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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.

On the tensor product of spins $j_1$ and $j_2$, the combined angular momentum operator is $\mathbf{J}={\mathbf{S}}^{(1)}\otimes I+I\otimes {\mathbf{S}}^{(2)}$. The Clebsch-Gordan decomposition iswhere $J$ increases by one and each irreducible spin representation occurs once. The dimensions agree: ${\sum }_J(2J+1)=(2j_1+1)(2j_2+1)$.

The uncoupled orthonormal basis diagonalizes $S_3^{(1)},S_3^{(2)}$; the coupled orthonormal basis diagonalizes ${\mathbf{J}}^2,J_3$. The Clebsch-Gordan coefficients are their change-of-basis entries, with conventionThey vanish unless $m_1+m_2=M$. Since both bases are orthonormal, this change-of-basis matrix obeys unitarity. Overall phases of the coupled basis vectors are fixed by a convention and do not change the energy eigenvalues.

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.

In the ordered product basis $(|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩)$, the Hamiltonian operator has matrixThe middle block has characteristic polynomial $(E+\alpha /4{)}^2-{\beta }^2-{\alpha }^2/4$. Hence the exact energy eigenvalues areThis is the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field, with its interaction and field parameters rescaled to the paper's units. For positive $\beta$ and $|\alpha |\ll \beta$, the Taylor expansion giveswhich agrees with the first-order result of part (c). For positive $\alpha$ and $|\beta |\ll \alpha$,These agree with the second-order triplet and singlet results. If $\alpha <0$, their association with the signs $\pm$ reverses; if $\beta <0$, the two antiparallel product-state labels reverse. The unordered exact spectrum and the perturbative expressions with their signed denominators remain valid.

Let $X$ exchange the two equal-spin factors, so $X^{†}=X=X^{-1}$. For the unperturbed Hamiltonian operator, addition of angular momentum givesFor nonzero $\alpha$, different $J$ have distinct energy eigenvalues, while each eigenspace has the $2J+1$ values of $M$. The stated symmetry of the Clebsch-Gordan coefficients gives the exchange parity of two coupled identical spins:However, $D=(S_3^{(1)}-S_3^{(2)})/\hslash$ changes sign: $XDX=-D$. For any two basis vectors in the same $J$ eigenspace, insertion of $X^2=I$ therefore givesThus the entire projected perturbation $P_J\beta D{P}_J$ is zero, including its off-diagonal matrix elements. Degenerate perturbation theory now proves that every first-order-in-$\beta$ energy correction vanishes. This is first-order cancellation for an exchange-odd perturbation; checking only the individual diagonal expectations would not by itself settle a degenerate problem.

Continue the transmission amplitude to complex wavenumber $q=i\kappa$ with $\kappa >0$, equivalently complex momentum $k=i\hslash \kappa$. Then $E=-{\hslash }^2{\kappa }^2/(2m)<0$. In the left exterior region the $A$ term grows at negative infinity and the $B$ term decays; in the right exterior region the $D$ term grows and the $C$ term decays. A bound state thus requires $A=D=0$, with nonzero $B,C$. The transfer matrix relation gives $M_{22}(i\kappa )=0$. Equivalently, a zero of the incoming coefficient is a bound-state pole of $t(q)$, whose numerator is nonzero at $q=i\kappa$.

The denominator found in part (c) givesFactorization produces two branches. To identify their parity, use an even interior wavefunction $A\cosh (\kappa x)$ or an odd interior wavefunction $A\sinh (\kappa x)$ for $|x|<a$, joined to a decaying exterior wavefunction at $a$. Continuity and the derivative jump give, respectively,Rearrangement gives the implicit energy eigenvalue equationsOnly strictly positive equation roots $\kappa$ are retained. With $\nu =-\mu >0$, the even equation root lies above $\nu$ and the odd equation root, when it exists, lies below $\nu$, so the even bound state has lower energy. The formal zero equation root in the odd equation does not itself establish a normalizable wavefunction.

Use the amplitudes and positive action integrals of part (a), and define the right-well phaseMatching at the inner classical turning point gives $C=A$, $D=2B$. Matching the exterior decaying wavefunction at the outer classical turning point givesThe trigonometric identitytherefore yieldsAll three pieces on $x>0$ are now fixed up to $E$ and $F$:The Airy function descriptions replace the apparent singularities at the classical turning points. For the permitted energy eigenvalues, part (d) extends this right-hand wavefunction to the left by even or odd parity; $F$ is then determined by wavefunction normalization. At arbitrary $E$, the decaying right-hand solution instead has a unique continuation by the Schrödinger equation, which need not decay at negative infinity.

