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A confined ground state requires $k>0$; take $\alpha >0$ so the trial wavefunction is normalizable. Its normalized form isThe Gaussian moment recurrence gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Also, integration by parts gives $⟨p^2⟩={\hslash }^2\int |f_{\alpha }^{\prime }{|}^{2}dx={\hslash }^2\alpha /2$. The Rayleigh quotient for the quartic oscillator is consequentlyDifferentiation yields ${\mathcal{E}}^{\prime }(\alpha )={\hslash }^2/(4m)-3k/(2{\alpha }^3)$, so the unique minimum is at ${\alpha }_{\ast }=(6mk/{\hslash }^2{)}^{1/3}$. Since the quotient diverges at both endpoints $0$ and $\infty$, this is the global minimum in the trial family. The Rayleigh-Ritz variational principle givesFor the first excited energy eigenstate, choose the odd trial familyThe even confining potential has an even, nodeless ground state, by the nodeless theorem for a one-dimensional ground state. By the Sturm oscillation theorem, its first excited state is odd with one node. The proposed family has the same parity operator eigenvalue and one node, and is orthogonal to the exact ground state, not merely to its Gaussian trial approximation. Minimization in the odd linear subspace therefore bounds the first excited energy from above. For completeness, its Rayleigh quotient is $3{\hslash }^2\beta /(4m)+15k/(4{\beta }^2)$, minimized at ${\beta }^3=10mk/{\hslash }^2$, giving $E_1\le (9{\hslash }^2/8m)(10mk/{\hslash }^2{)}^{1/3}$. These are Gaussian variational bounds for a quartic oscillator.

If $k=0$, there is no normalizable free particle ground state on the whole line, though the spectral infimum is zero. If $k<0$, the Hamiltonian operator is not bounded below, so a ground-energy bound of the requested kind does not exist.

LetThe normalized orbital energy eigenstates vanish outside the well and have one-particle energies $n^2\epsilon$. Define the normalized symmetric and antisymmetric combinations for $r\ne s$ byThe normalized spin functions are the singlet state ${\chi }_{00}=(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ and the triplet states ${\chi }_{11}=|\uparrow \uparrow ⟩$, ${\chi }_{10}=(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, ${\chi }_{1,-1}=|\downarrow \downarrow ⟩$. Their exchange energies are respectively $3J/4$ and $-J/4$.

For $0<J\ll \epsilon$, the three lowest distinct energy levels of the exchange-coupled spin-one-half particles in an infinite square well areThe spatially doubly occupied orbital can only occur with the singlet state. The next orbital sum is $1^2+2^2=5$, where either exchange symmetry is possible. The next unused sum is $2^2+2^2=8$, well separated under the small-coupling assumption. The paper's inequality is equivalent, up to a numerical factor, to $|J|\ll \epsilon$; the factor ${\hslash }^2$ in $J$ is essential because the spin operators carry units of angular momentum.

The sign of $\lambda$ is not explicitly specified. The table assumes the usual intended positive coupling in the displayed minus-sign convention. For $J<0$ with $|J|\ll \epsilon$, the same three sets of eigenfunctions occur, but the two rows at $5\epsilon$ exchange order. At $J=0$ those rows merge into a single energy level of multiplicity four; if three distinct energies are requested exactly at that point, the third is $8\epsilon$ with the single state ${\Phi }_{22}^{+}{\chi }_{00}$.

Each one-particle energy eigenstate has definite parity about the centre of the well, so $⟨n|x|n⟩=0$. Consequently, a doubly occupied orbital has $⟨x_1{x}_2⟩=0$. For distinct orbitals, the two off-diagonal exchange terms instead survive:The position matrix elements in a symmetric infinite square well, with the present width $a$, giveThe sign of $x_{12}$ depends on the orbital phase convention, while $D$ does not. The perturbation acts only on the spatial part and cannot mix orthogonal singlet states and triplet states. Within a triplet it is proportional to the identity on the spin factor. Thus first-order nondegenerate perturbation theory, with this trivial diagonal treatment of spin degeneracy, gives the shifts corresponding to the three rows in part (b):These formulae also apply for $J<0$, with the reordered levels described above. At $J=0$, they resolve the fourfold $5\epsilon$ level into a shifted triplet and singlet. The next doubly occupied orbital ${\Phi }_{22}^{+}$ still has zero first-order shift. The perturbative requirement is $|\alpha |a^2\ll \epsilon$ for mixing with remote orbital states; the spin-independent perturbation does not require division by the small singlet–triplet gap.

