimized quartic upper bound. These are different harmonic comparison potentials, not contradictory answers. Crossing potential curves cannot be ordered globally; the comparison must specify the frequency or Gaussian function width used.

Differentiate the Gaussian before substituting into the Hamiltonian:ThusAn energy eigenvalue must be independent of $x$, so the coefficient of $x^2$ vanishes. Normalizability requires $\alpha >0$, givingThe constant term is the quantum harmonic oscillator's zero-point energy; a Gaussian width ${\alpha }^{-1/2}$ balances its kinetic and potential contributions.

Relative to the ground state, the quantum harmonic oscillator levels are $vhc\tilde{\omega }$. Put ${\theta }_v=hc\tilde{\omega }/k$. Summing the geometric series gives the vibrational partition function $q=(1-e^{-{\theta }_v/T}{)}^{-1}$. Therefore the vibrational internal energy isThe omitted zero-point energy would add $Nhc\tilde{\omega }/2$ if absolute oscillator energies were used.

For one mole, $Nk=R$, so the three-level Morse oscillator result isFor the full quantum harmonic oscillator with the given wavenumber,The specified anharmonic calculation is about $1.7\%$ higher. Anharmonicity lowers the first excitation spacings and increases excited-state occupation. Both results remain small because $100\mathrm{K}$ is well below the vibrational temperature. The comparison also mixes a three-level anharmonic truncation with an infinite harmonic sum; omitted higher Morse levels would make a further small positive correction at this temperature.

Quantum tunnelling is penetration of, or transmission through, a finite potential barrier into a classically forbidden region, where the total energy is less than the potential energy. The wavefunction generally remains nonzero there, so the Born rule assigns a nonzero probability of detection. This does not require an increase in total energy; it is the classical relation between a definite position and a definite kinetic energy that cannot be imposed on the quantum state. For a bound quantum harmonic oscillator, the relevant quantity is the probability of finding the particle beyond its classical turning points; it is not an escaping transmitted beam.

For the reduced quantum harmonic oscillator, $V(q)=q^2/2$ and the ground-state energy is $E_0=1/2$. Thus the classical turning points satisfy $q^2/2=1/2$, giving $q=\pm 1$. Normalizing the wavefunction gives ${\varphi }_0(q)={\pi }^{-1/4}{e}^{-q^2/2}$, because ${\int }_{-\infty }^{\infty }{e}^{-q^2}dq=\sqrt{\pi }$. The Born rule therefore gives the harmonic-oscillator forbidden-region probabilityThe factor two includes both tails. Using the exact complementary error function gives $\mathrm{erfc}(1)=0.1572992$. A stationary real wavefunction has zero probability current, so this result describes forbidden-region occupancy, not continuous loss of the bound particle.

Write $g=e^{-q^2/2}$ and $f=2q^2-a$. Product differentiation for the reduced quantum harmonic oscillator givesFor an eigenfunction, this must equal $E(2q^2-a)g$ at every $q$. Matching the $q^2$ coefficients gives $E=5/2$, and matching constants then gives $5a/2=2+a/2$. Hence $a=1$ and $E=5/2$. This is the even $n=2$ oscillator state. Its normalized wavefunction is ${\varphi }_2=(2q^2-1)e^{-q^2/2}/(\sqrt{2}{\pi }^{1/4})$.

The Einstein model of a solid replaces the vibrations of $N$ identical atoms by $3N$ independent quantum harmonic oscillators with a common frequency ${\omega }_E$. Each atom has three displacement directions; the restoring force is harmonic, coupling between different atoms is replaced by a fixed local restoring potential, and anharmonicity and frequency dispersion are neglected. The crystal volume and ${\omega }_E$ are held fixed when calculating the vibrational heat capacity at constant volume.

The one-oscillator partition function isConsequently, with $T_E=\hslash {\omega }_E/k_B$,Differentiating at fixed volume givesThe zero-point energy does not contribute to this derivative. The result tends to the Dulong-Petit law $3N{k}_B$ at high temperature and vanishes exponentially at low temperature; the latter is a limitation of the single-frequency model for a real acoustic spectrum.

