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For the XXZ spin chain, write $\Delta =1+\alpha$ and replace each spin by a classical vector of length $S$. A bond is $J{S}^2{\mathbf{n}}_m^{\mathsf{T}}\mathrm{diag}(1,1,\Delta ){\mathbf{n}}_{m+1}$, whose minimum is $-J{S}^2\max (1,|\Delta |)$. For an even periodic chain all bonds can attain that bound simultaneously. The classical ground states of an XXZ spin chain therefore give $\alpha >0$ gives the two easy-axis Néel states ${\mathbf{S}}_m=(-1{)}^{m}S\hat{\mathbf{z}}$ and its reversal, with energy $-JN{S}^2(1+\alpha )$.

For $-2<\alpha <0$ the ground states are planar Néel configurations, ${\mathbf{S}}_m=(-1{)}^{m}S(\cos \phi ,\sin \phi ,0)$, with arbitrary $\phi$: their degeneracy is a circle and their energy is $-JN{S}^2$. This includes the easy-plane interval used later. Part (a), however, does not restrict negative $\alpha$ to that interval. For $\alpha <-2$, $\Delta <-1$ and the lowest states are instead the two uniform $z$-polarized ferromagnetic configurations, with energy $JN\Delta S^2$. At $\alpha =-2$, a rotation by $\pi$ about $z$ on alternating sites maps the classical energy to an isotropic ferromagnet: ${\mathbf{S}}_m=S((-1{)}^m{n}_x,(-1{)}^m{n}_y,n_z)$ for any unit vector $\mathbf{n}$ is a ground state. This special degeneracy is a sphere. At the other boundary, $\alpha =0$, the ordinary Néel direction is arbitrary on a sphere. These statements concern classical minimizing configurations, not the degeneracy of the exact finite-chain quantum ground state.

The normalized Hartree product state and its equivalent second quantization form areThe product is already symmetric under exchanging bosons; no additional $1/\sqrt{N!}$ belongs in the coordinate wavefunction. The factor in the second quantization expression instead normalizes repeated creation operators acting on the vacuum state. This is a fixed-number two-mode condensate, with every boson in the same normalized orbital.

Set $c=\cos \theta$ and $s=\sin \theta$. Its binomial distribution of occupations givesThe expectation value of the Hamiltonian operator is thereforeKeeping $N(N-1)$ rather than replacing it by $N^2$ retains the exact pair count of the trial quantum state. Equivalently, with $ϵ=w-\mu$, its interaction contribution is $wN(N-1)(c^4+s^4)$ and its one-particle contribution is $ϵN-tN\sin (2\theta )\cos \chi$.

Choose the nonredundant parameter range $0\le \theta \le \pi /2$, with $\chi$ modulo $2\pi$. For $t>0$, minimizing over the relative phase gives $\chi =0$. Putting $x=\sin (2\theta )\in [0,1]$, the remaining nonconstant energy is $-tNx-\frac{1}{2}wN(N-1)x^2$, which decreases with $x$ for the repulsive coupling $w>0$. HenceThe optimal orbital is the symmetric ${\varphi }_0=({\phi }_L+{\phi }_R)/\sqrt{2}$. Both the quantum tunnelling term and the repulsive interaction favor this balanced occupation within the Hartree product state family. This result uses a variational method: for strong repulsion relative to quantum tunnelling, the true many-body ground state can have smaller occupation fluctuations than this fixed-number two-mode condensate.

For a uniform continuum gas with pair interactions depending on relative positions, a Galilean boost changes every particle momentum by $p_0\hat{\mathbf{x}}$ and leaves the interaction energy unchanged. Write ${\hat{P}}_x={\sum }_j{\hat{p}}_{j,x}$. The transformed many-body Hamiltonian operator isThe resting ground state has $⟨{\hat{P}}_x⟩=0$, so the Galilean boost energy of an interacting gas isThe mass here is the bare particle mass, because this is an exact transformation of the microscopic kinetic energy. Substituting the Fermi-liquid effective mass $m^{\ast }$ would omit the change in interaction energy encoded by the Landau quasiparticle energy functional. This conclusion requires the Principle of Galilean relativity; a gas in a lattice or a fixed external potential energy landscape need not obey it.

