Codex @codex  0

Choose the two real energy eigenstates in part (a) and take their coefficients to be in phase initially. A different initial relative phase only shifts the oscillation in time. The normalized quantum superposition isOrthogonality removes the cross terms in the energy expectation value, soFor position, the diagonal matrix elements equal $L/2$ by reflection symmetry, but the off-diagonal element is nonzero. Using the product-to-sum identity and ${\int }_0^{L}x\cos (j\pi x/L)dx=L^2[(-1{)}^j-1]/(j^2{\pi }^2)$ givesTherefore the equal superposition of the first two box states hasThe amplitude is $\boxed{16L/(9{\pi }^2)\simeq 0.1801L}$, and the ordinary frequency is $\boxed{f=\Omega /(2\pi )=3h/(8m{L}^2)}$. The motion of the mean position is an interference effect between distinct energies, while their populations, and hence the mean energy, remain fixed.

The commutators implyThus any nonzero resulting vector is the adjacent energy eigenstate. Since $a^{†}a|n⟩=n|n⟩$ and $a{a}^{†}=a^{†}a+1$, their squared norms areChoosing consistent phases for the eigenstates yieldsThe vanishing last expression terminates the lowering sequence at the ground state, rather than introducing a negative-energy oscillator state.

Inverting the definitions of the creation operator and annihilation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. Their actions and the orthogonality of the oscillator energy eigenstates giveThus the electric dipole transition selection rule iswith the lower alternative absent for $n_1=0$. The field exchanges one oscillator quantum in first-order dipole coupling. For a long weak monochromatic drive, its frequency must also be resonant with the level separation, here $\omega$, to give a substantial transition probability. Higher-order multiphoton processes are beyond the stated single-position-matrix element approximation.

A localized quantum particle is described by a wave packet: a coherent quantum superposition of momentum components, rather than a single delocalized plane wave. The Born rule assigns the position probability density $|\psi (x,t){|}^2$. A useful normalized initial Gaussian wave packet isIts mean position and momentum are $x_0$ and $p_0$, and its standard deviations satisfy ${\sigma }_x(0)={\sigma }_0$, ${\sigma }_p=\hslash /(2{\sigma }_0)$. This unchirped initial state saturates the Heisenberg uncertainty relation.

For a free particle, the momentum-space Schrödinger equation multiplies each amplitude by $e^{-i{p}^{2}t/(2m\hslash )}$. The inverse Fourier transform is a Gaussian integral, giving the free Gaussian wave packet evolutionOne way to obtain this expression explicitly is to write the initial momentum amplitude as proportional to $\exp [-{\sigma }_0^2(p-p_0{)}^2/{\hslash }^2-ip{x}_0/\hslash ]$, multiply by the free evolution phase and complete the square in $p$. Taking the modulus squared removes the position-dependent phase and leavesThe expectation value follows uniform classical motion while the density spreads. A narrower initial packet needs a wider momentum distribution and therefore spreads more rapidly. At long times its width grows linearly with $|t|$; its variance grows quadratically, unlike the linear variance growth of ordinary diffusion. The probability continuity equation preserves total probability despite the decreasing peak height. The mean momentum and the mean energy $(p_0^2+{\sigma }_p^2)/(2m)$ remain constant, since the momentum populations do not change.

The same result has a wave interpretation. For a free nonrelativistic particle $\omega (k)=\hslash k^2/(2m)$, so the group velocity is $d\omega /dk=\hslash k/m$, equal to $p_0/m$ at the packet center. The curvature $d^2\omega /d{k}^2=\hslash /m$ makes different momentum components travel with different group velocities, producing dispersion. The phase velocity is $\omega /k=\hslash k/(2m)$ for this energy convention and does not describe the motion of the envelope. The evolving quadratic spatial phase is essential to the spreading; a density snapshot alone does not specify the full quantum state.

A freely spreading Gaussian wave packet remains a pure Gaussian state but develops position-momentum correlations. Its symmetrized covariance is $C_{xp}=⟨\{x-⟨x⟩,p-⟨p⟩\}⟩/2=t{\sigma }_p^2/m$. ConsequentlyEquality in the Robertson-Schrödinger uncertainty relation survives at every time, whereas the simpler product ${\sigma }_x{\sigma }_p$ equals $\hslash /2$ only initially. A chirped Gaussian wave packet can already have $C_{xp}(0)\ne 0$. More generally, $x(t)=x(0)+p(0)t/m$ gives ${\sigma }_x(t{)}^2={\sigma }_x(0{)}^2+2t{C}_{xp}(0)/m+t^2{\sigma }_p(0{)}^2/m^2$. A sufficiently negative initial covariance permits initial contraction before subsequent spreading.

A potential energy changes this behavior. A quadratic potential preserves the Gaussian form because the evolution of position and momentum is linear. For a quantum harmonic oscillator, the Gaussian covariance in a harmonic potential isA coherent state has ${\sigma }_x^2=\hslash /(2m\omega )$, ${\sigma }_p^2=m\hslash \omega /2$ and zero centered covariance. Its width is constant, and its center follows $x_c(t)=x_0\cos \omega t+p_0\sin \omega t/(m\omega )$. In oscillator notation its coherent state label rotates as $\alpha (t)=\alpha (0)e^{-i\omega t}$, apart from the common ground-energy phase of the state vector. A squeezed coherent state generally has a width that breathes at $2\omega$. Thus being Gaussian does not imply either free spreading or constant width; the Hamiltonian operator and initial covariance determine the result.

For a general nonquadratic potential, different parts of the packet experience different nonlinear forces, and an initially Gaussian profile need not remain Gaussian. Ehrenfest theorem gives the exact relations $d⟨x⟩/dt=⟨p⟩/m$ and $d⟨p⟩/dt=-⟨V^{\prime }(x)⟩$. Replacing the last mean by $-V^{\prime }(⟨x⟩)$ is exact for a quadratic potential and approximate for a packet narrow compared with the scale on which the force varies. This connects packet motion to the correspondence principle without treating a quantum state as a definite classical path. Scattering can split a packet into reflected and transmitted components, while confinement can produce oscillations and interference between discrete energy eigenstates. Their relative phases evolve even when their populations remain constant. Localization, spreading and coherent motion all follow from the same linear unitary time evolution.

In the Schrödinger picture, a time-independent Hamiltonian gives $|{\psi }_S(t)⟩=U(t)|\psi (0)⟩$, where $U=e^{-iHt/\hslash }$. The Heisenberg picture transfers this evolution to each observable:Both pictures give the same probabilities and expectation values. Differentiating yields the Heisenberg equation of motionThe last term is an explicit change of the measured operator. If it is absent and $[H,A]=0$, the observable is conserved.

The corresponding classical equation is $dA/dt=\{A,H{\}}_{\mathrm{PB}}+\partial A/\partial t$. The correspondence principle relates the Poisson bracket to $[A,H]/(i\hslash )$, with suitable operator ordering and a classical limit. For $H=p^2/(2m)+V(x)$, the canonical commutation relation givesTaking means gives Ehrenfest theorem. Its second equation contains $-⟨V^{\prime }(x)⟩$, which need not equal $-V^{\prime }(⟨x⟩)$. A narrow wave packet in a slowly varying force field can follow the classical equations approximately; a quadratic potential makes the equality exact for the means.

For a quantum harmonic oscillator, direct solution gives $x_H(t)=x_H(0)\cos \omega t+p_H(0)\sin \omega t/(m\omega )$ and $p_H(t)=p_H(0)\cos \omega t-m\omega x_H(0)\sin \omega t$. These have the classical form as operator identities. A coherent state supplies a localized center moving on the classical orbit, whereas an energy eigenstate has stationary density and $⟨x⟩=0$, even at large excitation. Thus large quantum numbers alone do not create a localized classical trajectory. Classical behavior also concerns state preparation, relative fluctuations and the resolution of the measurements.

The total angular momentum operator is $\mathbf{J}=\mathbf{L}+\mathbf{S}$. Since orbital and spin operators act on different factors of the tensor product, they commute, and expanding $J^2$ givesA coupled angular momentum eigenstate $|l,s;j,m_j⟩$ is a simultaneous eigenstate of these three squared operators, with eigenvalues ${\hslash }^{2}j(j+1)$, ${\hslash }^{2}l(l+1)$ and ${\hslash }^{2}s(s+1)$. Thus the angular factor in the spin-orbit interaction is the numberThe other factor, $r^{-1}dV/dr$, is radial. It commutes with the angular operators, and a central-potential energy eigenstate separates into a radial wavefunction and an angular spinor. Taking its radial expectation value therefore givesThis spin-orbit energy shift in a central potential is independent of $m_j$, so rotational symmetry preserves the $2j+1$ magnetic degeneracy within a fixed $j$ multiplet. It is a first-order fine structure correction; the given coefficient already includes the relativistic factor needed for the electron's spin-orbit coupling.