In the harmonic approximation, each in-plane displacement coordinate contributes one quadratic kinetic energy term and one quadratic restoring potential energy term. The equipartition theorem assigns $k_{B}T/2$ to each. There are two such coordinates per chemical atom, henceHere the fixed-size heat capacity is at constant area in this two-dimensional model. On cooling, phonon excitation freezes out and the heat capacity decreases. For each nonzero angular frequency, the quantum harmonic oscillator contributeswhich is below $k_B$ at finite $x$ and approaches zero as $T\to 0$. Two linearly dispersing in-plane acoustic phonon branches give the familiar two-dimensional Debye model scaling $C_v\propto T^2$ at sufficiently low temperature. Out-of-plane modes are excluded by the specified model.

For one electromagnetic normal mode, $a$ and $a^{†}$ are the annihilation operator and creation operator, with commutator $[a,a^{†}]=I$. Their action on normalized number states isThey remove or add one photon, whose energy is $\hslash \omega$. The vacuum state $|0⟩$ is normalized and obeys $a|0⟩=0$. It has no photons, but retains zero-point energy $\hslash \omega /2$ for the quantum harmonic oscillator Hamiltonian $H=\hslash \omega (a^{†}a+1/2)$. The coherent state $|\alpha ⟩=D(\alpha )|0⟩$ is a displaced vacuum state.

The expectation value of an observable is the mean outcome of repeated measurements on identically prepared systems in the normalized state. For a self-adjoint $Q$ it is real and equals $⟨\psi ,Q\psi ⟩$. Assume the state belongs to the domains needed for the second moments and the commutator below, so all these quantities exist.

For real $\lambda$, positivity of the squared norm givesThe commutator expectation is purely imaginary, so this is a real quadratic in $\lambda$. If $⟨P^2⟩>0$, its minimum is nonnegative exactly when its discriminant is nonpositive. If $⟨P^2⟩=0$, nonnegativity for all real $\lambda$ forces the linear coefficient to vanish, yielding the same conclusion. ThusReplace the observables by their centred versions $Q-⟨Q⟩I$, $P-⟨P⟩I$. Their commutator is unchanged, and their second moments are the variances $(\Delta Q{)}^2,(\Delta P{)}^2$. Taking square roots gives the Robertson uncertainty principleFor the quantum harmonic oscillator, $p=-i\hslash {\partial }_x$ gives $[x,p]\psi =i\hslash \psi$. The uncentred inequality above consequently gives $⟨x^2⟩⟨p^2⟩\ge {\hslash }^2/4$. The arithmetic-geometric mean inequality now yieldsThis is a bound for every normalized state of finite oscillator energy, not just an eigenstate. It is attained by the centred Gaussian ground state, for which $⟨p^2⟩=m^2{\omega }^2⟨x^2⟩$ and the uncertainty product equals $\hslash /2$.

Write the Hamiltonian operator as $\hat{H}={\hat{p}}^2/(2m)+V(\hat{q})$, with $[\hat{q},\hat{p}]=i$, and define the quantum-mechanical propagator by $U(\beta ,\alpha ;T)=⟨\beta |e^{-iT\hat{H}}|\alpha ⟩$. Use $N$ intervals of length $\epsilon =T/N$, endpoint coordinates $q_0=\alpha ,q_N=\beta$, and the Trotter product formula. Inserting the completeness relations of the position eigenstates between the short-time factors givesNormalize momentum eigenstates by $⟨q|p⟩=e^{ipq}/\sqrt{2\pi }$. The Gaussian integral over the intermediate momentum isThe square-root branch is fixed by continuation from $\mathrm{Im}\epsilon <0$. Consequently the precise time-sliced configuration-space path integral isThis formula defines the notation $\int \mathcal{D}q{e}^{iS[q]}$; it is not an unnormalized infinite product of ordinary integrals. For real time take the boundary value from $T-i\delta$, $\delta >0$, or the corresponding oscillatory-distribution limit. For a self-adjoint, bounded-below Hamiltonian operator under the usual product-formula hypotheses, the operator limit supplies the definition even where the coordinate kernel is a distribution. The discrete exponent approaches the action with kinetic term $m{\dot{q}}^2/2$, and the prefactor is part of the functional measure. A different consistent slicing gives the same kernel for this separated kinetic-plus-potential Hamiltonian.

For the quantum harmonic oscillator, expand the action about a classical path $x$, writing $q=x+\xi$ with homogeneous Dirichlet boundary conditions on $\xi$. The only mixed term isThe boundary term vanishes and the Euler-Lagrange equation sets the remaining integrand to zero. Thus $S[x+\xi ]=S[x]+S[\xi ]$ exactly, rather than only to quadratic approximation. Translation of each integrated coordinate has unit Jacobian determinant; in the continuum measure convention of the question this is $\mathcal{D}q=\mathcal{D}\xi$. The classical-path factorization of a quadratic path integral follows:The fluctuation action, measure and boundary values depend on $T,m,\omega$ but not on $\alpha ,\beta$. For real $T$ with $\sin \omega T\ne 0$ the boundary-value classical path exists uniquely. At the conjugate times $\omega T=k\pi$, generic endpoints admit no such path and the kernel is a delta distribution; the formula is interpreted by the same regulated boundary limit, not as an ordinary finite prefactor times a nonexistent classical solution.

Take the momentum operator $p=-i\hslash d/dx$ and sufficiently regular real $f$. Acting on a test wave function, the commutator is $[p,f]=-i\hslash f^{\prime }$. Expanding the factorized quantum Hamiltonian givesso $V(x)=(f(x{)}^2-\hslash f^{\prime }(x))/(2m)$, which is real.

Set $Q=p-if$. Its zero-mode equation iswith solutionIt satisfies $H{\psi }_0=Q^{†}Q{\psi }_0/(2m)=0$ as a differential equation. It is an actual quantum state only when it is square-integrable. If $f(x)=c{x}^n+\cdots$, $c\ne 0$, its antiderivative has leading term $c{x}^{n+1}/(n+1)$. For even $n$, this leading term has opposite signs at the two ends of the line, so the wave function grows at one end and cannot be normalized. For odd $n$, it has the same sign at both ends. Hence normalizability requires odd $n$, and in fact holds exactly when $n$ is odd and $c>0$. A negative leading coefficient makes it grow at both ends; $f=0$ gives a nonnormalizable constant. These are the normalizable zero mode of a polynomial factorized Hamiltonian conditions.

For a normalized state in the operator domain, integration by parts, with the appropriate vanishing boundary term, givesThus every energy eigenvalue is nonnegative. Equality forces $Q\psi =0$ almost everywhere, whose first-order equation has the one-dimensional solution space above. Whenever its zero mode is normalizable, every energy eigenstate not proportional to ${\psi }_0$ has strictly positive energy. If the formal zero mode is not normalizable, there is no zero-energy eigenstate; the positivity statement still applies to every normalizable eigenstate. This argument concerns eigenvalues and does not assert a strictly positive bound on an arbitrary continuous spectrum.

For the quantum harmonic oscillator, take $f(x)=m\omega x$ with $m,\omega >0$. ThenIts zero mode is normalizable and has no competitor of lower energy. Using the Gaussian integral for its normalization gives the concise answerup to an arbitrary constant phase.

For a separated weighted density, write $P_n=e^{\gamma t}p(\sigma )$. Its eigenvalue equation isSet $p=\psi \exp [-{\sigma }^2/(2\kappa \tau )]$. Differentiation eliminates the first derivative of $\psi$, giving the tilted Ornstein-Uhlenbeck oscillator transformationComplete the square and introduce the dimensionless variable $x=(\sigma -\kappa n{\tau }^2)/\sqrt{\kappa \tau }$. The equation becomesComparing with the quantum harmonic oscillator gives $2E=1+\kappa n^2{\tau }^3-2\gamma \tau$. Its energies $E_m=m+1/2$ therefore correspond to ${\gamma }_m=\kappa n^2{\tau }^2/2-m/\tau$. The dominant nonnegative separated mode is the ground state, which is nodeless. ConsequentlyFor an independent check, $\log [\tilde{B}(t)/B_0]={\int }_0^t\tilde{\sigma }(s)ds$ is a centered Gaussian random variable with varianceThe exponential moment of a Gaussian linear functional gives $\mathbb{E}[B^n]=B_0^n\exp [n^{2}V(t)/2]$, whose long-time growth rate is the same ${\gamma }_n$. This exact Ornstein-Uhlenbeck multiplicative amplification formula also distinguishes the finite-time transient from the asymptotic exponential law. A general transient can contain sign-changing excited eigenfunctions in its expansion while the total weighted density remains nonnegative.

For the quantum harmonic oscillator, split the two expectation values into their variances and squared means:The last two terms are nonnegative, soApply the arithmetic-geometric mean inequality to the two remaining nonnegative terms, then the Heisenberg uncertainty principle:The argument applies to any normalized state with finite oscillator energy; the stated stationary state is a particular case. Equality requires zero position and momentum means, minimum quantum uncertainty, and balanced kinetic and potential contributions, as in the oscillator ground state.

