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Put $\gamma =2m\hslash \omega$. If ${\psi }_n$ is a normalized energy eigenstate at $E_n$, the factorization from part (a) givesIt follows inductively that $v_n=(a_{+}{)}^n{\psi }_0$ has squared norm ${\gamma }^{n}n!$. The commutator with $H$ establishes its energy eigenvalue at each step. Henceis normalized and has $E_n=(n+\frac{1}{2})\hslash \omega$. Distinct levels are orthogonal because $H$ is a self-adjoint operator. In particular,

To decompose the specified initial wavefunction, retain this exact phase convention for the creation and annihilation operators. Their difference givesSince $a_{-}{\psi }_0=0$ and $a_{-}{a}_{+}{\psi }_0=\gamma {\psi }_0$,This quadratic excitation of an oscillator ground state, $\Psi (x,0)=C_0^{-1}{X}^2{\psi }_0$, contains only the $n=0$ and $n=2$ energy eigenstates. The printed initial function has an unspecified overall normalization, so the Born rule requires division by its squared norm. After wavefunction normalization the state isThe minus sign is consistent with the paper's convention: its $a_{+}$ is $i$ times the usual real-coordinate raising operator multiplied by $\sqrt{\gamma }$, so its ${\psi }_2$ has the opposite sign to the usual real second harmonic oscillator eigenfunction. The energy measurement probabilities are independent of this choice of phase.

Therefore the complete answer isEvery other energy has probability zero. At any later time the two coefficients only acquire their separate dynamical phases, so these probabilities remain the same. A direct check of the normalization isThis also confirms the factor of three in the squared norm of the two-component expansion.

The coefficient of the momentum eigenstate ${\psi }_2$ is $e^{-i\Omega t}/\sqrt{6}$. The Born rule therefore gives $\mathrm{Pr}(P=2\hslash )=1/6$. Conditional on this outcome, the projective measurement prepares the post-measurement state $e^{-i\Omega t}{\psi }_2$, equivalent to ${\psi }_2$ up to global phase.

Since $H{\psi }_2=(4{\hslash }^2/(2m)){\psi }_2$, the subsequent energy measurement has conditional probability one of giving the specified energy eigenvalue. The probability for the entire sequence is consequentlyAny intervening unitary time evolution under the same Hamiltonian operator preserves this energy eigenstate, up to global phase.

The ground state is the single energy eigenstate ${\psi }_0$. Its coefficient in the normalized wavefunction is $\sqrt{2/3}$, so the Born rule givesThis probability is independent of $t$ because unitary time evolution changes only the dynamical phases of the energy eigenstates.

In the ordered orthonormal basis $({\psi }_1,{\psi }_2)$, the Hamiltonian operator has matrixwhere ${\sigma }_1$ is a Pauli matrix. Its symmetric and antisymmetric eigenvectors giveThe energy eigenstates ${\chi }_{\pm }$ are normalized and orthogonal; each energy eigenvalue is nondegenerate. Their full stationary states are $e^{-i\omega t}{\chi }_{+}$ and $e^{i\omega t}{\chi }_{-}$. Since they form an orthonormal basis, the general solution to the Time-dependent Schrödinger equation isFor a normalized quantum state, $|c_{+}{|}^2+|c_{-}{|}^2=1$; the constants are determined by its initial inner products with ${\chi }_{\pm }$.

Under unitary time evolution, each energy eigenstate acquires its own dynamical phase. Since $E_2=E_{-2}=E_{\star }$, the wavefunction isAt $t=0$, the identity $1-\cos (2x/R)=2{\sin }^2(x/R)$ recovers the normalized initial wavefunction.

Using ${\sin }^{2}u=\frac{1}{2}-\frac{1}{4}{e}^{2iu}-\frac{1}{4}{e}^{-2iu}$, the energy eigenstate decomposition isOnly the energy eigenvalues zero and $E_{\star }=2{\hslash }^2/(m{R}^2)$ occur. The last two terms are distinct momentum eigenstates in the same energy eigenspace.

The Hamiltonian operator separates into two commuting one-dimensional quantum harmonic oscillators. WriteProducts of harmonic oscillator eigenfunctions giveHere $H_n$ is the physicists' Hermite polynomial, so $H_0=1$, $H_1(u)=2u$ and $H_2(u)=4u^2-2$. The first six energy eigenstates, ordered with multiplicity, areFor this two-dimensional oscillator level ordering, the excitation energies obeyEvery state with $n_1+n_2\ge 3$ costs at least $3\hslash {\omega }_1$, so none has been missed. Equalities within the list occur when ${\omega }_2={\omega }_1=\omega$. Then the two-dimensional isotropic harmonic oscillator hasThe quantum degeneracy counts $n_1=0,\dots ,N$ with $n_2=N-n_1$. The six listed states occupy three distinct energy levels, of quantum degeneracies one, two and three.

