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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 $V(x)$ is potential energy, $E$ is the stationary state's total energy, and $\psi (x)$ is its spatial probability amplitude. The wave function has time factor $e^{-iEt/\hslash }$; $|\psi {|}^2$ is a position probability density for a normalized state. Stationary plane-wave scattering solutions are idealizations described by probability-current ratios, or components of a normalizable wave packet.

Above an upward step, a classical particle is fully transmitted with reduced speed. Quantum mechanically the abrupt wavelength change generally produces both reflected and transmitted amplitudes even though both regions are classically allowed. With unit incident amplitude and no wave arriving from the right,The reflected component has signed wavevector $-k_1$ and the same wavenumber magnitude as the incident component. A finite potential jump with constant mass and no delta interaction requires continuity of $\psi$ and ${\psi }^{\prime }$:The probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ gives probability reflection and transmission $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(k_2/k_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The amplitude $t$ may exceed one without violating conservation of total probability, because the transmitted group speed is smaller.

For $0<E<V_0$, let $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The bounded right-hand solution is $t{e}^{-\kappa x}$, an evanescent wave tail with amplitude penetration scale $1/\kappa$. It carries no outgoing probability current in an infinitely long forbidden half-space, so reflection is total. This differs from transmission through a finite potential barrier, where a second boundary can support quantum tunnelling.

For a downward step, replace $k_2$ by $\sqrt{2m(E+V_0)}/\hslash >k_1$. The same boundary conditions give a nonzero negative $r$, corresponding to a reflected amplitude with phase reversal. If reflection were forbidden, $r=0$ would require both $t=1$ and $k_1=k_2$, impossible for a nonzero step. Reflection therefore does not require a classical potential barrier; the discontinuity of wavenumber mismatch is enough.

For the specified electron energies, $k_2/k_1=\sqrt{40/100}=\sqrt{0.4}$, so the requested amplitude coefficients areThe corresponding probability coefficients are approximately $\mathcal{R}=0.05069$ and $\mathcal{T}=0.94931$. electron mass cancels from these ratios; there is no need to invent an additional material effective mass.

Use a stationary wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$ solving the time-independent Schrödinger equation. For energy above the positive step potential, the positive wavenumbers areIncidence only from the left permits a right-moving incident plane wave and a left-moving reflected plane wave for $x\le 0$, but only a right-moving transmitted plane wave for $x>0$:The omitted right-incident component would describe a different scattering experiment. Wavefunction matching at a finite potential step requires continuity of both $\psi$ and $d\psi /dx$ for a finite jump and the same mass on either side:Eliminate $B=C-A$ from the second condition:The first ratio is a transmission amplitude; a transmitted probability current also depends on its different propagation velocity.

The probability current of the incident wave is $j_{\mathrm{in}}=\hslash k_1|A{|}^2/m$, and that of the reflected wave is $j_{\mathrm{ref}}=-\hslash k_1|B{|}^2/m$. The quantum reflection probability is the positive ratio of reflected to incident flux magnitudes:For comparison, the transmitted probability current is $j_{\mathrm{tr}}=\hslash k_2|C{|}^2/m$, so the transmission probability isThese satisfy $R+T=1$, as required by conservation of probability current for a real, nonabsorbing potential energy. In particular, $|C/A{|}^2$ alone is not the transmitted flux fraction when $k_2\ne k_1$.

Use adjacent segments $[-\ell ,0]$ and $[0,\ell ]$, with $\ell =2\pi /k_1$. Integrate the threshold probability density from part (d). The left segment contains two complete density periods, so its average ${\cos }^2$ is $1/2$:The right-hand density is constant:The common stationary-state normalization cancels, yieldingThe larger local probability weight on the right is compatible with zero transmitted probability current at threshold: density and flux measure different quantities.

For a downward step potential, $V_0=-|V_0|$, the region on the right is classically allowed at every positive incident energy. The same matching calculation applies withAs $E\to 0^{+}$, $k_1\to 0$ and $k_2\to k_{\ast }=\sqrt{2m|V_0|}/\hslash >0$. Henceand the flux coefficients satisfyThis quantum reflection at a downward step occurs even though a classical particle would simply accelerate into the lower-potential region. The abrupt change in wavenumber causes strong reflected-wave interference.