The Hartree-Fock method approximates the ground state wavefunction by a single Slater determinant, minimizing the expectation value of the Hamiltonian over its occupied orbitals. It retains fermionic exchange symmetry exactly within that determinant but omits electron correlation beyond a single Slater determinant.

At $N=M$, the two spin occupations per site are $x$ and $1-x$. With $\tau =|t|>0$, the trial energy per site isIts derivative is $e^{\prime }(x)=-4\tau \cos (\pi x)+V[4x-2-(2/\pi )\sin (2\pi x)]$. Since $x=0$ is a boundary point, its local minimum test is one-sided: $e^{\prime }(0)=-4\tau -2V$. At $V=-2\tau$, where this derivative vanishes, its Taylor series begins $e(x)-e(0)=(2{\pi }^2\tau /3)x^3+O(x^4)>0$ for small positive $x$. HenceThe same applies to $x=1$ by spin interchange.

The symmetry $e(x)=e(1-x)$ makes $x=1/2$ a stationary point for every $V$. Here $e^{\prime \prime }(1/2)=4\pi \tau +8V$, so the second-derivative test settles all cases except equality. Writing $y=x-1/2$, at $V=-\pi \tau /2$ the Taylor series is $e(1/2+y)-e(1/2)=({\pi }^3\tau /6)y^4+O(y^6)$. ThusThe minimizing branches interpolate continuously. To find all of them, put $u=\pi (1/2-x)$ on $0<x<1/2$. An interior stationary point obeysThe right side decreases strictly from $-\pi /2$ to $-2$, sinceThe positivity follows from $u-\sin u\cos u={\int }_0^{u}2{\sin }^{2}vdv$. There is therefore exactly one such stationary point on each side of $1/2$ when $-2<V/\tau <-\pi /2$. The signs of $e^{\prime }$ show that both are global minima, whereas $1/2$ is then a local maximum. The complete answer isThe pair merges through a supercritical pitchfork bifurcation at $V/\tau =-\pi /2$. Comparing only $e(0)$ and $e(1/2)$ would miss the intervening ground states with partial spin polarization.

For $V\gg |t|$, the leading objective is to avoid adjacent electrons of the same spin. The unpolarized Fermi sea, although optimal within the preceding restricted family, has interaction energy per site $V(1/2-2/{\pi }^2)>0$. Its cost grows linearly with $V$, while states with no forbidden neighbouring pairs have zero interaction energy.

There are two independent spinless fermion t-V chains, one for each spin. On an even ring, avoiding all neighbouring pairs allows at most $M/2$ particles in each chain. Since the total is $M$, both bounds must be saturated, and each chain must alternate occupied and empty sites. Their two independent choices give four limiting occupation patterns:The first two are Néel states with one electron per site; the last two are charge density waves alternating doubly occupied and empty sites. Opposite-spin double occupation costs nothing in this particular Hamiltonian. Each species has alternating occupation in all four patterns.

These are the exact zero-interaction ground states at $t=0$, and the limiting ordered configurations as $V/|t|\to \infty$. At large finite $V/|t|$, virtual hopping admixes configurations with forbidden pairs, producing an energy correction of order $-M|t{|}^2/V$. A finite ring can have superpositions of the ordered configurations, with possible quantum tunnelling splittings; four exactly degenerate product ground states are not implied at finite hopping. The four ordered choices describe the strong-coupling thermodynamic limit, with spontaneous symmetry breaking understood.

Measure single-particle energy from the quantum harmonic oscillator ground state energy $\hslash \omega$, and write $\tilde{\mu }={\mu }_{\mathrm{physical}}-\hslash \omega \le 0$. The Bose-Einstein distribution and the supplied density of states giveThis excited-state population increases with the chemical potential and has the finite saturation valueAny additional particles must occupy the ground state: this is two-dimensional harmonic-trap Bose condensation. The critical temperature and the condensate fraction are thereforeHere $N$ is the total particle number, the integral of the inhomogeneous number density over the trap; there is no fixed uniform area in which $N$ itself is a number density. Equivalently $N\kappa ={\pi }^{2}m(k_B{T}_c{)}^2/(6{\hslash }^2)$ relates the trap population to the critical temperature. Harmonic confinement changes the low-energy density of states and permits this saturation in two spatial dimensions.

