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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.

In the time-independent Schrödinger equation, $-{\hslash }^2{\psi }^{\prime \prime }/(2m)$ is the kinetic-energy operator acting on the state, $V\psi$ its potential-energy contribution and $E\psi$ the energy-eigenvalue term. The complex wave function is a probability amplitude: the Born rule assigns density $|\psi {|}^2$ for a normalized single-particle state. A plane-wave beam uses an incident-amplitude or flux normalization instead of a unit-integral wavefunction.

For a free plane wave, ${\psi }^{\prime \prime }=-k^2\psi$, so $\boxed{E={\hslash }^2{k}^2/(2m)}$ and its time factor obeys $\hslash \omega =E$. Above the upward potential step, the bounded subthreshold solution has decay constantFor the PDF's unit incident amplitude, write $\psi =e^{ikx}+R{e}^{-ikx}$ on the left and $\psi =T{e}^{-\alpha x}$ on the right, omitting their common time phase. Continuity of both $\psi$ and ${\psi }^{\prime }$ at this finite step givesConsequentlyThe evanescent amplitude versus transmitted current distinction matters: the second quantity is a squared amplitude, not a transmitted-flux probability. The decaying tail has zero probability current, despite nonzero density and the possible value $|T{|}^2>1$.

Using $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$ and the electron mass and elementary charge specified above, the energy difference1eV gives $\alpha =5.12\mathrm{n}{\mathrm{m}}^{-1}$. Thus at the stated positionThis is relative to the unit incident density in the source figure, not an absolute normalized probability per metre. ^[ref]

Here $V(x,y)$ is the potential energy and $E$ is the stationary energy eigenvalue. The Schrödinger equation is the eigenvalue equation for kinetic plus potential energy. Its boundary conditions select the allowed energies. For a normalized wavefunction, the Born rule assigns probability $|\psi (x,y){|}^{2}dxdy$ to a small area, and $\int |\psi {|}^{2}dxdy=1$. A stationary state's full time dependence is $\psi (x,y)e^{-iEt/\hslash }$; its probability density is time-independent.

The infinite walls impose $\psi =0$ on all four boundaries. For a separated solution $\psi =X(x)Y(y)$ in the zero-potential interior,The first term depends only on $x$ and the second only on $y$, so each is constant, $-k_x^2$ and $-k_y^2$. Vanishing at $x=0$ and $y=0$ removes the cosine terms. Vanishing at the other boundaries givesThe normalized two-dimensional infinite square well eigenbasis and energies areNormalization follows from the two sine-squared integrals $a/2$ and $b/2$. Zero quantum numbers would give an identically zero function. These separated products form a basis, not every possible wavefunction: a general state is their superposition, and a degenerate energy eigenspace can have nonseparable stationary combinations of products with that same energy.

For $a=1\mathrm{nm}$ and $b=2\mathrm{nm}$, the ground state is $(1,1)$ and the nearest excitation is $(1,2)$, along the longer side. The next quantum along the shorter side costs more energy. ThusThis transition has nonzero dipole matrix element for an electric-field component along $y$. Light polarized strictly along $x$ would not drive it, so the minimum assumes an allowed polarization rather than an arbitrary fixed one.

For the requested degeneracy, keep $b>a$. The first two states are necessarily $(1,1)$ and $(1,2)$, since $E_{21}-E_{12}=3{\pi }^2{\hslash }^2(a^{-2}-b^{-2})/(2m)>0$. The third level is the smaller of $E_{21}$ and $E_{13}$: states with both indices at least two lie above $E_{21}$, and higher states on the $n_x=1$ sequence lie above $E_{13}$. Make these two candidates equal:The third-level degeneracy in a rectangular quantum box then has lowest distinct energies $11/8$, $5/2$ and $35/8$ in units ${\pi }^2{\hslash }^2/(2m_e{a}^2)$, with multiplicities one, one and two. The third-level pair is exactly $(2,1)$ and $(1,3)$.

Equation and interpretation. The supplied equation is the time-independent Schrödinger equation for a nonrelativistic particle with mass $m$ in a prescribed potential energy $V(x)$. Equivalently,It is an eigenvalue problem for the Hamiltonian operator. A stationary energy state has full wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. By the Born rule, $|\psi (x){|}^{2}dx$ is the probability of finding the particle in a small interval $dx$, with wavefunction normalization ${\int }_{-\infty }^{\infty }|\psi {|}^{2}dx=1$.