Both this limit and part (d) give total reflection, but their probability amplitudes and spatial densities differ. At the upward-step threshold, $B/A\to +1$ and $C/A\to 2$: an antinode meets the origin and the right-hand density stays finite while its propagation velocity vanishes. Here $B/A\to -1$ and $C/A\to 0$: the incident and reflected waves nearly cancel at the origin, and the transmitted density vanishes while the right-hand propagation velocity remains finite. In both descriptions the incident probability amplitude is held fixed for comparison.

The coefficients above are limiting flux ratios for $E>0$. At exactly $E=0$ the incident probability current vanishes, so directly dividing zero-energy fluxes would not define a scattering probability.

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.

Classically, a particle with $0<E<V_0$ is completely reflected and never enters the raised region. In quantum mechanics the wavefunction penetrates the classically forbidden side as an evanescent wave. Define the tunnelling decay constantBoundedness at positive infinity excludes $e^{+\kappa x}$, soMatching the wavefunction and its derivative givesThe reflected probability amplitude has unit modulus, hence quantum reflection probability one. The decaying tail carries zero probability current: this semi-infinite potential step has no transmitted beam at infinity, unlike a finite potential barrier. The sketch shows an exact matched example rather than attempting to assign a single real value to a complex wavefunction.

Because the probability density decays as $|C{|}^2{e}^{-2\kappa x}$, the requested $1/e$ penetration distance isFor a proton, take $m_p=1.67262\times 10^{-27}\mathrm{kg}$, $V_0-E=10\mathrm{MeV}=1.60218\times 10^{-12}\mathrm{J}$ and $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe wavefunction amplitude decay length $1/\kappa =1.44\mathrm{fm}$ is twice this value. The nonrelativistic equation supplied in the paper is appropriate for this estimate since $10\mathrm{MeV}$ is small compared with the proton rest energy, approximately $938\mathrm{MeV}$. If the original PDF's unbarred $h$ is instead read literally as the Planck constant, the displayed differential equation gives $d_{\mathrm{literal}}=2\pi d=4.53\mathrm{fm}$. This is a valid calculation for that literal coefficient, but $0.720\mathrm{fm}$ is the physical estimate using the standard Schrödinger equation.

A scanning tunneling microscope brings a tip and surface, both electrical conductors, close enough to leave only a small vacuum gap. An applied electric potential difference makes occupied electron states on one side able to feed unoccupied states on the other. Although the gap is a classically forbidden potential barrier, quantum tunnelling through its finite width produces an electric current.

For an approximate rectangular potential barrier of gap width $d$, the transmission probability, and thus the electric current at fixed small bias, varies approximately as $e^{-2\kappa d}$. This strong distance dependence makes the device sensitive to changes in separation on the scale of a chemical atom. Feedback control moves the tip vertically to keep the electric current constant during a scan; the required height changes form an image of the surface, with contrast also depending on its electronic states. The finite gap permits transmitted probability current, whereas the semi-infinite step in part (b) does not.

For a real potential, the Time-dependent Schrödinger equation isMultiply this equation by ${\psi }^{\ast }$, multiply its conjugate by $\psi$, and subtract. The potential terms cancel and the product rule converts the remaining terms into a divergence. ConsequentlyThis is the probability current and its continuity equation.

Put $R(r)=\sin (kr)/r$. The supplied radial Laplacian formula gives ${\nabla }^{2}R=-k^{2}R$ for $r>0$. Also $i\hslash {\partial }_t\psi =E\psi$. Therefore the free-particle equation is satisfied whenThe apparent singularity at zero is removable: $R(r)=k-k^3{r}^2/6+\cdots$ extends smoothly there and satisfies the same equation. Since $R$ is real and the time dependence is only a common phase, ${\psi }^{\ast }\nabla \psi =R\nabla R$ is real, so $\mathbf{j}=0$. The density $R^2$ is time-independent, proving the continuity equation explicitly. This is a standing spherical wave with zero net radial current.