Use the canonical commutation relationsWriting $c_j=\sqrt{m{\omega }_j/(2\hslash )}$, $d_j=1/\sqrt{2m\hslash {\omega }_j}$, direct expansion of the commutator givesThe Fock vacuum ${\psi }_{0,0}$ is annihilated by both $a_{j,-}$. Successive creation operators generateHence, in the order of part (a), the six energy eigenstates are proportional toIf the Fock vacuum has unit norm, dividing the general expression by $\sqrt{n_1!n_2!}$ gives normalized number states. Normalization is not required here, but using this convention makes the later change of basis unambiguous.

The orbital angular momentum $L_3=X_1{P}_2-X_2{P}_1$ generates rotations in the plane. The canonical commutation relations giveUse the commutator product rule $[A,B^2]=[A,B]B+B[A,B]$. Since different coordinates commute,Thus, for arbitrary positive angular frequencies,At ${\omega }_1={\omega }_2$, this becomes $\boxed{[L_3,H]=0}$. The rotational symmetry of the two-dimensional isotropic harmonic oscillator allows a basis of simultaneous energy eigenstates and orbital angular momentum eigenstates.

Choose the first three energy eigenstates consistently as ${\psi }_{0,0}$, ${\psi }_{1,0}=a_{1,+}{\psi }_{0,0}$ and ${\psi }_{0,1}=a_{2,+}{\psi }_{0,0}$. The Fock vacuum is annihilated by both lowering operators, soFor example $a_{1,-}{a}_{1,+}{\psi }_{0,0}={\psi }_{0,0}$, while $a_{2,-}{\psi }_{1,0}=0$, giving the middle formula. Thus the simultaneous eigenstates can be chosen asTheir pairs of eigenvalues $(H,L_3)$ are respectively $(\hslash \omega ,0)$, $(2\hslash \omega ,+\hslash )$ and $(2\hslash \omega ,-\hslash )$. The factor $1/\sqrt{2}$ normalizes the combinations when the original three states are normalized. Mixing only inside the degenerate first-excited energy eigenspace preserves the energy eigenvalue while diagonalizing $L_3$.

Inside the infinite square well, $V=0$, and the time-independent Schrödinger equation isFor $E>0$, write $k^2=2mE/{\hslash }^2$ and $\psi =A\sin kx+B\cos kx$. The first Dirichlet boundary condition gives $B=0$, and the second requires $kL=n\pi$ for a nonzero state. The case $n=0$ is identically zero; negative $n$ only changes an overall quantum phase. For $E=0$ a linear solution satisfying both walls is zero, and for $E<0$ the hyperbolic solutions likewise cannot satisfy both walls nontrivially.

Wavefunction normalization, ${\int }_0^L|\psi {|}^{2}dx=1$, fixes $|A|=\sqrt{2/L}$. The orthonormal energy eigenstates and energy eigenvalues are thereforeEach state vanishes outside the well. Its stationary time-dependent form is ${\Psi }_n(x,t)={\psi }_n(x)e^{-i{E}_{n}t/\hslash }$. General states are normalized superpositions of these modes, rather than additional single energy eigenstates.

The two probabilities concern different quantum measurements, so there is no requirement that they match or that one bound the other. The spatial interval $[0,L/3]$ contains little of the state: its probability density is smallest near zero and increases towards $L$. By contrast, the positive ground state extends across the whole well. It has substantial same-sign overlap with the positive linear wavefunction, with no cancellation between lobes in the overlap integral. The right panel of the figure in solution illustrates these two amplitudes and marks the small position interval.

Consequently a large global ground-state probability can coexist with a small probability of finding the particle in the left third. This is not because the ground state is spatially confined to that third; a ground-state particle can be found throughout the well. In operator language the two events are projections onto a spatial subspace and onto one energy eigenstate, respectively.

Replacing $L$ by the angular momentum operator gives the Hamiltonian operatorThe time-dependent Schrödinger equation is thereforeFor a stationary state, separate variables as $\Psi (\varphi ,t)=\psi (\varphi )T(t)$. Dividing by the product separates a function of $t$ from one of $\varphi$:Thus $T=e^{-iEt/\hslash }$ up to a constant, and the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2I)=E\psi$.

Diagonalize the commuting angular momentum operator at the same time: $-i\hslash {\psi }^{\prime }=\ell \psi$ has solutions $A{e}^{i\ell \varphi /\hslash }$. A single-valued wavefunction requires $e^{2\pi i\ell /\hslash }=1$, hence $\ell =n\hslash$ with $n\in \mathbb{Z}$. Normalizing with the angular probability measure $d\varphi$ givesThese form an orthonormal basis under ${\int }_0^{2\pi }{\psi }_m^{\ast }{\psi }_{n}d\varphi ={\delta }_{mn}$. The periodic boundary conditions apply to both the wavefunction and its first derivative, ensuring the physical self-adjoint Hamiltonian operator. For probability per arc length $ds=rd\varphi$, the normalization instead reads ${\psi }_n=e^{in\varphi }/\sqrt{2\pi r}$; the two conventions describe the same probabilities. General states are superpositions of these energy eigenstates.