A universal quantum gate set can implement any unitary operator on any finite number of qubits, exactly when arbitrary continuous gate parameters are allowed, or to arbitrary accuracy with a suitable finite gate alphabet. More precisely, for every target $V$ and $ϵ>0$ there is a finite quantum circuit $W$ with ${\min }_{\gamma }\Vert W-e^{i\gamma }V\Vert <ϵ$ in operator norm. This ignores only an overall global phase of the complete operation. Universality lets a fixed collection of controllable elementary operations run arbitrary quantum algorithms rather than requiring a different physical interaction for each algorithm.

All single-qubit unitary gates, together with the controlled-NOT gate between selectable pairs, form an exact universal quantum gate set. One way to see sufficiency is to eliminate entries of a finite unitary matrix by two-level rotations between computational basis vectors; these rotations decompose into controlled single-qubit rotations, which in turn decompose into single-qubit gates and controlled-NOT gates. Single-qubit gates alone preserve product states and cannot supply general quantum entanglement. A finite approximate universal example is the Hadamard gate, the T gate $T=\mathrm{diag}(1,e^{i\pi /4})$, and the controlled-NOT gate; finite exact words over a finite alphabet cannot represent every continuously parameterized unitary operator exactly.

Distinguish the Hamiltonian operator $H_{\mathrm{int}}$ from the Hadamard gate $H_{\mathrm{Had}}$. For the printed collisional phase gate,As written, $\Omega$ has units of energy. The Schrödinger equation givesbecause only $|01⟩$ has a nonzero energy eigenvalue. In the ordered computational basis $(00,01,10,11)$, defineIf an interaction strength is instead quoted as an angular frequency ${\Omega }_{\mathrm{f}}$, the Hamiltonian operator is $\hslash {\Omega }_{\mathrm{f}}|01⟩⟨01|$ and $\varphi ={\Omega }_{\mathrm{f}}\tau$. An alternative definition with $e^{+i\varphi }$ would have $\varphi =-\Omega \tau /\hslash$. The minus sign above follows the usual time-evolution operator convention. At $\varphi =\pi$ these conventions give the same $U_{\pi }$.

For the quantum circuit, use $X$ on the first qubit to move the selective minus sign from $01$ to $11$:Conjugating the target of this Controlled-Z gate by the Hadamard gate gives the required controlled-NOT gate:The rightmost matrix acts first. Thus the network applies $X_1,H_2$, then $U_{\pi }$, then $X_1,H_2$, and maps $|a,b⟩$ to $|a,b\oplus a⟩$, where $\oplus$ is exclusive or.

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.

The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$, with attractive Coulomb potential $V(r)=-\kappa /r$, where $\kappa =e^2/(4\pi {\epsilon }_0)$. Separating $\psi =R(r)Y_{\ell }^m(\theta ,\phi )$ uses the spherical harmonic angular eigenvalue $-\ell (\ell +1)$ of the unit-sphere Laplacian. The radial kinetic contribution is $-{\hslash }^2(R^{\prime \prime }+2R^{\prime }/r)/(2m)$, the angular kinetic contribution is the centrifugal term ${\hslash }^2\ell (\ell +1)R/(2m{r}^2)$, and the remaining term is the attractive potential energy. The right side is the energy eigenvalue times the radial wavefunction. Here the stated $m$ neglects nuclear recoil; including it replaces the electron mass by the reduced mass.

For $\ell =0$, substitution of the ground-state exponential givesIts $1/r$ coefficient must vanish and its constant coefficient determines the energy, soHere $a_0={\hslash }^2/(m\kappa )$ is the Bohr radius.

For $R_2=N_2(r+b)e^{-\beta r}$, direct differentiation yieldsCancel the nonzero exponential in the radial Schrödinger equation and compare coefficients of $r$, the constant term, and $1/r$. They give, respectively,ConsequentlyThe positive decay constants ensure normalizability; the second radial wavefunction has one positive-radius node at $2a_0$, as appropriate to the first radial excitation.