The unperturbed infinite square well is invariant under the parity operator. Its one-dimensional energy eigenvalues are nondegenerate, so each energy eigenstate has definite parity. Its position expectation value therefore vanishes: $|{\varphi }_n(x){|}^2$ is even, while $x$ is odd. The first-order nondegenerate perturbation theory correction is consequentlyThe electric field does mix opposite-parity quantum states, and second-order nondegenerate perturbation theory contains two factors of $eEx$. Its leading contribution is thus proportional to $E^2$: this is a quadratic Stark effect. More generally, applying the parity operator changes $H(E)$ into $H(-E)$, so their corresponding energy eigenvalues agree. An analytic expansion about zero electric field contains only even powers of $E$. We assume $e|E|a$ is small compared with the unperturbed excitation scale.

Use zero-based labels to agree with ${\varphi }_0$ in the later parts. The normalized infinite square well energy eigenstates and energy eigenvalues areThe position matrix elements in a symmetric infinite square well, obtained by integrating the product of the two wavefunctions against $x$, areThe integral in the question uses positive integer labels; replacing them by $k+1,n+1$ is essential here. Only opposite-parity states, $k+n$ odd, contribute, as also follows from the parity operator selection rule.

Second-order nondegenerate perturbation theory now gives the requested infinite sum:In fully explicit form this isThe symbol $m$ here denotes the electron mass, not a summation index. For the ground state $n=0$, every denominator in the first form is negative, so the ground energy level moves downward. For excited states, contributions from lower and higher energy levels have opposite signs.

The electron's electric dipole moment is $\mathbf{d}=-e⟨x⟩\hat{\mathbf{x}}$. It vanishes in the unperturbed ground energy eigenstate. The first correction to the wavefunction is linear in the electric field and has the opposite parity; its interference with the unperturbed state therefore gives a nonzero position expectation value of first order.

In detail, the normalized ground quantum state, to the order needed, isThe normalization correction starts at second order and does not affect the linear term below. The two cross terms giveThe absence of an $E^2$ term follows from the parity operator: $⟨x⟩$ changes sign under field reversal. Comparing with the second-order nondegenerate perturbation theory energy sum yieldsThus the induced electric dipole moment points along the electric field, with positive electric polarizability. The induced dipole moment and quadratic energy shift obey $\delta {ϵ}_0^{(2)}=-{\alpha }_{\mathrm{pol}}{E}^2/2$. This factor of one half accounts for the energy needed to induce the dipole. As a check, the Hellmann–Feynman theorem gives $d_x=-\partial {ϵ}_0/\partial E$, with the same result.

Normalize the trial quantum state as $(|{\varphi }_0⟩+c|{\varphi }_1⟩)/\sqrt{1+c^2}$. With the phases used above, the off-diagonal position matrix element is $x_{01}=-32a/(9{\pi }^2)$. DefineThe diagonal perturbation matrix elements vanish by parity, so the Rayleigh quotient isThe stationarity equation is $W+\Delta c-W{c}^2=0$. The minimum continuously connected to the unperturbed ground energy eigenstate hasEquivalently, diagonalize the Hamiltonian operator matrix in this two-dimensional trial space. Its lower eigenvalue isThereforeThis retains exactly the $k=1$ term of the full ground-state second-order nondegenerate perturbation theory sum. Every omitted term is negative, so the full quadratic correction is more negative than this restricted one. This is consistent with the Rayleigh-Ritz variational principle: minimizing over a two-state subspace gives an upper bound on the exact ground energy. The bound applies to the unexpanded two-state minimum; a truncated series is to be compared order by order.

Insert the complete orthonormal basis of energy eigenstates on either side of the Boltzmann factor. The spectral theorem givesAt low temperature, the gap $\hslash \omega$ suppresses excited energy eigenstates. The harmonic oscillator eigenfunction of the ground state isThereforeThe common zero-point energy factor cancels. Dividing the kernel by its trace would give the position probability density of the canonical ensemble; at zero temperature this tends to $|{\phi }_0(x){|}^2$, whose variance is $\hslash /(2m\omega )$. Thus zero temperature projects onto the ground state and retains its quantum spread, rather than concentrating at the minimum of the classical potential energy.

There is also a check on the first correction. With $q=\sqrt{m\omega /\hslash }x$, the first excited harmonic oscillator eigenfunction satisfies $|{\phi }_1(x){|}^2=2q^2|{\phi }_0(x){|}^2$ and ${\phi }_1(0)=0$. Hence the spectral theorem gives $R=e^{-q^2}[1+2q^2{e}^{-z}+O(e^{-2z})]$, in agreement with part (c).