The emitted photon energy is the difference of the two energy levels:Using $E_{\gamma }=hc/\lambda$ givesNo numerical mass or radius is supplied, so this is the determinate result. It is the wavelength conditional on the transition occurring. An isolated particle on a circle with no coupling would remain in its energy eigenstate; a charged particle coupled to electromagnetic radiation can emit subject to its transition selection rules. The $3\to 2$ change has $\Delta n=-1$, compatible with an electric-dipole angular factor for a charged ring particle.

Each energy eigenstate acquires its own phase under the time-independent Hamiltonian operator:Orthonormality removes all off-diagonal terms from the expected value of the energy, givingReturn up to an overall phase requires the relative phase to return to one. For $E_1\ne E_2$, the shortest positive recurrence time isIf the two energies coincide, the whole superposition is stationary up to a common phase for every time; there is then no smallest strictly positive recurrence time. A half-period changes the relative sign and produces the orthogonal antisymmetric state rather than the original symmetric state.

The one-dimensional Time-dependent Schrödinger equation isThe left side generates time evolution; the two terms on the right are the kinetic and potential parts of the Hamiltonian operator. For a time-independent potential and an energy eigenstate, set $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$. Cancelling the time factor gives the time-independent Schrödinger equation, $-{\hslash }^2{\varphi }^{\prime \prime }/(2m)+V\varphi =E\varphi$.

By the Born rule, $|\psi {|}^{2}dx$ is the position probability in an interval of width $dx$, with $\int |\psi {|}^{2}dx=1$ for a normalized particle. Therefore $\boxed{[\psi ]={\mathrm{length}}^{-1/2}}$ in one dimension. An ideal scattering plane wave instead uses flux normalization and is not itself a normalizable single-particle state on the whole line.

An initially occupied, isolated instantaneous energy eigenstate follows its changing eigenspace, acquiring a dynamical phase and generally a Berry phase. For smooth nondegenerate evolution, a useful local smallness condition isA nonvanishing spectral gap and suitably smooth, slow driving throughout the evolution are needed; the local inequality alone is not a universal long-time error bound. Gap closure invalidates the simple nondegenerate approximation. A separated degenerate eigenspace may mix internally even when transitions out of it are suppressed.

Complete the square in the potential energy. With $x_F=F/(M{\omega }^2)$,The quantum harmonic oscillator is translated without changing its level spacing. If ${\phi }_n(x)$ is its normalized unforced energy eigenstate, thenEquivalently, $|n;F⟩=e^{-i{x}_{F}p/\hslash }|n⟩$. The ground state position is centered at $x_F$, and its position uncertainty remains $\sqrt{\hslash /(2M\omega )}$.

For a time-independent Hamiltonian operator, the time-evolution operator is $U(T)=e^{-iHT/\hslash }$, with $T=t-t^{\prime }$. Insert its resolution of identity in normalized energy eigenstates:Since $⟨x|\alpha ⟩={\phi }_{\alpha }(x)$, the retarded quantum propagator in the question isThe Heaviside step function imposes the chosen retarded convention. If the Hamiltonian operator has a continuous spectrum, the spectral sum includes its continuum integral; it must not be restricted to bound states. The spectral resolution of a quantum propagator also shows that $K$ tends to $\delta (x-x^{\prime })$ as $T\downarrow 0$. With the step factor it obeys $(i\hslash {\partial }_t-H)K=i\hslash \delta (T)\delta (x-x^{\prime })$; the corresponding Green function with a unit delta source is $K/(i\hslash )$.

Define the signed dimensionless displacementThe spatial translation operator is $T(x_0)=e^{-ip{x}_0/\hslash }$, and the phases of the new energy eigenstates are fixed by $|n^{\prime }⟩=T(x_0)|n⟩$. Their expansion coefficients are thereforeSince $p=i\sqrt{m\hslash \omega /2}(a^{†}-a)$,Set $A=-\lambda a^{†}$ and $B=\lambda a$. Their commutator is $[A,B]={\lambda }^{2}I$, which commutes with both operators. Thus the stated special case of the Baker--Campbell--Hausdorff formula is applicable and givesBecause $a|0⟩=0$,Taking the overlaps with $⟨n|$ therefore provesThis is the coherent state after shifting an oscillator equilibrium. Its normalization follows from $e^{-{\lambda }^2}{\sum }_n{\lambda }^{2n}/n!=1$.

There is a sign qualification in the printed expansion: $(m\omega x_0^2/(2\hslash ){)}^{n/2}=|\lambda {|}^n$ agrees with $(-\lambda {)}^n$ when $x_0\le 0$, equivalently $\gamma \ge 0$. For a general signed displacement $x_0>0$, the odd coefficients have the opposite sign in the specified translated basis. The squared displacement alone fixes the occupation probabilities, not these coherent relative phases. The two-exponential identity also requires the central-commutator condition used above; it is not valid in that simple form for arbitrary operators.