These formulas are leading large-$N$ results with $k_B{T}_c\gg \hslash \omega$. A sharp transition belongs to the trapped-gas thermodynamic limit $N\to \infty$, $\omega \to 0$ at fixed $N{\omega }^2$; a finite trap has finite-size rounding. In particular, finite $N_0$ requires $\tilde{\mu }=-k_{B}T\log (1+1/N_0)<0$, approaching zero in the condensation limit.

The Hartree product state underlying the Gross–Pitaevskii equation has the factorized one-body coherence of a Hartree condensate. Choosing its nodeless ground state orbital real givesIn particular, the Thomas–Fermi approximation for a condensate from part (d) predictsThe curve starts at $\mu /g$, decreases with zero slope at the origin, and reaches zero with an infinite slope at $R$. Restoring the kinetic energy in the full Gross–Pitaevskii equation rounds that edge and produces a nonzero tail. This corrects the Thomas–Fermi approximation for a condensate, but the resulting one-body density matrix still factorizes.

The additional missing physics is quantum depletion. Interactions scatter pairs of bosons into orbitals outside the Bose-Einstein condensate even at zero temperature. In terms of the true Bose-Einstein condensate orbital ${\phi }_0$ the exact one-body density matrix decomposes asThe single-orbital Hartree product state sets $N_0=N$ and omits ${\rho }_{\mathrm{ex}}$ altogether. In a weakly interacting Bose gas, Bogoliubov theory adds a comparatively short-range noncondensed contribution near the origin, together with reduced weight in the broad Bose-Einstein condensate component. The central value remains the total number density, not just the condensate density. Its numerical shift relative to the original Gross–Pitaevskii equation solution also depends on the interaction-induced redistribution of number density; it is not fixed by the condensate fraction alone.

Take $m,\omega >0$ and use the usual common invariant domain for the position operator, momentum operator and their products. Since $X$ and $P$ are self-adjoint,Expanding the commutator and using $[P,X]=-i\hslash I$ givesThe two products areConsequently the quantum harmonic oscillator Hamiltonian operator isThe identity $[AB,C]=A[B,C]+[A,C]B$ now givesThus $H(a_{\pm }\psi )=(E\pm \hslash \omega )a_{\pm }\psi$. Each ladder image is an eigenstate if it is nonzero; a zero vector is not an eigenstate. The distinction is needed for lowering the ground state. Their squared norms areThese follow by moving the operator to the other side of the inner product with its adjoint operator. The first inequality implies $E\ge \hslash \omega /2$ for every eigenstate.

For a lowest-energy quantum state, $a_{-}{\psi }_0$ must vanish, since otherwise it would be an eigenstate with lower energy. The norm formula then forces $E_0=\hslash \omega /2$. Raising never vanishes and producesTo see that there are no other eigenvalues, repeatedly lower any eigenstate. If $E/(\hslash \omega )-1/2$ were not a nonnegative integer, there would be a final nonzero lowered quantum state with energy strictly between $\hslash \omega /2$ and $3\hslash \omega /2$. Its lower image would have positive norm but energy below the lower bound, a contradiction. Thus every eigenstate reaches the ground eigenspace after an integer number of lowerings. On a positive level, lowering is injective by its strictly positive norm factor. The assumed one-dimensional ground eigenspace therefore makes every eigenspace one-dimensional, so the displayed raised quantum states give all eigenstates up to nonzero multiples. If ${\psi }_0$ is normalized, their norms are $\Vert {\psi }_n{\Vert }^2=n!$.

For $k\ge 1$, $(a_{+}{)}^k\psi$ has eigenvalue $E+k\hslash \omega$, distinct from $E$. eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal: applying self-adjointness to their inner product gives $(E^{\prime }-E)⟨\psi ,\varphi ⟩=0$. ThereforeThe expectation of $(a_{-}{)}^k$ also vanishes, either by lowering and orthogonality or by conjugating the adjoint relation.