Expand the wavefunction in unperturbed energy eigenstates,The $c_j$ are interaction picture probability amplitudes, so $|c_j{|}^2$ are level populations and $|c_1{|}^2+|c_2{|}^2=1$ in this closed two-level system. A classical field ${\mathbf{E}}_0\cos \omega t$ supplies the electric-dipole interaction $-\mathbf{d}\cdot {\mathbf{E}}_0\cos \omega t$. Choosing the relative phase of the energy eigenstates makes $\Omega =-⟨1|\mathbf{d}\cdot {\mathbf{E}}_0|2⟩/\hslash$ real. This is the resonant Rabi angular frequency.

Combining the field's two exponentials with the unperturbed quantum phases produces terms oscillating at $\omega -{\omega }_0=\delta$ and at $\omega +{\omega }_0$. The rotating-wave approximation drops the latter, leaving exactly the supplied equations with coupling $\Omega /2$. It requires $|\Omega |,|\delta |\ll {\omega }_0$ and a field envelope slow enough on the carrier timescale. Spontaneous emission and dephasing must be negligible during the sequence. For a ground-state hyperfine level transition, the analogous direct drive is a magnetic dipole transition; the same equations hold with the magnetic matrix element replacing the electric one.

Choose the relative phases of the two energy eigenstates so that $d=⟨{\psi }_2|{\hat{D}}_0|{\psi }_1⟩$ is real and nonnegative. Define the resonant Rabi angular frequency $\Omega =d/\hslash$, the Rabi detuning $\delta ={\omega }_0-\omega$, and $\Sigma ={\omega }_0+\omega$. Substitution into the Schrödinger equation, followed by projection onto the two eigenstates, gives the exact interaction picture equationsKeep both probability amplitudes dynamical: replacing $c_1$ by one would incorrectly impose weak excitation and miss complete population transfer.

For $|\delta |,\Omega \ll {\omega }_0$, apply the rotating-wave approximation to discard the rapidly oscillating terms at $\Sigma$. Differentiating the second remaining equation and using the first givesThe initial conditions are $c_2(0)=0$ and ${\dot{c}}_2(0)=-i\Omega /2$. With ${\Omega }_R=\sqrt{{\Omega }^2+{\delta }^2}$, their solution isConsequently,This result permits order-one excitation, including full population transfer on resonance, even though the drive is small compared with the optical angular frequency.

For an average from switch-on to time $\tau$, integrate the squared sine:On resonance this limiting average is $1/2$, although the instantaneous probability repeatedly reaches one. The zero-coupling, zero-Rabi detuning case is understood separately as $P_2=0$.

Near resonance alone does not make this an exact result for arbitrarily strong driving. The stated sinusoidal Hamiltonian also needs the rotating-wave approximation. A concrete counterexample is $\omega ={\omega }_0$ and $\Omega ={\omega }_0$: the exact equations give $P_2(t)={\Omega }^2{t}^2+O(t^4)$ at switch-on, while the claimed rotating-wave sinusoid gives ${\Omega }^2{t}^2/4+O(t^4)$. Thus “without assuming weak radiation” can mean without assuming small excitation, but cannot remove the drive-to-carrier scale separation. The infinite-time average above belongs to the rotating-wave model; omitted corrections can accumulate over sufficiently long times.

Linearity and the shared $l=1$ angular eigenvalue giveUsing the relative phases in part (c),Thus each real orbital remains an energy eigenstate with $E=-1/8$ hartree and has definite orbital angular momentum magnitude $\sqrt{2}\hslash$, but neither is an eigenfunction of ${\hat{l}}_z$. A measurement of $l_z$ gives $+\hslash$ or $-\hslash$, each with probability $1/2$. Its mean is zero and its variance is ${\hslash }^2$; a zero mean does not mean that a measurement gives zero.

Write $Q=c^{†}c+d^{†}d$ and $P=cd+d^{†}{c}^{†}$. Expanding the number operators and restoring the displayed order with the canonical anticommutation relations givesThe minus sign in the first line follows from $c^{†}{d}^{†}+dc=-(cd+d^{†}{c}^{†})$. Substitution into the Hamiltonian operator yieldsLet $E=\sqrt{{\epsilon }^2+{\Delta }^2}>0$. A convenient real Bogoliubov transformation hasThese obey $u^2+v^2=1$, $u^2-v^2=\epsilon /E$ and $2uv=\Delta /E$. Equivalently $u=\cos \theta$, $v=\sin \theta$ with $0<\theta <\pi /4$ and $\tan (2\theta )=\Delta /\epsilon$. The coefficient of $P$ vanishes, while the coefficient of $Q$ is $E$. The constant isThus the diagonalized two-mode fermion pairing Hamiltonian isThe two transformed fermionic occupation numbers independently take values zero or one. The energy eigenvalues and their eigenspace dimensions are thereforeThe positive quasiparticle energy $E$ makes the transformed vacuum state the unique ground state. The constant $\mu$ is essential for the correct ground-state energy.

As a separate sign and spectrum check, use the original fermionic Fock space basis. Both singly occupied states are energy eigenstates with energy $\epsilon$, since the pairing terms annihilate them. In the ordered even-fermion parity basis $(|0⟩,a^{†}{b}^{†}|0⟩)$,The off-diagonal signs follow from $b^{†}{a}^{†}=-a^{†}{b}^{†}$ and $ab{a}^{†}{b}^{†}|0⟩=-|0⟩$. Its characteristic polynomial is $z^2-2\epsilon z-{\Delta }^2$, giving the remaining two eigenvalues $\epsilon \pm E$, in agreement with the Bogoliubov transformation.

Write $|\uparrow ⟩,|\downarrow ⟩$ for the two $S_3$ eigenstates, of eigenvalues $\pm \hslash /2$. The spin-spin interaction can be evaluated usingFor $|\beta |\gg |\alpha |$, take $H_0=\beta (S_3^{(1)}-S_3^{(2)})/\hslash$ and $V=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$. The unperturbed energy eigenvalues are $\beta ,-\beta ,0,0$ on $|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩$, respectively. For an isolated normalized energy eigenstate $|n⟩$, first-order nondegenerate perturbation theory gives $\Delta E_n^{(1)}=⟨n|V|n⟩$. The two antiparallel spins have expectation $-\alpha /4$, soFor the zero-energy degenerate eigenspace, degenerate perturbation theory requires diagonalizing the restriction of $V$. On the ordered basis $(|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩)$ it is $(\alpha /4)I$: the ladder operators cannot take one parallel state to the other. Indeed neither parallel state couples to anything else under $H$, so the remaining two levels are exactly $\alpha /4$, retaining their twofold quantum degeneracy.

For $|\alpha |\gg |\beta |$, take $H_0=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$ and $V=\beta D$, where $D=(S_3^{(1)}-S_3^{(2)})/\hslash$. The spin-one-half triplet state basis and spin-one-half singlet state areThe identity ${\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}=({\mathbf{J}}^2-3{\hslash }^2/2)/2$ gives unperturbed energy eigenvalues $E_t^{(0)}=\alpha /4$ and $E_s^{(0)}=-3\alpha /4$. Direct action givesConsequently the restriction of $V$ to the entire triplet degenerate eigenspace is zero, so no first-order splitting is missed. The singlet expectation is also zero. In second-order nondegenerate perturbation theory, $\Delta E_n^{(2)}={\sum }_{r\ne n}|⟨r|V|n⟩{|}^2/(E_n^{(0)}-E_r^{(0)})$. Within the triplet, the corresponding second-order effective matrix is ${\beta }^2|t_0⟩⟨t_0|/\alpha$, already diagonal in this basis. Thus, through second order,The last two expressions have higher corrections beginning at $O({\beta }^4/|\alpha {|}^3)$, as the exact calculation in part (d) also verifies. For positive $\alpha$, coupling to the lower singlet raises the triplet's $M=0$ level and lowers the singlet; the signs are precisely the energy denominators above.

Infinite walls impose the Dirichlet boundary conditions $\psi (-a)=\psi (a)=0$. Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E\psi$. Its nonzero solutions satisfying both endpoint conditions have wave numbers $k_n=n\pi /(2a)$, $n\ge 1$. Sincethe normalized energy eigenstates are the sine modes with prefactor $1/\sqrt{a}$. They are orthonormal by sine orthogonality and complete by the Fourier sine series on an interval of length $2a$. Differentiating twice gives

The initial wavefunction has unit norm, since ${\int }_0^a(2/a){\sin }^2(\pi x/a)dx=1$. On its nonzero half, $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E_{\ast }\psi$ with $E_{\ast }={\hslash }^2{\pi }^2/(2m{a}^2)$. A small domain qualification is important: the derivative jumps at zero, so the state is not in the operator domain of the Dirichlet Hamiltonian operator. Its finite energy expectation value is nevertheless defined by the quadratic form of a positive quantum Hamiltonian. The energy form of a half-well sine state givesThis is also the rigorous weak interpretation of the printed $\int \psi H\psi$. Indeed the distributional second derivative is ${\psi }^{\prime \prime }=-(\pi /a{)}^2\psi +\sqrt{2/a}(\pi /a){\delta }_0$ inside the well. The Dirac delta term pairs to zero because $\psi (0)=0$. One should not instead conclude that this state is an energy eigenstate from its equation on just the positive half.

Its coefficients in the full-well energy eigenstates areFor even $n$, sine orthogonality gives $a_2=-1/\sqrt{2}$ and all other even coefficients zero. For $n=2p+1$, set $u=x/a$ and use $\sin [n\pi (u+1)/2]=(-1{)}^p\cos (n\pi u/2)$. Product-to-sum integration then givesSince these coefficients are $O(n^{-2})$, ${\sum }_n{E}_n|a_n{|}^2$ converges, confirming that the state has finite energy. With $E_n=n^2{E}_1$, the spectral expression isThe kinetic-energy quadratic form equals this spectral sum: expanding the weak derivative in the corresponding orthonormal cosine modes and using Parseval's identity gives $\int |{\psi }^{\prime }{|}^2={\sum }_n{k}_n^2|a_n{|}^2$. Equating it to the directly integrated value $4E_1$ provesIn contrast, ${\sum }_n{E}_n^2|a_n{|}^2$ diverges, consistently showing why $H\psi$ is not an ordinary square-integrable vector even though the energy expectation is finite.

Expand the wavefunction in the complete orthonormal energy eigenstates. The Time-dependent Schrödinger equation makes each coefficient satisfy $i\hslash {\dot{a}}_n(t)=E_n{a}_n(t)$, soThe series is interpreted in the Hilbert space norm. Its norm is conserved because the phase factors have modulus one and the energy eigenstates are orthonormal. For initial data in the domain of the Hamiltonian operator, this is a strong solution of the equation; for any square-summable coefficients it gives the corresponding unitary time evolution.

In the two-dimensional span of the first two energy eigenstates, the measured observable has matrix $S=\left(\begin{array}{cc}7& 24\\ 24& -7\end{array}\right)$. Its square is $625I$, and its trace is zero, giving eigenvalues $25,-25$. The orthogonal complement has eigenvalue zero. Thus the full spectrum is $\boxed{\{-25,0,25\}}$.

The normalized eigenstates for $-25$ and $25$ are respectivelyThe lowest measurement outcome prepares ${\chi }_{-}$ up to an irrelevant overall phase. The subsequent unitary time evolution givesBy the Born rule, the probability of measuring the lowest eigenvalue again is the squared projection onto its one-dimensional eigenspace:This is the two-level quantum return probability with $q=9/25$. At $t=0$ the probability is one; if the two energies happen to be equal it remains one for every time.

Interpret the box as an infinite square well, with zero potential on $0<x<a$ and impenetrable walls. The wavefunction vanishes at both walls. The time-independent Schrödinger equation inside is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E\psi$. Multiplying by ${\psi }^{\ast }$ and integration by parts shows $E\int |\psi {|}^{2}dx=({\hslash }^2/(2m))\int |{\psi }^{\prime }{|}^{2}dx>0$ for any nonzero state with the wall conditions.

For $E>0$, put $k^2=2mE/{\hslash }^2$. The condition at zero gives $\psi =A\sin (kx)$, and the condition at $a$ requires $\sin (ka)=0$. The zero value of $k$ gives only the zero wavefunction, so the allowed energy eigenvalues areThe corresponding normalized energy eigenstates are $\sqrt{2/a}\sin (n\pi x/a)$. Each level is nondegenerate in this one-dimensional problem. A nonzero constant interior potential would add that constant to all energies.

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.

The two given energy eigenstates are orthonormal. Resolve the initial state as $\psi (0)=\cos \theta {\psi }_0-\sin \theta {\psi }_1$. The Schrödinger equation therefore givesThe second component, which is the probability amplitude for the requested final state, is $\sin \theta \cos \theta (e^{-i{E}_{0}t/\hslash }-e^{-i{E}_{1}t/\hslash })$. By the Born rule,This includes $P(t)=0$ when the energy eigenvalues coincide or when the initial state is itself an energy eigenstate.