If the entire energy gap is carried by one photon, neglecting atomic recoil, conservation of energy gives the requested frequencyThere is a physical qualification: these are the $2s$ and $1s$ states, so the usual single-photon electric-dipole transition is forbidden by the electric-dipole selection rules for hydrogen. Explicitly, both angular factors are $Y_0^0$, and $\int |Y_0^0{|}^2\hat{\mathbf{r}}d\Omega =0$, so every component of $⟨1s|\mathbf{r}|2s⟩$ vanishes. The leading isolated-atom decay is instead two-photon emission, for which ${\nu }_1+{\nu }_2={\nu }_{\mathrm{gap}}$ and individual frequencies are not fixed. This metastability and the two-photon channel are documented in NIST's hydrogen spectral-data compilation. Thus the boxed answer is the energy-gap frequency under the printed single-photon assumption, rather than an assertion of an allowed electric-dipole line.

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.

Inside the infinite square well, the time-independent Schrödinger equation is $-{\hslash }^2{u}^{\prime \prime }/(2m)=Eu$ with $u(-a)=u(a)=0$. For $E<0$ the exponential solutions cannot satisfy both conditions except trivially, and for $E=0$ neither can a linear function. For $E>0$ put $k=\sqrt{2mE}/\hslash$. The first boundary condition selects $u=C\sin (k(x+a))$, and the second gives $\sin (2ak)=0$. Thuson $[-a,a]$, with $u_n=0$ outside. Their squared integrals are one, and the sine orthogonality identity gives an orthonormal set. The sine Fourier basis on the interval is complete: odd extension to an interval of length $4a$ reduces this to completeness of the Fourier series basis. Hence these are a complete set of normalized energy eigenstates.

Choosing the harmless sign of the second energy eigenstate so that $v_1=a^{-1/2}\cos (\pi x/(2a))$ and $v_2=a^{-1/2}\sin (\pi x/a)$, the specified wave function is $v_1/\sqrt{5}+2v_2/\sqrt{5}$. The Born rule therefore giveswith no other possible measured energies. The two energy eigenvalues are $E_1={\hslash }^2{\pi }^2/(8m{a}^2)$ and $E_2=4E_1$.

After the first measurement gives the ground state, the old state is $v_1$. Doubling the width puts the walls at $\pm 2a$, and the new normalized ground state is $w_1=(2a{)}^{-1/2}\cos (\pi x/(4a))$. The stipulated unchanged state is zero outside the original interval, so the probability amplitude isSquaring givesThe ground-state overlap after sudden expansion of a square well stays unchanged during subsequent evolution in the fixed new well, because the energy-basis coefficients acquire only phases. The answer is conditional on the already observed old ground state; its earlier probability $1/5$ is not another factor in this conditional question.

The canonical commutation relations are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hslash {\delta }_{ij}$. Substitution into the definitions of the creation and annihilation operators gives

The normalized ground state is defined byRepeated use of $[A_i,A_i^{†}]=1$ gives the Cartesian number state of the three-dimensional isotropic harmonic oscillatorwith energy eigenvalue

Inverting the ladder-operator definitions yieldsThe antisymmetry of the Levi-Civita symbol then cancels the two-creation and two-annihilation terms in $\mathbf{L}=\mathbf{X}\times \mathbf{P}$, leaving the orbital angular momentum in oscillator ladder operatorsWrite $|j⟩=A_j^{†}|0⟩$. ThenSince the orbital angular momentum eigenvalue is $\ell (\ell +1){\hslash }^2$, every first-excited state hasFinally,Therefore the normalized magnetic quantum number $m=+1$ state can be chosen aswhich obeys $L_z|\psi ⟩=\hslash |\psi ⟩$ and $\Vert |\psi ⟩\Vert =1$.

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.

A boson is an identical particle whose total multiparticle quantum state is symmetric under exchange of any two particles, whereas a fermion has a total state that is antisymmetric under every exchange. The Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions.

For three distinguishable spin-one particles, the spin Hilbert space has dimensionLetbe the total angular momentum operator. Since each particle haswe haveThe Hamiltonian operator is consequentlyOn the total-spin sector with quantum number $J$, its energy eigenvalue is

The Clebsch-Gordan decomposition may be performed by first coupling particles $1$ and $2$. Their intermediate spin is $j_{12}=0,1,2$, and coupling the third spin givesThus the three spin-one angular-momentum decomposition contains total spin $J=0,1,2,3$ with multiplicities $1,3,2,1$. Multiplying each multiplicity by the multiplet dimension $2J+1$ givesThe degeneracies sum to $27$, as required.

When the particles are indistinguishable, their integer spin makes them bosons. Their common spatial wavefunction is symmetric, so their spin wavefunction must also lie in the symmetric three-spin-one subspace. Its decomposition isHence only the $J=3$ and $J=1$ levels remain, now with one copy of each multiplet: