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The result $16ϵ$ projects the state onto the corresponding nondegenerate energy eigenstate. Immediately after an ideal measurement,for an unobservable overall phase $\chi$. Subsequent isolated evolution adds $e^{-16iϵt/\hslash }$ but does not mix in other energy eigenstates. A repeat ideal energy measurement gives $16ϵ$ with probability one, provided the hamiltonian is unchanged and no intervening disturbance occurs.

Only $n=1,2,3$ have $E_n<10c$. The printed squared coefficients sum to $S=0.1^2+0.4^2+0.3^2+0.86^2=0.9996$, rather than exactly one. Normalizing those numbers givesTreating the coefficients as rounded values of an already normalized state gives the intended approximately $0.26$ answer. Probabilities are sums of squared amplitudes because the energy eigenstates are orthonormal.

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus isIt changes width and height with time, so this is not a stationary state. Its integral is the constant $\sqrt{\pi \hslash /m}$; multiplying the wavefunction by $(m/(\pi \hslash ){)}^{1/4}$ normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance $\hslash (1+t^2)/(2m)$. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

Put $s=1+it$. Logarithmic differentiation givesThereforeso the appropriate potential energy is $\boxed{V(x)=0}$. The wavefunction never vanishes, so no other time-independent potential gives the same equation for this precise phase choice.

The squared modulus isIt changes width and height with time, so this is not a stationary state. Its integral is the constant $\sqrt{\pi \hslash /m}$; multiplying the wavefunction by $(m/(\pi \hslash ){)}^{1/4}$ normalizes it. The normalized Gaussian wave packet stays centred at zero but spreads, with position variance $\hslash (1+t^2)/(2m)$. It represents a localized free particle with a spread of momenta, rather than an energy eigenstate. Free Gaussian wave packet evolution broadens the packet even with zero mean momentum.

The parity operator acts as $P\psi (x)=\psi (-x)$, with $P^2=I$. For an even potential energy $[H,P]=0$, so energy eigenstates can be chosen even or odd, with parity eigenvalues $+1$ and $-1$. One-dimensional bound states are nondegenerate; the ground state is nodeless and even.

For the finite square well, square integrability excludes the growing exterior exponential, while even parity excludes the interior sine. Thus the interior solution is proportional to $\cos kx$, with matching decaying tails andContinuity at $a$ gives $B\cos ka=A{e}^{-qa}$ and $-Bk\sin ka=-qA{e}^{-qa}$. Their ratio yieldsThe conditions at $-a$ follow from the same parity.

For a narrow weak well, $ka\ll 1$, hence $q\simeq k^{2}a\simeq 2m{V}_{0}a/{\hslash }^2$ andThis is self-consistent because $|E_0|/V_0\simeq 2m{V}_0{a}^2/{\hslash }^2\ll 1$. It is the shallow finite-square-well energy, equivalently the attractive delta potential limit with integrated strength $2a{V}_0$. An arbitrarily weak attractive well of this form still has its even bound ground state.

The first excited state must be odd. Replace the interior cosine by $\sin kx$ and choose exterior tails with opposite signs. The matching condition becomes $q=-k\cot ka$. Its first root requires $\pi /2<ka<\pi$. At the binding threshold $q\to 0$, sois necessary and sufficient for the first odd bound state. Equality gives a zero-energy nonnormalizable threshold solution, not an additional bound state.

Writing $a=cx+idp$, where $c=\sqrt{m\omega /(2\hslash )}$ and $d=1/\sqrt{2m\omega \hslash }$, the canonical commutation relation givesConsequently $[a^{†}a,a]=-a$ and $\boxed{[H,a]=-\hslash \omega a}$. If $a|{\varphi }_0⟩$ were nonzero, this commutator would make it an energy eigenstate with energy $E_0-\hslash \omega$, contradicting the lowest-energy property. Therefore $a|{\varphi }_0⟩=0$.

The commutator product rule and induction giveActing on the vacuum and shifting the sum then shows that the unnormalized bosonic coherent state $|\tilde{\psi }⟩=e^{\alpha a^{†}}|0⟩$ satisfies $a|\tilde{\psi }⟩=\alpha |\tilde{\psi }⟩$.