Infinite barriers. Infinite exterior potential energy excludes any finite-energy probability outside the interval. The infinite square well consequently has $\psi =0$ outside and the hard-wall boundary conditionsThe interior wavefunction must meet the zero exterior amplitude continuously. Its derivative need not be continuous across an infinite wall; imposing both zero value and zero derivative would eliminate every nonzero interior solution.

Ground state. Inside the well $V=0$, so ${\psi }^{\prime \prime }+k^2\psi =0$, with $k=\sqrt{2mE}/\hslash$. Its general solution is $A\sin kx+B\cos kx$. The left wall gives $B=0$, and the right wall gives $ka=n\pi$ for $n=1,2,\dots$. Thus the discrete energy eigenvalues are $E_n=n^2{\pi }^2{\hslash }^2/(2m{a}^2)$. The smallest, $n=1$, is the ground state. Sinceits normalized wavefunction, choosing an irrelevant overall phase to make it positive inside, is

Finite barriers. For a finite step and constant particle mass, both $\psi$ and ${\psi }^{\prime }$ are continuous at $x=0,a$. Wavefunction matching at a finite potential step follows by integrating the Schrödinger equation across a vanishingly small neighborhood of the step: a bounded potential cannot generate a finite jump in ${\psi }^{\prime }$. The bound-state exterior solutions decay rather than vanish identically:Normalizability excludes the growing exterior exponentials. The equivalent interior boundary conditions are ${\psi }^{\prime }(0)=\kappa \psi (0)$ and ${\psi }^{\prime }(a)=-\kappa \psi (a)$; neither boundary value is zero for the nontrivial ground state.

The ground state of a finite square well remains symmetric about $a/2$ and has no nodes. It bends smoothly into exponential tails at both walls, extending over a larger region than the infinite-well state. Its maximum after normalization is lower for the deep well considered here. When $V_w\gg E$, the penetration length $1/\kappa$ is short, so the two shapes are close except near the walls and in the exterior tails.

The finite curve is an illustrative exact matched solution with $V_w=10E_{1,\infty }$; no numerical wall height is specified by the question.

Energy comparison. The finite barriers relax the confinement, allowing a less rapidly varying interior wavefunction and therefore smaller kinetic energy. More explicitly, the even interior solution is proportional to $\cos [k(x-a/2)]$, and matching to the tails gives $k\tan (ka/2)=\kappa$. The lowest root lies at $0<ka<\pi$, while the infinite well has $ka=\pi$. HenceAs $V_w\to \infty$, the tails disappear and the finite-well energy approaches the infinite-well value from below. This comparison uses the same zero of potential energy inside both wells.

The time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

$V(x)$ is the particle's potential energy as a function of position, and $E$ is its total energy eigenvalue in this time-independent Schrödinger equation. Solving it with the boundary conditions determines the allowed stationary states and energy levels.

The wavefunction is a probability amplitude: the Born rule gives position probability density $|\psi (x){|}^2$, normalized to unit integral for one particle. $\psi$ itself need not be real or positive and is not a classical matter-density profile. The full stationary-state time dependence is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, leaving $|\Psi {|}^2$ time independent.

Include Alice's detector and environment before taking a partial trace. Ideal measurement produces the global stateOrthogonality of the two record states removes their cross terms from the reduced state of the particles:This nonselective quantum measurement has changed the initially entangled singlet into a separable, classically anticorrelated reduced state. Its entropy is one bit, although the full particles-plus-records state is still pure under the assumed unitary dynamics. The reduced density matrix is an improper mixture; tracing the environment does not by itself assert a literal collapse of that global state.

The Born rule gives Alice outcomes zero and one with equal probability. Conditional on a recorded result, the selective quantum measurement assigns $|01⟩$ or $|10⟩$ to the particles. The unitary measurement coupling and orthogonal records explain decoherence in this basis; they do not supply an additional rule selecting a unique outcome. That is part of the measurement problem. In a collapse postulate the individual outcome is stochastic; other interpretations account for outcome experience differently while retaining these operational probabilities.

When Bob makes the same kind of measurement, the global state becomesTracing both laboratories gives the same reduced matrix, ${\rho }_{AB}^{(2)}={\rho }_{AB}^{(1)}$. Bob has added a correlated record, not restored the lost reduced coherence. If Alice's outcome is known, Bob's opposite outcome is certain in this ideal common basis. Without that record, his marginal is $I/2$, so his outcome remains equiprobable. Alice cannot choose her random result or use this correlation for faster-than-light signalling, consistently with quantum no-signalling.