The spatial translation operator is $U(\alpha )=e^{-i\alpha \hat{p}/\hslash }$. From $[\hat{x},\hat{p}]=i\hslash$ and the preceding central-commutator result,Thus $\hat{x}U|x⟩=U\hat{x}|x⟩+\alpha U|x⟩=(x+\alpha )U|x⟩$, consistent with $U|x⟩=|x+\alpha ⟩$. Since $⟨x|U=⟨x-\alpha |$, the translated wavefunction is $\boxed{\psi (x-\alpha )}$.

For the quantum harmonic oscillator, the creation and annihilation operators give $\hat{p}=i\sqrt{m\hslash \omega /2}(a^{†}-a)$. Put $s=\alpha \sqrt{m\omega /(2\hslash )}$. The central commutator of $s{a}^{†}$ and $-sa$ is $s^2$, soAs $a|0⟩=0$ and $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$,Taking position wavefunctions provesThe coefficient squares sum to $e^{-s^2}{\sum }_n{s}^{2n}/n!=1$, checking normalization of this coherent state.

The ladder operators obey $[a,a^{†}]=1$ and $N=a^{†}a$ has $[N,a]=-a$, $[N,a^{†}]=a^{†}$. Since $⟨\psi ,N\psi ⟩=\Vert a\psi {\Vert }^2\ge 0$, its eigenvalues are nonnegative. If $N|\nu ⟩=\nu |\nu ⟩$, lowering gives eigenvalue $\nu -1$ and squared norm $\nu$. Repeated lowering cannot produce a negative nonzero eigenstate; the norm product $\nu (\nu -1)\cdots$ forces $\nu$ to be an integer. The final lowered state has $a|0⟩=0$. With a unique ground state all normalized states are

For time-independent perturbation theory, write $H^{\prime }=H+\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$. The standard first correction is $\lambda ⟨n|W|n⟩$ and the second is ${\lambda }^2{\sum }_{j\ne n}|⟨j|W|n⟩{|}^2/(E_n-E_j)$. The first vanishes. Only $j=n-2,n+2$ occur, with ladder coefficients $\sqrt{n(n-1)}$ and $\sqrt{(n+1)(n+2)}$, respectively. ThereforeThe downward term is zero for $n<2$.

Inverting the given position/momentum expressions givesThus $\boxed{\alpha =1-2\lambda ,\beta =1+2\lambda }$. For $|\lambda |<1/2$ both are positive, and this is a quantum harmonic oscillator of effective mass $m/\alpha$ and frequency $\omega \sqrt{\alpha \beta }$. Its exact energies areExpansion gives $E_n^{\prime }=\hslash \omega (n+1/2)[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with both perturbative corrections.

For potential period $a$, a Bloch state has the form ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$, or equivalently ${\psi }_k(x+a)=e^{ika}{\psi }_k(x)$. Quasimomentum is defined modulo the reciprocal-lattice vector $2\pi /a$. A Brillouin zone is one fundamental interval for that identification, usually $[-\pi /a,\pi /a)$. Here $a=2\pi$, so it is $[-1/2,1/2)$.

For small $V_0>0$, the first free-particle degeneracy occurs at the zone edge $k=1/2$, between plane waves $e^{ix/2}$ and $e^{-ix/2}$ of energy $1/4$. In this two-dimensional subspace, the diagonal matrix elements of the potential vanish. Its Fourier coefficient at wave number one is $-V_0/2$; the $\cos 2x$ term cannot connect these two waves at first order. Thus degenerate perturbation theory gives the matrixwith energies $1/4\pm V_0/2+O(V_0^2)$. Near the edge the two uncoupled energies have opposite slopes, and diagonalization gives the avoided crossing, so the lowest band gap is

For large $V_0$, expand about each deepest minimum $x=2\pi j$:The local quantum harmonic oscillator $-d^2/d{x}^2+(5V_0/2)x^2$ has levels $(2n+1)\sqrt{5V_0/2}$. Its localization width is $O(V_0^{-1/4})$, so the quartic term changes these low levels by $O(1)$. Tunnelling between successive deep wells gives exponentially narrow energy bands. The secondary minima at $x=(2j+1)\pi$ have potential zero, and their levels lie far above the first two deep-well levels, whose energies are $-2V_0+O(\sqrt{V_0})$. Therefore the gap between the first and second bands is the first oscillator spacing to leading order:The $O(1)$ anharmonic shifts and exponentially small bandwidths do not change this leading estimate.