The displayed series lacks its wavefunction normalization factor. Since $(a^{†}{)}^n|0⟩=\sqrt{n!}|n⟩$, its squared norm is ${\sum }_n|\alpha {|}^{2n}/n!=e^{|\alpha {|}^2}$. The normalized coherent state needed for the stated expectation values isUse $⟨a⟩=\alpha$, $⟨a^{†}⟩={\alpha }^{\ast }$, $⟨a^{†}a⟩=|\alpha {|}^2$ and $⟨a{a}^{†}⟩=|\alpha {|}^2+1$. For $x=x_0(a+a^{†})$, $x_0=\sqrt{\hslash /(2m\omega )}$,The displacement changes the mean position but not its ground-state uncertainty. Unnormalized bras and kets give the same physical expectation values only after division by their squared norm.

For a time-independent Hamiltonian operator, a symmetry operator $U$ satisfies $UH{U}^{†}=H$, or $[U,H]=0$. Thus applying $U$ to an energy eigenstate preserves its energy. Commuting Hermitian observables can be diagonalized within each degenerate energy eigenspace, allowing states to be labelled by symmetry quantum numbers.

A continuous unitary symmetry $U(\epsilon )=e^{-i\epsilon Q/\hslash }$ has Hermitian generator $Q$. Differentiating the invariance condition gives $[H,Q]=0$. The Heisenberg equation of motion then yieldswhen $Q$ has no explicit time dependence. spatial translations connect to momentum conservation, rotations to angular-momentum conservation, and time-translation invariance to energy conservation. A central potential energy conserves all components of angular momentum, but they cannot all be assigned sharp simultaneous values because they do not commute; $H,L^2,L_z$ are a compatible choice. Discrete reflection symmetry gives parity labels without a continuously variable generator. A quantum degeneracy can be protected by a symmetry, but a quantum degeneracy alone does not identify its physical origin.

The Time-dependent Schrödinger equation $i\hslash {\partial }_t|\psi ⟩=H(t)|\psi ⟩$ governs unitary evolution of a closed system. If $H$ is independent of time, expand in its energy eigenstates:An individual energy eigenstate changes only by a phase, so its probability density and time-independent-observable expectations are stationary. Superpositions can have interference terms oscillating at $(E_n-E_m)/\hslash$; fixed energy populations do not imply that every observable is constant.

For general $H(t)$, $U(t,t_0)=\mathcal{T}\exp [-(i/\hslash )\int H(t^{\prime })d{t}^{\prime }]$ requires time ordering unless Quantum Hamiltonians at different times commute. The equivalent Heisenberg description evolves observables rather than the state, giving $d⟨A⟩/dt=(i/\hslash )⟨[H,A]⟩+⟨{\partial }_{t}A⟩$. Explicit time dependence can transfer energy from an applied perturbation.

For a weak perturbation $V(t)$ to a known $H_0$, the first-order transition amplitude from $n$ to $m\ne n$ isIts squared modulus gives the leading transition probability while depletion remains small. A nearly monochromatic perturbation emphasizes energy differences near its quantum of frequency; a continuum and suitable long-time limit yield Fermi's golden rule. These approximations have stated weak-coupling and time-resolution limits; they do not replace exact unitary dynamics in every regime.

In the small-angle approximation the physical pendulum is a harmonic oscillator with $\omega =2\pi /T$. Its classical energy is $E=m{\omega }^2{A}^2/2=3.55\times 10^{-3}\mathrm{J}$. Equating this to $(n+1/2)\hslash \omega$ givesAn oscillating bob is not a single stationary energy eigenstate. An ideal coherent state has a Poisson distribution of oscillator occupations centred on this mean, with relative spread $1/\sqrt{\bar{n}}$, exceptionally small here. This quantum occupation scale of a classical oscillator fixes the excitation scale without requiring the real, damped clock to be a pure coherent state.

Apply the radial momentum operator to a smooth radial test function. More generally $[f(r),p_r]=i\hslash f^{\prime }(r)$, giving the first two labelled identities. Expanding the square gives the third:The kinetic-energy decomposition is $p^2=p_r^2+L^2/r^2$. A central potential commutes with $L^2$ and $L_z$, so choose simultaneous angular eigenfunctions $Y_{lm}$, with $L^2=l(l+1){\hslash }^2$ and $l=0,1,\dots$. The radial Hamiltonian consequently has the stated centrifugal term. The radial inner product uses measure $r^{2}dr$; adjoints below are taken on compatible regular bound-state domains. The formal radial momentum is not thereby asserted to be a self-adjoint observable on a half-line.

Set $q=l+1$, $W=-q/r+1/(qa)$ and $D=i{p}_r/\hslash$. Since $D^{†}=-D$ formally and $[D,W]=W^{\prime }=q/r^2$,The hydrogen radial Hamiltonian factorization follows by matching the Coulomb coefficient, which fixes the Bohr radiusFor the second pair of labelled relations, the opposite product changes the sign of $W^{\prime }$, henceUsing $[A,A^{†}A]=[A,A^{†}]A$ in the factorization then givesThis Coulomb angular-momentum intertwining means applying $A_l$ to an energy eigenstate either gives zero or an eigenstate of $H_{l+1}$ with the same energy. For a normalized negative-energy state,At fixed $E<0$ this would become negative if raising $l$ continued indefinitely. The bound-state ladder termination by norm positivity occurs at some $L$ with $A_L\psi =0$, givingThe terminating radial function is proportional to $r^L{e}^{-r/[(L+1)a]}$, regular and normalizable; adjoint intertwiners generate lower $l$ values at fixed $n$. Positive-energy scattering states are not subject to this terminating negative-energy argument. Here $m$ is the mass in the supplied Hamiltonian; including nuclear motion replaces it by the reduced mass.

Use $[x,p]=i\hslash$. Acting on a smooth test wavefunction, $[f(x),p]\psi =f(-i\hslash {\psi }^{\prime })+i\hslash (f\psi {)}^{\prime }=i\hslash f^{\prime }\psi$. Take $f$ real and use the self-adjoint operator domain with vanishing integration-by-parts boundary terms. Writing $Q=p-if$ givesThe energy form is $⟨\psi ,H\psi ⟩=\Vert Q\psi {\Vert }^2/(2m)\ge 0$, so every normalizable energy eigenstate has $E\ge 0$. If a normalizable zero-energy state exists, its nonnegative norm must vanish, $Q{\psi }_0=0$. HenceZero energy is an additional existence assumption, not a consequence of factorization alone.

For ${\psi }_1=Q^{†}{\psi }_0$, the commutator $[Q,Q^{†}]=2\hslash f^{\prime }$ givesThus $\alpha (x)=-i{\hslash }^2{f}^{\prime \prime }(x)/m$ and $\beta (x)=\hslash f^{\prime }(x)/m$ in the requested relation. Generally these are multiplication functions, not constant coefficients, and ${\psi }_1$ is not automatically an energy eigenstate.

If $f(x)=c{x}^n+\cdots$, its integral has leading term $c{x}^{n+1}/(n+1)$. Decay at both infinities on the full real line requires odd $n$ and positive leading coefficient $c$. Odd degree alone is not sufficient. For $f=kx$ take $k=m\omega >0$. Then $H=H_{\mathrm{SHO}}-\hslash \omega /2$, $\alpha =0$ and $\beta =\hslash \omega$. After normalization and removal of an irrelevant phase,The corresponding factorized-Hamiltonian energies are zero and $\hslash \omega$; the constant shift must be restored for the stated oscillator Hamiltonian.

For the quantum harmonic oscillator, put ${\ell }_0=\sqrt{\hslash /(m\omega )}$ and $\xi =x/{\ell }_0$. The normalized energy eigenstates arewhere $H_n$ is a Hermite polynomial. Successive energies are equally spaced; the ground state retains zero-point energy $\hslash \omega /2$. The $n$th wavefunction has $n$ nodes of a wavefunction and parity $(-1{)}^n$. In particular, the first three shapes are proportional to $e^{-{\xi }^2/2}$, $\xi e^{-{\xi }^2/2}$, and $(2{\xi }^2-1)e^{-{\xi }^2/2}$. The classical turning points satisfy $x=\pm {\ell }_0\sqrt{2n+1}$; the wavefunctions extend beyond them and decay, rather than ending abruptly.