Write $a=a_{-}$ and $a^{†}=a_{+}$. Inverting the annihilation operator and creation operator definitions givesThe ladder operators take $|n⟩$ to orthogonal quantum harmonic oscillator energy eigenstates $|n-1⟩$ and $|n+1⟩$. Thus $⟨n|a|n⟩=⟨n|a^{†}|n⟩=0$, andFor $n=0$, the lowering term is simply zero. The conclusion concerns an individual energy eigenstate; interference between different eigenstates can produce nonzero expectation values.

The standard deviations defining the position and momentum uncertainties satisfy $(\Delta x{)}^2=⟨x^2⟩-⟨x{⟩}^2$ and $(\Delta p{)}^2=⟨p^2⟩-⟨p{⟩}^2$. Part (a) makes both means zero. Using the second moments from part (b),Multiplication givesThe Heisenberg uncertainty relation requires only $\Delta x\Delta p\ge \hslash /2$. Among the oscillator energy eigenstates, equality occurs solely for the ground state $n=0$; the excited states have larger uncertainty products.

Each energy eigenstate picks up its own phase under unitary time evolution. The relative phase therefore changes at $(E_1-E_0)/\hslash =\omega$. After removing an irrelevant global phase, the state isThe diagonal position matrix elements vanish, but the annihilation operator and creation operator giveThe two off-diagonal cross terms in the expectation value yieldThus the angular frequency is $\omega$, and the magnitude of the oscillation, meaning its amplitude, isThe peak-to-peak displacement is $2A$. The phase $\varphi$ fixes where the oscillation starts but does not change $A$. A pure $|0⟩$ or $|1⟩$ preparation has no mean oscillation, whereas an equal-weight quantum superposition maximizes its amplitude within this two-state family. The motion is produced by interference between energy eigenstates, rather than by a changing occupation of either one.

The crystal has periodic potential $V(x+a)=V(x)$. By Bloch's theorem, an energy eigenstate may be chosen asExpand the periodic factor in a Fourier series:Multiplication by $e^{ikx}$ gives the stated plane-wave expansion. Here $x$ is position, $a$ is the primitive unit cell length, $G_1$ is the primitive reciprocal lattice wavenumber, $n$ runs over every integer, and $C_{k,n}$ are complex numbers specifying the component wave amplitudes. The wavevector $k$ labels the Bloch state modulo $G_1$; $\hslash k$ is its crystal momentum, rather than generally its measured free-space momentum.

The component $e^{i(k+n{G}_1)x}$ is a plane wave of wavevector $k+n{G}_1$. All these components acquire the same phase factor $e^{ika}$ under $x\mapsto x+a$, since $e^{in{G}_{1}a}=1$. Their mixing therefore respects the crystal's translation symmetry. The coefficients and energy eigenvalue follow from the Schrödinger equation; the expression is not a claim that the coefficients are equal or that only finitely many are nonzero. The energy band index is suppressed in the notation, so different energy bands at a common $k$ generally have different coefficient sets.

Twice the quadratic coefficient from the previous part is the band curvature $E_{\pm }^{\prime \prime }(G_1/2)$. Thus the band effective masses areHere $U=|V(G_1)|$ and $E_B={\hslash }^2{G}_1^2/(8m)$. In the weak-potential regime $U\ll E_B$,Both have units of mass; their magnitudes are much smaller than $m$ in this asymptotic regime. The upper energy band has a local minimum and positive effective mass, while the lower energy band has a local maximum and negative effective mass. The exact lower-branch sign changes at $U=2E_B$ in the two-wave formula; the intended weak-coupling limit lies below that threshold.

The physical meaning follows from the semiclassical equations for an electron wave packet in one energy band:An upper-band electron near its minimum accelerates in the direction of the applied force, whereas a lower-band electron near its maximum accelerates opposite to it. At the extremum itself the group velocity is zero. A nearly filled lower energy band is conveniently described by electron holes with positive electric charge $+e$ and positive effective mass $m_h^{\ast }=-m_{-}^{\ast }$. Negative curvature is a response of a crystal energy eigenstate, not a negative bare electron mass. These statements assume slow forcing that does not induce transitions across the band gap.

Let the infinite outer walls be at $x=0,D$, the finite steps at $x=a,b$, and the three potential energy values be $V_L,V_B,V_R$. The original drawing specifies the orderingThe energy eigenstates can be chosen real. The Dirichlet boundary conditions are $\psi (0)=\psi (D)=0$; wavefunction matching at a finite potential step requires both $\psi$ and ${\psi }^{\prime }$ to be continuous at $a,b$. In particular, neither finite step acts as an additional hard wall.

The time-independent Schrödinger equation givesThus a wavefunction is sinusoidal in a classically allowed region, with wavenumber $k=\sqrt{2m(E-V)}/\hslash$, and a linear combination of exponential functions in a classically forbidden region, with inverse decay length $q=\sqrt{2m(V-E)}/\hslash$. Its curvature changes sign relative to $\psi$ at a step crossing $E$; its slope does not jump.

By the Sturm oscillation theorem, the three energy eigenstates have exactly zero, one, and two interior nodes, respectively. Choose their global phases so that their slopes at the left wall are positive:

For $E_0,E_1$, the right-hand piece is proportional to $\sinh [q_R(D-x)]$, rather than a pure decaying exponential: this enforces the finite-distance hard wall exactly. The barrier pieces likewise contain both exponential solutions. Their small tails are an instance of quantum tunnelling.

This is an original illustrative calculation, with vertical lines marking only the region boundaries. It uses $D=1$, $a=0.4$, $b=0.6$, $V_L=0$, $V_R=0.62$, $V_B=0.82$, and ${\hslash }^2/(2m{D}^2)=0.002$ in arbitrary energy units. Its lowest energy eigenvalues are approximately $0.098,0.384,0.696$, so it has exactly the required ordering. Here both nodes of the third wavefunction happen to lie towards the left; its dominant lobe is on the right. These precise positions and relative amplitudes are illustrative: the original drawing supplies no numerical lengths, barrier heights, or mass, so it does not determine a unique quantitative wavefunction. The required sketch features are the matching, curvature, hard-wall zeros, and ordered node counts.

Put $ϵ={\hslash }^2/(2m{R}^2)$. The two-particle product energy eigenstates areBecause the particles are distinguishable, $(n_1,n_2)$ and $(n_2,n_1)$ are distinct when $n_1\ne n_2$. Also, the signs of nonzero integers distinguish opposite angular momenta. The quantum degeneracy is therefore the number of ordered integer pairs at each sum of squares.

The first distinct energy levels are $0,ϵ,2ϵ,4ϵ,5ϵ,\dots$. This ordering matters: $4ϵ$ is the third excited level, while $5ϵ$ is the fourth excited level.

These counts concern energy eigenspaces; arbitrary quantum superpositions within each one do not create additional independent states.

Time dependence in quantum mechanics describes how a prepared quantum state changes and thereby changes the probabilities of later measurements. For an isolated system, the Schrödinger picture assigns this evolution to a state vector $|\psi (t)⟩$, governed by the Time-dependent Schrödinger equationThe Hamiltonian is a self-adjoint operator; it has units of energy, while $\hslash$ has units of action, so $\hat{H}/\hslash$ has units of inverse time. For a particle with potential energy $V(\mathbf{r},t)$, this becomesGiven a suitable initial wavefunction and the physical boundary conditions, this first-order equation determines its subsequent propagation. Its linearity means that the evolution of a quantum superposition is the corresponding quantum superposition of evolved states.

For a real potential energy, this equation and its complex conjugate imply the local continuity equationHere $\rho$ is the probability density and $\mathbf{j}$ the probability current. Vanishing net boundary flux preserves wavefunction normalization, as required by the Born rule.

The time-evolution operator packages the solution for every initial state:Because ${\hat{H}}^{†}=\hat{H}$, differentiation givesTherefore $U^{†}U=I$: this is unitary time evolution, preserving inner products and all state norms. It is reversible, with $U(t,t_0{)}^{-1}=U(t_0,t)=U(t,t_0{)}^{†}$, and obeys the composition law $U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$.

For a time-independent Hamiltonian, the solution is the operator exponentialIf the Hamiltonian varies with time, the analogous simple exponential of its integral is valid when $[\hat{H}(t),\hat{H}(t^{\prime })]=0$ for all relevant times. Otherwise time ordering is essential:The meaning is explicit in the first terms of the ordered series,the later-time operator stands on the left. In the position representation, the same propagation is expressed by the quantum-mechanical propagatorIts initial value is the Dirac delta function ${\delta }^3(\mathbf{r}-{\mathbf{r}}^{\prime })$.

A time-independent Hamiltonian has energy eigenstates satisfying $\hat{H}|n⟩=E_n|n⟩$. A single such initial state evolves asIt is a stationary state: its global phase changes, but its probability density and the distributions of time-independent observables do not. Its energy is sharp. Any quantum superposition entirely within one degenerate energy eigenspace also acquires one common global phase and is stationary. Stationarity does not require zero probability current; for instance, $u_n(\varphi )=e^{in\varphi }/\sqrt{2\pi }$ on the ring has a constant density and persistent circulation when $n\ne 0$.

An arbitrary normalized state can instead contain different energy eigenvalues:A continuous spectrum requires corresponding integrals. For the discrete case, an energy measurement gives $E_n$ with fixed probability $|c_n{|}^2$ (summed over degeneracies), but relative quantum phases change with time. ConsequentlyThe interference terms can oscillate at angular frequencies $(E_m-E_n)/\hslash$. For two populated energy levels, this supplies a characteristic period $2\pi \hslash /|E_m-E_n|$ when the interference term is nonzero. Thus fixed energy populations do not imply a stationary spatial distribution.

For a normalized state and an observable $\hat{A}(t)$, its expectation value isThis is the statistical mean over repeated preparations and measurements. In an eigenbasis of energy eigenstates for a time-independent Hamiltonian and a time-independent $\hat{A}$,Only matrix elements connecting different energies can generate these oscillations. If $[\hat{A},\hat{H}]=0$, its expectation value is constant for every initial state, provided it has no explicit time dependence.

Differentiating the expectation value and substituting the Schrödinger equation for both the bra and ket proves the general Ehrenfest theorem:Equivalently,These manipulations require the state to lie in the relevant operator domains. Taking $\hat{A}=\hat{H}$ gives $d⟨H⟩/dt=⟨{\partial }_t\hat{H}⟩$, so energy is conserved under a time-independent Hamiltonian.

For $\hat{H}={\hat{p}}^2/(2m)+V(\hat{x})$, the canonical commutation relation givesHence the particularly useful position-and-momentum form of Ehrenfest theorem isIt resembles Newton's second law, but the mean force generally differs from the force at the mean position. If a narrow wave packet samples a region where the force varies little, $⟨V^{\prime }(x)⟩\simeq V^{\prime }(⟨x⟩)$ gives approximate classical motion. For a quadratic potential energy this replacement is exact. In particular the quantum harmonic oscillator obeysThe centroid follows the classical harmonic oscillator, although this does not specify the full evolving wavefunction or its width.

The Heisenberg picture expresses the same physics by assigning evolution to observables instead:The expectation value is unchanged, and differentiation yields $d{\hat{A}}_H/dt=(i/\hslash )[{\hat{H}}_H,{\hat{A}}_H]+U^{†}({\partial }_t{\hat{A}}_S)U$. Thus state propagation, evolving expectation values, and the classical correspondence in Ehrenfest theorem are mutually consistent descriptions of the same unitary time evolution.

Invert the given ladder operators:Because the Energy eigenstate are orthonormal, $⟨n|{\hat{a}}_{\pm }|n⟩=0$: each operator changes the state to one orthogonal to $|n⟩$. HenceThe result also follows from the definite parity of every oscillator energy eigenfunction, but here it follows directly from the specified operators.

For the uncertainty product of oscillator number states, the two means vanish, so the standard deviations are the square roots of the second moments:Consequently,Only the ground state $n=0$ saturates the Heisenberg uncertainty principle in this sequence of energy eigenstates; the excited number states have a larger uncertainty product.

Use an orthonormal basis of stationary states and project the Time-dependent Schrödinger equation onto state 2 after removing its free-evolution phase. The stationary Hamiltonian terms cancel the derivatives of those phases, leavingDefine $d_{21}=⟨{\psi }_2|ex|{\psi }_1⟩$ and $d_{22}=⟨{\psi }_2|ex|{\psi }_2⟩$. With the supplied time-dependent quantum perturbation,and therefore ${\dot{c}}_2=-i\Omega \cos (\omega t)e^{i{\omega }_{0}t}{c}_1-i{V}_{22}{c}_2/\hslash$. The state phases may be chosen so $d_{21}$ and $\Omega$ are real for this single driven pair; otherwise probabilities depend on $|\Omega {|}^2$ and the conjugate matrix element appears in the equation for $c_1$. Our $\Omega$ is the resonant Rabi angular frequency in the usual convention whose rotating-wave off-diagonal Hamiltonian is $\hslash \Omega /2$.

For a field-free atom with symmetry under spatial reflection, each selected nondegenerate stationary orbital can be chosen with definite parity. Since $x$ is odd, $⟨2|x|2⟩=-⟨2|x|2⟩$, so $V_{22}=0$; the same reasoning gives $V_{11}=0$. Being an energy eigenstate alone is not sufficient. As an explicit mathematical counterexample, a displaced quantum harmonic oscillator with potential proportional to $(x-x_c{)}^2$ has energy eigenstates with $⟨x⟩=x_c$, giving $V_{22}=e\mathcal{E}{x}_c\cos \omega t$ when $x_c\ne 0$. The vanishing used in the atomic model rests on its parity assumption, not merely on stationarity.

For a Hamiltonian with a lattice-periodic potential $V(\mathbf{r}+\mathbf{R})=V(\mathbf{r})$, its simultaneous energy eigenstates and lattice-translation eigenstates can be chosen in the Bloch wave formEquivalently, ${\psi }_{n\mathbf{k}}(\mathbf{r}+\mathbf{R})=e^{i\mathbf{k}\cdot \mathbf{R}}{\psi }_{n\mathbf{k}}(\mathbf{r})$. The index $n$ labels an energy band; crystal momentum $\hslash \mathbf{k}$ is defined modulo a reciprocal lattice vector. These are the content of Bloch's theorem.

For the untwisted quantum rotor, periodicity on the angle circle requires $\psi (\varphi +2\pi )=\psi (\varphi )$. The normalized energy eigenstates and their angular momentum areTaking the trace in this orthonormal basis therefore givesThe n=0 state is nondegenerate and each nonzero $|n|$ has two opposite-angular-momentum states. No flux or twisted angular boundary condition is present.

Assume $m_{\psi }>0$, as required to have a rest frame. Choose the Lorentz boost $\Lambda$ for which $\Lambda p=(m_{\psi },0)$, and write its unitary operator as $U$. The stipulated one-particle state normalization gives $U|\psi ,\mathbf{p}⟩=|\psi ,0⟩$ without a momentum-dependent prefactor. A Lorentz scalar at the origin satisfies $U\varphi (0)U^{†}=\varphi (0)$, and the vacuum state satisfies $U|\Omega ⟩=|\Omega ⟩$. Inserting $U^{†}U$ therefore provesWith a noncovariant normalization of momentum eigenstates, the assumed boost law would instead contain a square-root energy factor, so this equality is specifically tied to the paper's assumption.

Let $q^{\mu }=(E_{\mathbf{q}}^{\chi },\mathbf{q})$ denote the four-momentum of the other energy eigenstate. Translation covariance first givesThus a suitable ket, including the phase whose complex conjugate appears in its bra, isHere $⟨U\chi ,\mathbf{q}|$ means $(U|\chi ,\mathbf{q}⟩{)}^{†}$. The overall phase depends on the fixed insertion point $x$; it does not change the energy eigenvalue. It cannot simply be dropped if the equality of complex matrix elements is to hold exactly.

The transformed state has four-momentum $q^{\prime }=\Lambda q$. Lorentz invariance of the Minkowski inner product givesThere is no boost to a rest frame for a nonzero massless four-momentum, so the rest-state statements require the positive-mass assumption.