These common-basis measurements alone do not distinguish quantum predictions from a local source sending opposite preassigned bits. To test that explanation, prepare fresh singlets and let the observers independently choose among incompatible measurement axes before recording each result. For singlet spin measurements along unit vectors $\mathbf{a},\mathbf{b}$, quantum mechanics predicts $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$. The CHSH inequality bounds $|S|\le 2$ for Bell-local models with measurement-independent settings. Choosinggives $S=E(\mathbf{a},\mathbf{b})+E(\mathbf{a},{\mathbf{b}}^{\prime })+E({\mathbf{a}}^{\prime },\mathbf{b})-E({\mathbf{a}}^{\prime },{\mathbf{b}}^{\prime })=-2\sqrt{2}$. A suitable Bell test constrains that class of explanations. Independently chosen settings, control of detection bias and separation of the measurement events matter when interpreting an actual experiment.

If Alice first performs the fixed-axis irreversible measurement described above, the resulting mixed state has $E(\mathbf{a},\mathbf{b})=-a_z{b}_z$ and cannot supply that singlet violation. The incompatible settings must belong to the Bell measurement itself. Moreover, Bell violations do not choose among all interpretations of quantum mechanics: interpretations sharing the Born predictions remain observationally equivalent for this experiment. Objective-collapse modifications could instead require additional interference tests with specified departures from unitary dynamics.

Set $N=2^n$, $a=N^{-1/2}=\sin t$ and $c=\cos t$, with $N>1$. The ordered vectors $|w⟩,|s⟩$ are normalized but nonorthogonal, since $⟨w|s⟩=a$. Their coordinate Gram matrix is thereforeAn operator matrix in this basis is obtained from the coefficients in the images of its basis vectors, not by treating those vectors as orthogonal measurement outcomes. Directly applying the two reflection operators givesTheir columns are consequentlyThese are physical unitary operators even though the coordinate matrices are not unitary in the ordinary Euclidean coordinate norm: they preserve $S$, so $U^{†}SU=S$.

The Grover iteration in a nonorthogonal basis isIt has determinant one and trace $2-4a^2=2\cos 2t$, hence eigenvalues $e^{\pm 2it}$. For example, the invertible matrix of eigenvectorssatisfiesThis diagonalization is exact for $N>1$; no large-$n$ approximation is needed.

To extract an actual probability using the Born rule, introduce the orthonormal unmarked superposition $|b⟩=(|s⟩-a|w⟩)/c$. In the ordered basis $|w⟩,|b⟩$, the same iteration is the rotationAfter $r$ iterations,Equivalently the original coordinates are $[\sin (2rt)/c,\cos ((2r+1)t)/c{]}^T$. The overlap with $|w⟩$ is the first coordinate plus $a$ times the second, recovering the same amplitude. Squaring just the first coordinate would be incorrect in a nonorthogonal basis.

For large $n$, $t\simeq N^{-1/2}$, and dropping the initial angle $t$ gives the requested approximation $P_r\simeq {\sin }^2(2rt)$. The first near-unit success occurs at the integer closest to $\pi /(4t)-1/2$:One can choose $r_{\mathrm{opt}}=\lfloor \pi /(4t)\rfloor$, with either neighbor equivalent in a tie. Further iterations overshoot and reduce success rather than monotonically improving it. The special $N=1$ problem needs no search and the two-vector coordinate construction becomes singular.

For an unstructured search, a reversible Boolean quantum oracle implements a phase change on the marked entry, while the diffusion reflection about a uniform state is known. Repeating the pair of reflections amplifies the marked amplitude and measurement yields a candidate that can be checked by another oracle query. The bounded-error query count is $O(\sqrt{N})$ for one marked item, compared with classical $\Theta (N)$ queries. It is a quadratic oracle-query speedup: state preparation, the cost of implementing the oracle and access to a classical database still matter. The method does not reveal an unknown marked value without the promised query access.^[ref]

Consistent histories assigns probabilities to suitably compatible sequences of quantum alternatives without making an external measuring apparatus fundamental. At successive times choose complete orthogonal sets of projection operators. In the Heisenberg picture, a history has class operator $C_{\alpha }=P_{{\alpha }_n}(t_n)\cdots P_{{\alpha }_1}(t_1)$. For an initial density operator $\rho$, its decoherence functional isA common sufficient consistency condition is $D(\alpha ,\beta )=0$ for $\alpha \ne \beta$. Then $p(\alpha )=D(\alpha ,\alpha )$ is nonnegative and coarse-grained alternatives have additive probabilities: the interference cross-terms vanish. The weaker requirement on real cross-terms can suffice for individual probability sum rules. Predictions must stay within one compatible framework; mutually incompatible history families cannot simply be combined into one classical account.

The many-worlds interpretation instead treats unitary quantum evolution as universally valid and interprets the correlated outcome branches as components of one global state. A measurement interaction givesWhen the environment records become nearly orthogonal, quantum decoherence suppresses observable interference between stable apparatus records. Each relative state contains an observer correlated with a definite record; no additional physical collapse is postulated. Branch structure is an emergent approximation, rather than a unique literal splitting at every arbitrary choice of basis.

Both approaches use the ordinary unitary formalism and decoherence, but their interpretations of histories and branches differ. Decoherence alone does not prove that one unique branch is selected, nor does the unitary equation by itself settle every assumption needed to justify Born rule probabilities. Neither interpretation predicts ordinary interference experiments incorrectly merely by interpreting the same state differently.

A spin measurement along unit vector $\mathbf{a}$ has outcome projector ${\Pi }_s(\mathbf{a})=(I+s\mathbf{a}\cdot \mathbf{\sigma })/2$, with $s=\pm 1$. The spin singlet state obeys $⟨{\sigma }_i\otimes I⟩=⟨I\otimes {\sigma }_i⟩=0$ and $⟨{\sigma }_i\otimes {\sigma }_j⟩=-{\delta }_{ij}$. Applying the Born rule,Thus $P_{++}=P_{--}=(1-\mathbf{a}\cdot \mathbf{b})/4$ and $P_{+-}=P_{-+}=(1+\mathbf{a}\cdot \mathbf{b})/4$. Substitution into part (a) givesEqual axes give perfect anticorrelation, while orthogonal axes give zero correlation. Each local marginal remains uniform.

A qubit has state $\alpha |0⟩+\beta |1⟩$ with $|\alpha {|}^2+|\beta {|}^2=1$. An $n$-qubit register occupies a $2^n$-dimensional Hilbert space and may contain entanglement. This large state space does not mean that a measurement reads out all its complex amplitudes: computational-basis outcomes occur with the Born rule probabilities.

A quantum circuit prepares a register, applies quantum gates, and measures selected outputs, possibly using classical feed-forward. Closed gates are unitary operators and therefore reversible; initialization, measurement and discarding can be treated using a larger system and environment. Single-qubit rotations together with an entangling two-qubit gate give a universal quantum gate set. The CNOT gate acts as $|x,y⟩\mapsto |x,y\oplus x⟩$.

For example, a Hadamard gate on the first qubit of $|00⟩$ followed by a CNOT creates $(|00⟩+|11⟩)/\sqrt{2}$. A Hadamard produces amplitudes with relative signs, so later gates can make unwanted alternatives interfere destructively and useful alternatives reinforce. Quantum algorithms organize this interference to improve a specified computational task; superposition alone does not supply an arbitrary speedup or permit reading every branch.

A finite discrete universal set such as Clifford gates together with a non-Clifford phase gate approximates general unitaries to chosen precision. Runtime includes the number of gates, repetitions and classical processing. Measurement outcomes may require multiple runs to estimate a probability or amplify success.

Quantum decoherence and imperfect gates limit an implementation. Quantum error correction encodes logical information into several physical qubits and measures error syndromes without directly learning the encoded amplitudes. Under appropriate noise and control assumptions, fault tolerance permits reliable longer computations; simply maintaining an isolated ideal state is not a realistic account of a large processor.

For a nonzero quantum state $\psi$ and a self-adjoint operator $A$, the expectation value and standard deviation of its measurement areThe squared deviation is also called the variance of the observable. All displayed products are understood on quantum states for which the relevant moments exist. Self-adjointness gives $\overline{⟨\psi ,A\psi ⟩}=⟨A\psi ,\psi ⟩=⟨\psi ,A\psi ⟩$, so $a$ is real. Expanding the squared norm gives the variance expression. A norm is nonnegative and is zero exactly for the zero vector, hence ${\Delta }_{\psi }(A)=0$ precisely when $A\psi =a\psi$. Thus zero dispersion is equivalent to being an eigenstate.