The wavefunctions are plotted with arbitrary vertical scale about their energy levels. Their signed excursions are amplitudes, not additional energies or probability densities.

Stationary states. Inside the infinite square well, the time-independent Schrödinger equation has sinusoidal solutions. The boundary conditions $\varphi (0)=\varphi (a)=0$ select $k_{n}a=n\pi$; normalization then givesEach wavefunction vanishes outside the well. There is no nonzero $n=0$ state.

Immediate energy measurement after expansion. For the one-sided sudden expansion of an infinite square well, take the original wall at $x=0$ as fixed, so the expanded well occupies $0<x<2a$. In the sudden approximation, the initial wavefunction remains $\sqrt{2/a}\sin (\pi x/a)$ on $0<x<a$, extended by zero elsewhere. The new ground state is ${\chi }_1=a^{-1/2}\sin (\pi x/(2a))$. Its probability amplitude isThe Born rule therefore givesThe motion of the walls matters: a symmetric expansion about the old midpoint would instead give $64/(9{\pi }^2)$. The stated width alone does not explicitly identify the final endpoints. The first result uses the natural continuation of the specified left boundary; if both boundaries are moved, the final location must be supplied. For a new well $(b,b+2a)$ containing $(0,a)$, the overlap is $c_1(b)=(4\sqrt{2}/(3\pi ))[\cos (\pi b/(2a))-\sin (\pi b/(2a))]$.

Evolution of one energy eigenstate. For a time-independent Hamiltonian operator, write $|\psi (t)⟩=f(t)|{\varphi }_n⟩$. Substitution into the Time-dependent Schrödinger equation gives $i\hslash \dot{f}=E_{n}f$, soOnly a global phase changes, leaving all measurement probabilities unchanged.

Delayed energy measurement. Expand the unchanged initial state in the complete new energy eigenbasis,For the fixed-left-wall geometry, direct overlap integration givesorthogonality isolates the new ground-state amplitude as $c_1{e}^{-i{E}_1^{\prime }t/\hslash }$. Its phase cancels on squaring, so the ground-state probability remains $32/(9{\pi }^2)$ for every delay, provided the expanded well remains fixed and no intervening measurement or environment changes the state. The position probability density generally evolves because different eigenstates acquire different relative phases.

Factorization. Put $\alpha =\sqrt{m\omega /(2\hslash )}$ and use the canonical commutation relation $[x,p]=i\hslash$:It follows thatThe $-1/2$ arises from quantum operator ordering; treating $x$ and $p$ as commuting would lose the zero-point energy.

commutators. Direct use of $[x,p]=i\hslash$ gives $[a,a^{†}]=1$. The product rule for commutators then yields $[a^{†}a,a]=-a$ and $[a^{†}a,a^{†}]=a^{†}$, hence

Raising the energy. If $H|\varphi ⟩=E|\varphi ⟩$, thenThus the creation operator produces an energy eigenstate with energy $E+\hslash \omega$. It is nonzero: for a normalized physical oscillator eigenstate, $\Vert a^{†}\varphi {\Vert }^2=⟨a{a}^{†}⟩=E/(\hslash \omega )+1/2>0$. Normalize by this norm if a normalized state is wanted.

Lowest wavefunctions and orthogonality. In the position representation in quantum mechanics,The annihilation condition therefore gives ${\psi }_0^{\prime }=-(m\omega /\hslash )x{\psi }_0$. Applying $a^{†}$ to its solution givesTheir inner product is proportional to ${\int }_{-\infty }^{\infty }x{e}^{-m\omega x^2/\hslash }dx=0$, since the integrand is odd. These are the even and odd lowest wavefunctions shown in A2, with energies $\hslash \omega /2$ and $3\hslash \omega /2$.

A density operator describes both a quantum state prepared in one pure state and a statistical preparation whose members occupy different pure states. If normalized vectors $|{\psi }_i⟩$ occur with probabilities $p_i$, defineThe resulting operator is Hermitian, positive semidefinite and has trace one. Positivity follows because $⟨v|\rho |v⟩={\sum }_i{p}_i|⟨{\psi }_i|v⟩{|}^2\ge 0$. Conversely, the spectral theorem gives such a representation of any finite-dimensional operator with these properties, using its eigenvectors and eigenvalues. The matrix representing $\rho$ depends on the chosen basis; the operator and its predictions do not.

One operator gives every measurement probability. For a projective measurement with orthogonal projections ${\Pi }_a$ and outcomes $a$,For example, the first formula follows by averaging $⟨{\psi }_i|{\Pi }_a|{\psi }_i⟩$ over the preparation. The trace expression is basis independent. A general measurement replaces ${\Pi }_a$ by positive measurement effects that sum to the identity, without changing the probability formula. Thus $\rho$ specifies the statistics accessible to measurements on that system.

The distinction between pure and mixed states is especially transparent in the eigenvalues. A pure state has $\rho =|\psi ⟩⟨\psi |$, ${\rho }^2=\rho$, and $\mathrm{Tr}{\rho }^2=1$. A mixed state has at least two positive eigenvalues and $\mathrm{Tr}{\rho }^2<1$. For a qubit,pure states lie on the Bloch sphere; mixed states lie inside the Bloch ball. The maximally mixed qubit $\rho =I/2$ can be prepared by choosing equally between the two $z$ spin eigenstates, or equally between the two $x$ spin eigenstates. These ensembles are different preparation procedures but give the same density operator. No measurement on the resulting qubit alone can identify which ensemble was used.

quantum coherence must be distinguished from merely having two populated levels. In the $z$ basis, the state $|+x⟩$ hasBoth give equal probabilities for the two $z$ outcomes. A measurement along $x$ gives the positive result with probability one for the first state and one half for the second. The off-diagonal entries encode the relative phase information responsible for this difference. Whether a matrix entry is off diagonal is basis dependent; purity is not.

For an isolated system, each component evolves with the same unitary time-evolution operator. Therefore $\rho (t)=U(t)\rho (0)U^{†}(t)$ andUnitary evolution preserves the eigenvalues and purity of $\rho$. A density operator commuting with a time-independent Hamiltonian operator is stationary. A thermal equilibrium example is the canonical ensemble ${\rho }_{\beta }=e^{-\beta H}/Z$, where $Z=\mathrm{Tr}(e^{-\beta H})$ and $\beta =1/(k_{\mathrm{B}}T)$. expectation values then reproduce quantum statistical mechanics without assigning the system a definite energy eigenstate on every run.

A mixed state can also arise without classical randomness in the joint preparation. For a bipartite quantum system, the reduced density operator ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ gives every measurement probability accessible to subsystem $A$. For the Bell state $(|00⟩+|11⟩)/\sqrt{2}$, the joint state is pure, whereas ${\rho }_A=I/2$. Its local mixedness reflects entanglement with the inaccessible system; the partial trace discards correlations that matter for joint measurements. More generally, interaction with an environment can suppress off-diagonal entries in a selected basis, producing decoherence while the combined system still evolves unitarily.

Finally, state evolution conditional on a measurement outcome differs from isolated evolution. An ideal projective measurement yielding $a$ updates the state by the Lüders rule $\rho \mapsto {\Pi }_a\rho {\Pi }_a/P(a)$. Discarding the outcome gives ${\sum }_a{\Pi }_a\rho {\Pi }_a$, which generally removes coherence between different eigenspaces. The Von Neumann entropy $-\mathrm{Tr}(\rho \ln \rho )$ quantifies mixedness: it vanishes for pure states and reaches $\ln d$ for $I/d$. Multiplying by $k_{\mathrm{B}}$ gives thermodynamic entropy units. These uses make the density operator a common language for preparation uncertainty, measurement, thermal states and entangled subsystems, while retaining their distinct physical origins.

For a nonrelativistic spinless particle in a scalar potential, the Time-dependent Schrödinger equation isAn initial wavefunction together with boundary conditions specifies the evolution. The Hamiltonian operator must be self-adjoint on the appropriate domain for closed-system evolution to be unitary. The equation is linear, so a superposition of solutions is another solution. Its complex wavefunction predicts probability amplitudes; the Born rule separately supplies the connection to measurement probabilities.