Solving the ladder definitions for $X$ and $P$ givesso $T-V=\frac{1}{2}\hslash \omega ((a_{+}{)}^2+(a_{-}{)}^2)$. Both terms have zero expectation by the preceding orthogonality. Since $T+V=H$, the oscillator form of the quantum virial theorem isExpectations here are divided by $\Vert \psi {\Vert }^2$ if the eigenstate is not normalized; no normalization assumption was used in the ladder argument.

In the scalar two-particle model, the fermionic exchange symmetry and bosonic exchange symmetry conditions areFor particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi (x_1,x_2)=\Psi (x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi =E\Psi$ implies $H(S\Psi )=E(S\Psi )$. The componentsare respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero quantum state, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.

The normal modes of two harmonically coupled identical particles provide the separation. The coordinate transformation to $y_{+},y_{-}$ is orthogonal, with absolute Jacobian one. Exchanging the particles leaves $y_{+}$ fixed and changes $y_{-}$ to $-y_{-}$. Hence bosonic quantum states are even and fermionic quantum states odd in the relative coordinate. The chain rule givesAlso $x_1^2+x_2^2=y_{+}^2+y_{-}^2$ and $(x_1-x_2{)}^2=2y_{-}^2$. Combining the two trapping potentials with the interaction therefore gives the separated Hamiltonian operatorThese are independent oscillators of frequencies $\omega$ and $\omega \cos \alpha$. Both frequencies are positive in the stated range. Their product eigenstates have energiesThe parity of an oscillator level is $(-1{)}^n$. Consequently bosons allow even $n_{-}$ and fermions odd $n_{-}$, while $n_{+}$ is unrestricted.

The bosonic ground state uses $n_{+}=n_{-}=0$, and the fermionic ground state uses $n_{+}=0,n_{-}=1$. The oscillator length in the relative mode isUsing the normalized Gaussian for each mode givesIndeed ${\varphi }_{\tau }(y)=\sqrt{2}y{\psi }_{\tau }(y)/\tau$ by differentiating the Gaussian, so it is the normalized odd first excited state. The ground energies are $E_{\mathrm{B}}=\hslash \omega (1+\cos \alpha )/2$ and $E_{\mathrm{F}}=\hslash \omega (1+3\cos \alpha )/2$.

Choose normalized factors for a normalized product quantum state; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,If the supplied factors are not individually normalized, the general expression has an additional factor $\Vert \psi {\Vert }^2$; choosing normalized factors is the convention in the displayed formula.

For the ground states, put $\rho (y)=|{\psi }_{\tau }(y){|}^2=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$. Then $|{\varphi }_{\tau }(y){|}^2=2y^2\rho (y)/{\tau }^2$ and ${\rho }^{\prime }=-2y\rho /{\tau }^2$. Thus, for differentiable $f$ with suitable Gaussian integrability and vanishing boundary terms,This is integration by parts, and every polynomial satisfies the required endpoint conditions. The relative-coordinate moments in even and odd oscillator ground states follow by substituting $f(y)=y^n$:For odd $n$, both expectations vanish by parity, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular ${\mathbb{E}}_{\mathrm{B}}[y_{-}^2]={\tau }^2/2$ and ${\mathbb{E}}_{\mathrm{F}}[y_{-}^2]=3{\tau }^2/2$, so the fermionic root-mean-square particle separation is $\sqrt{3}$ times the bosonic value. The fermionic probability density vanishes at coincidence $y_{-}=0$, expressing Pauli exclusion principle in this spatial symmetry sector. This is an exchange constraint on the wave function, not an additional classical force. As $\alpha$ approaches $\pi /2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator ground state.

Use an expansion for the eigenstate continuing the nondegenerate level, with intermediate normalization $⟨{\psi }_0,\psi (ϵ)⟩=1$:Insert these expressions into the eigenvalue equation. At first order,Taking the inner product with ${\psi }_0$ eliminates the left side by self-adjointness. ThusFor $j\ne 0$, projection onto ${\psi }_j$ instead givesAt second order, projecting the eigenvalue equation onto ${\psi }_0$ gives $E^{(2)}=⟨{\psi }_0,H^{\prime }\chi ⟩$. Expand $\chi$ in the orthonormal energy basis and use self-adjointness of $H^{\prime }$:Nondegeneracy ensures the denominators for this level do not vanish; convergence and validity of the expansion are among the assumed perturbation-theory hypotheses.