For bounded $H$, the operator exponential is the norm-convergent power seriesFor a general unbounded self-adjoint Hamiltonian operator, the spectral theorem defines the same notation byThis definition does not require the power series to converge on every quantum state. complex conjugation of the spectral multiplier gives $U_t^{\ast }=U_{-t}$, and multiplication of the phases gives $U_{-t}{U}_t=I$. Thus $U_t$ is unitary, preserves the quantum state norm, and has $U_0=I$.

For initial data in the domain of $H$, differentiating the spectral expression yields ${\partial }_t{U}_t{\psi }_0=-(i/\hslash )H{U}_t{\psi }_0$. Consequentlysolves the Schrödinger equation with the prescribed initial quantum state. Uniqueness follows by differentiating $U_{-t}{\psi }_t$: the two terms cancel, leaving the constant ${\psi }_0$.

In the two-dimensional energy basis, expand the initial quantum state as ${\psi }_0=c_1{\Psi }_1+c_2{\Psi }_2$. Each energy eigenvector receives its own phase under the operator exponential, givingThis is the general solution because every initial vector has a unique basis expansion.

In that orthonormal basis the other observable has the real symmetric matrixIts conjugate transpose equals itself, so it is self-adjoint. Also $A^2=I$ and $\mathrm{tr}A=0$, giving eigenvalues $+1$ and $-1$. Direct substitution gives the normalized orthogonal eigenvectorsAn irrelevant global phase may multiply either vector. With the chosen lower eigenstate initially,The Born rule says that the probability for the higher eigenstate is $|⟨v_{+},{\psi }_t⟩{|}^2$. Its amplitude is $\sin \alpha \cos \alpha (e^{-it{E}_2/\hslash }-e^{-it{E}_1/\hslash })$. Using $|e^{-iu}-e^{-iv}{|}^2=4{\sin }^2((u-v)/2)$ givesThis is the transition probability for a rotated two-level observable. Its time dependence comes from the relative phase between two energy eigenstates. If their energies coincide, or if the measured observable has the energy eigenvectors as its eigenvectors, this transition probability is zero at all times.

For a conditional spin measurement of identical fermions addressing the $z$ component in one spatial mode, the Born rule givesConditioned on this result, the mode spin state is $|{\uparrow }_a{\downarrow }_b⟩$; the other outcome produces $|{\downarrow }_a{\uparrow }_b⟩$. The spin product is not a spin-exchange eigenstate. The measurement has selected an addressed mode and recorded an outcome, so this conditional spin description distinguishes the measured subsystem.

The full electron wavefunction still obeys fermionic exchange symmetry. For example, the conditional spin-up outcome in spatial mode $a$ is represented in particle coordinates byIt is antisymmetric under full particle exchange. Thus the loss of a definite symmetry of the conditional spin product does not mean that measurement turns identical electrons into distinguishable particles. If the outcome is discarded, the mode spin state is instead the equal mixture of the two opposite-spin products. The measurement must specify which spatial mode is addressed for the conditional statement to be operationally meaningful.

The initial quantum state is the equal superposition of the first two box states, $({\psi }_1+{\psi }_2)/\sqrt{2}$. The orthonormality of the two energy eigenstates makes its squared norm equal to one. Their separate stationary state phases giveWrite $\Omega =(E_2-E_1)/\hslash =3{\pi }^2\hslash /(2m{a}^2)$. The probability density isEach squared sine integrates to $a/4$ on the left half. For the quantum interference term, the product-to-sum identity givesHence the Born rule givesThe quantum interference moves probability between the halves even though each component separately is a stationary state.

The requested energy eigenvalue is $E_2$. By the Born rule, its probability amplitude is the inner product $c_2=⟨{\psi }_2,\Psi (0)⟩$. Put $u=\pi x/a$ and use $e^{iu}\sin u=\frac{1}{2}\sin 2u+i{\sin }^{2}u$:The first integral is $\pi /2$; the second is zero under $u\mapsto \pi -u$. ThereforeFor the time-independent Hamiltonian operator, expand in its orthonormal basis of energy eigenstates:Unitary time evolution changes $c_2$ to $c_2{e}^{-i{E}_{2}t/\hslash }$, whose modulus is unchanged. Thus the same energy measurement probability holds at every time, even though relative quantum phases and other measurements can change.