For a time-independent potential, separation $\psi (\mathbf{r},t)=\varphi (\mathbf{r})e^{-iEt/\hslash }$ gives the time-independent Schrödinger equation $H\varphi =E\varphi$. Normalizability and boundary conditions select the allowed energies of bound states; scattering states instead use incoming and outgoing boundary conditions and generalized normalization. A general state is a superposition of energy eigenstates, with a spectral integral as well if there is a continuous spectrum. Different energies accumulate different phases, allowing time-dependent position densities and oscillatory expectation values even though a single energy eigenstate has a stationary density.

For a real scalar potential, combining the equation with its complex conjugate yields probability conservation:With no current escaping the boundary, $\int |\psi {|}^2{d}^{3}r$ is constant. Matching conditions at a finite potential step follow from the differential equation and enforce the flux balance used in B6.

The Schrödinger equation is a dynamical postulate of nonrelativistic quantum mechanics, not a consequence of classical mechanics alone. The substitutions $E\mapsto i\hslash {\partial }_t$ and $\mathbf{p}\mapsto -i\hslash \nabla$ motivate its free-particle form from the dispersion relation, but do not derive the measurement postulates. The scalar equation omits intrinsic spin and relativistic particle creation. Spinful or relativistic problems require the corresponding extended theory. Nor does its deterministic unitary evolution, by itself, prescribe conditioning on a particular observed measurement outcome.

Each energy eigenstate evolves by its own quantum phase $e^{-i{E}_{m}t/\hslash }$. The two $m=\pm 2$ components have the same energy, while the $m=0$ quantum phase stays one. ThusIn angle representation this isAt $t=0$, $1+\cos 2\varphi =2{\cos }^2\varphi$ recovers the initial state. The fixed coefficient magnitudes keep the angular momentum and energy probabilities constant even though the density in angle changes.

Put $\Omega =\lambda B$, including its sign. The initial vector is the $+\hslash /2$ eigenstate of $S_z$; it is not an energy eigenstate of $H=\Omega S_x$. In terms of the spin eigenstates found above, $|+z⟩=(|+x⟩+|-x⟩)/\sqrt{2}$. Each energy eigenstate acquires its own unitary time evolution phase:Equivalently, $e^{-iHt/\hslash }=I\cos (\Omega t/2)-i{\sigma }_x\sin (\Omega t/2)$ follows from ${\sigma }_x^2=I$. The Born rule givesThe expectation value of the spin angular momentum is $(\hslash /2)(0,-\sin \Omega t,\cos \Omega t)$, a rotation around the $x$ axis. Its direction follows the specified sign of the Hamiltonian; changing the magnetic coupling sign reverses the precession.

The normalized radial probability density givesConsequently the expectation value of the potential energy isHere the last equality uses $Z{e}^2/(4\pi {ϵ}_0)={\hslash }^2\beta /m_e$. Since this is an energy eigenstate, $E=⟨T⟩+⟨V⟩$, and so $\boxed{⟨T⟩=-E}$. The kinetic energy is positive because $E<0$, in agreement with the virial theorem for a Coulomb potential.

The nuclear beta decay is fast compared with the electronic orbital timescale. In the sudden approximation, the electronic wavefunction does not have time to change during the increase from $Z_i=1$ to $Z_f=2$. The subsequent probability of a particular daughter energy eigenstate is therefore its squared overlap with the initial state.

For the final $1s$ state, write ${\beta }_i=1/a_0$ and ${\beta }_f=2/a_0$. The sudden hydrogenic charge change gives the amplitudeThusThe final $2p$ states have spherical harmonics $Y_{1m}$, whereas the initial state is proportional to $Y_{00}$. Their angular overlap $\int Y_{1m}^{\ast }{Y}_{00}d\Omega$ vanishes by orthogonality. Accordingly $\boxed{P_{2p}=0}$ in the ideal spherically symmetric sudden-change model. Nuclear recoil or interactions with the emitted particles can produce small corrections. This is an overlap selection rule for a sudden change, not a photon-induced electric dipole transition.