For the cosine perturbation of the ground level in a one-dimensional box, write $u=\pi x/a$. The product-to-sum identity givesso, with the given common normalization of the sine eigenfunctions,The diagonal matrix element is $-ϵ/2$, and the only nonzero off-diagonal matrix element out of the ground state is $⟨{\varphi }_3,V{\varphi }_1⟩=ϵ/2$. Let $E_{\mathrm{box}}={\pi }^2{\hslash }^2/(2m{a}^2)$. Since ${\mathcal{E}}_3=9E_{\mathrm{box}}$, the ground energy isThe second-order term is negative, as expected for a nondegenerate ground level: every excited-quantum state denominator is negative. The first-order shift is also negative for positive $ϵ$, because the ground-quantum state density is largest near the center, where this perturbing potential is negative.

Choose an orthonormal eigenbasis $H_0|n⟩=E_n|n⟩$ and the phase convention $⟨0|{\psi }^{\prime }⟩=0$. Comparing coefficients of the eigenvalue equation at first order yieldsProjecting onto $|0⟩$ and the other eigenvectors givesProjection of the next-order equation onto $|0⟩$ givesThese are coefficients of the printed series, so $E^{\prime \prime }$ denotes the quadratic coefficient rather than the second derivative itself. Without a phase convention an arbitrary purely imaginary multiple of ${\psi }_0$ can be added to the first vector correction. Continuous spectrum, if present, requires the corresponding spectral integral; the oscillator has a discrete complete eigenbasis.

Put $K=m{\omega }^2/2$ and let $g(x)=(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}$ denote the ground-state density. Then $H^{\prime }=K(a^2-x^2)1_{[-a,a]}$ and $p={\int }_{-a}^{a}g(x)dx$. Because $g^{\prime }=-xg/{\sigma }^2$, integration by parts yieldsThus the two first-order shifts areThe first is the supplied ground-state expression; the second follows because the excited-state density is $x^{2}g/{\sigma }^2$, so $E_1^{\prime }=K(a^2{I}_2-I_4)/{\sigma }^2$.

Subtracting, and using $K{\sigma }^2=\hslash \omega /4$, givesThe bracket is zero at $a=0$, tends to one at infinity, and has derivative $2a^{2}g(a)/{\sigma }^2\ge 0$. Therefore the first-order predicted level gap isFor every $\lambda >2$ it becomes negative for all sufficiently large $a$, crossing when $p-2ag=2/\lambda$. It is still positive at $a=0$: the stated conclusion must be read with the large-$a$ regime, rather than for every interval width.

For $\lambda >1$ the inner potential curves downward, placing a central barrier between two symmetric minima at $\pm a$. The even unperturbed ground state has its greatest density at the central barrier, while the odd first excited state has a node there and is less penalized. This explains the relative first-order shifts. A negative predicted gap signals failure of that perturbation approximation, not inversion of the exact lowest two states: the true one-dimensional ground state remains nodeless and even, with the odd state above it. At a high barrier the actual low levels instead form a close double-well pair. The apparent level crossing from a finite-interval oscillator perturbation is useful precisely as a diagnostic of the approximation's range.

Let $U(t)=e^{-iHt/\hslash }$ and take an observable with no explicit time dependence. In the Schrödinger picture, ${\psi }_S(t)=U(t){\psi }_0$ and $A_S=A$. In the Heisenberg picture, ${\psi }_H={\psi }_0$ and $A_H(t)=U(t{)}^{†}AU(t)$. For the interaction picture choose a time-independent splitting $H=H_0+V$, put $U_0=e^{-i{H}_{0}t/\hslash }$, and defineThen $i\hslash {\dot{\psi }}_I=V_I{\psi }_I$ and ${\dot{A}}_I=(i/\hslash )[H_0,A_I]$. Unitarity gives the identical expectation in all three descriptions:Different choices of the split interpolate between pictures; $H_0=H$ makes the state constant, while $H_0=0$ gives the Schrödinger description.

Differentiate $U^{†}AU$ to obtain the Heisenberg equation of motionFor an explicitly time-dependent observable there is additionally $U^{†}({\partial }_{t}A)U$. Thus $H_H=H$ is constant. In a normalized energy eigenstate, $⟨[H,A_H]⟩=E⟨A_H⟩-E⟨A_H⟩=0$, so the expectation of its Heisenberg derivative vanishes whenever the products are defined.

For the harmonic oscillator, the canonical commutator givesThe last identity follows from the operator product rule without commuting $X$ and $P$. Although $XP$ is not itself self-adjoint, the same calculation can be applied to the self-adjoint dilation observable $(XP+PX)/2$ and gives the same derivative.

Stationarity of the three expectations gives $⟨P⟩=⟨X⟩=0$ and $⟨P^2⟩=m^2{\omega }^2⟨X^2⟩$. Substitution into $E=⟨P^2⟩/(2m)+m{\omega }^2⟨X^2⟩/2$ yields the quantum virial theoremThe vanishing means make the variances equal to the second moments, soThis satisfies the position-momentum Heisenberg uncertainty relation, with equality only for the ground state.

For bound hydrogen atom states, the radial wave function is regular and finite at the origin, and decays at infinity so that $4\pi {\int }_0^{\infty }{r}^2|\Psi (r){|}^{2}dr<\infty$. Equivalently the reduced radial function $w(r)=r\Psi (r)$ has $w(0)=0$ and is square integrable on the half-line. A singular $1/r$ branch is not an admissible regular Coulomb eigenfunction; its derivative energy diverges and it introduces an origin singularity absent from the Hamiltonian domain.

Write $\kappa =e^2/(4\pi {ϵ}_0)$, so $a={\hslash }^2/(m\kappa )$. Multiplication by $r$ turns the radial equation intoSince $w=a{e}^{-\lambda R}f(R)$ and $r=aR$, substitution givesChoose $\boxed{\lambda =\mu a>0}$ to remove the constant coefficient and factor out the decaying exponential. For $f={\sum }_{k\ge 0}{a}_k{R}^{k+c}$, the lowest power has coefficient $c(c-1)a_0$, so $\boxed{c(c-1)=0}$. The $c=0$ behavior would give $\Psi \sim a_0/R$, violating regularity; choose $c=1$.

Matching successive powers then yieldsA nonterminating tail eventually has one sign, and its coefficient ratio is asymptotic to $2\lambda /k$. More precisely, for any $0<\delta <\lambda$, the ratio is eventually at least $(2\lambda -\delta )/(k+1)$. Its same-sign tail consequently dominates a positive constant times the exponential series of rate $2\lambda -\delta$, up to finitely many terms. The factor $e^{-\lambda R}$ cannot then make $\Psi$ decay. Thus square integrability forces termination.

If the last nonzero term is $a_{n-1}{R}^n$, termination requires $\lambda n=1$. ThereforeFor $n=1$, $\lambda =1$, $f$ is proportional to $R$, andThe radial volume element is $4\pi r^{2}dr$, so the mean radius is the ratio of the radius-weighted norm integral to the norm integral. With $r=aR$, the common angular and normalization factors cancel:In the ground state $\boxed{⟨r⟩=3a/2}$.

A permutation $\pi$ of identical particles multiplies a fermionic state by $\mathrm{sgn}(\pi )$; a bosonic state is invariant. Particle exchange keeps $y_{+}$ fixed and sends $y_{-}$ to $-y_{-}$. Consequently the spatial wave function is even in $y_{-}$ for the stated bosonic model and odd for the stated fermionic model. This assumes no additional spin variables in the supplied model; exchange of full particle states is the underlying rule.

The coordinate change is orthogonal, with unit Jacobian determinant, so the kinetic energy becomeson the square $|y_{+}|,|y_{-}|<a$ with zero boundary values. Product states ${\psi }_r(y_{+}){\psi }_s(y_{-})$ have energyThe one-dimensional eigenfunction is even for odd $s$ and odd for even $s$. Bosons therefore allow odd $s$, whereas fermions require even $s$. The lowest bosonic pair is $(r,s)=(1,1)$; its first excited pair is $(2,1)$. The lowest fermionic pair is $(1,2)$. ThusThe printed bosonic ground-energy claim omits the factor ${\pi }^2/4$; the value above follows directly from the supplied one-dimensional spectrum and kinetic energy.