In the position representation the Time-dependent Schrödinger equation isFor a normalized state define probability density and probability current byMultiplying the equation by $\overline{\psi }$ and its complex conjugate by $\psi$, then subtracting, cancels the real potential. The resulting identity isThus $\boxed{{\partial }_t\rho +{\partial }_{x}j=0}$. Integrating over an interval gives $d{\int }_b^c\rho dx/dt=j(b)-j(c)$: probability changes only through flux across the endpoints. This is local conservation of quantum probability.

Continuity with the zero exterior imposes zero endpoint values on the infinite square well. Solving the stationary equation with these boundary conditions giveswith zero extension outside. The integral of ${\sin }^2(n\pi x/a)$ is $a/2$, which fixes the normalization. The initial coefficients are $3/5$ and $4/5$, whose squared moduli sum to one. Each energy component acquires its stationary phase:At the midpoint, ${\Psi }_1=\sqrt{2/a}$, ${\Psi }_1^{\prime }=0$, ${\Psi }_2=0$, ${\Psi }_2^{\prime }=-(2\pi /a)\sqrt{2/a}$. HenceTaking its imaginary part proves

For the Schrödinger equation with a real scalar potential and no vector potential, the probability current isTogether with the probability density $\rho =|\Psi {|}^2$, it satisfies the continuity equation ${\partial }_t\rho +\nabla \cdot \mathbf{j}=0$. Substituting ${\partial }_x\Psi =ik\Psi$ givesA complex wavefunction requires both its real and imaginary parts in the sketch. Choosing the irrelevant constant phase so that $A$ is real, they are $A\cos kx$ and $A\sin kx$, while the probability density is horizontal. It is the spatial phase gradient of a Schrodinger wavefunction ${\partial }_x\arg \Psi =k$ that determines the sign of the probability current. In particular, $e^{ikx}$ and $e^{-ikx}$ have the same real part but opposite probability currents, so the slope of the real part cannot determine the direction.

A nonzero plane wave on the whole real line is not square integrable. The displayed density is understood with box normalization, generalized normalization, or as a local description of a broad wave packet; it does not define a normalized probability distribution on the whole line.

The Time-dependent Schrödinger equation, with the Hamiltonian for a particle of mass $m$ in a real potential energy $V$, isA stationary state is an energy eigenstate: its wavefunction evolves by a single global phase,Its probability density is independent of time. For a normalized wavefunction, the probability density and probability current areSubtracting the complex conjugate of the Time-dependent Schrödinger equation, multiplied by $\Psi$, from the original equation multiplied by ${\Psi }^{\ast }$ cancels the real potential energy terms and gives the continuity equation ${\partial }_t\rho +{\partial }_{x}j=0$. Thus $j$ measures signed probability flow in the positive $x$ direction.

At the midpoint the second sine vanishes, whereas the derivative of the first sine vanishes. Thus the wavefunction and its spatial derivative simplify toSubstitution in the probability current givesThe continuity equation, integrated over the left half, gives $P_{\mathrm{L}}^{\prime }(t)=j(0,t)-j(a/2,t)$. The Dirichlet boundary condition makes $j(0,t)=0$, and the preceding probability formula givesPositive midpoint current removes probability from the left half. This is precisely the midpoint current of a two-mode box state.

Write $k=\pi /a$, $A=\sqrt{2/a}$ and ${\Psi }_0=A{e}^{ikx}\sin kx$. Its derivative isIntegration by parts with the Dirichlet boundary conditions gives the energy expectationIt is constant in time, since the time-independent Hamiltonian operator has fixed energy measurement probabilities: $⟨H{⟩}_t={\sum }_n{E}_n|c_n{|}^2$.

For the midpoint parity operator, $R{\Psi }_0=-A{e}^{-ikx}\sin kx$. Hence, again putting $u=kx$,Indeed the imaginary part integrates to zero by reflection and ${\int }_0^{\pi }\cos 2u{\sin }^{2}udu=-\pi /4$. This expectation value is also constant in time, because the bounded self-adjoint operator $R$ commutes with $H$ and hence with its time-evolution operator. Equivalently, $⟨R{⟩}_t={\sum }_n(-1{)}^{n+1}|c_n{|}^2$.

For the momentum operator, direct evaluation givesThe real sine-cosine integral vanishes. This expectation value is not constant in time: the infinite square well walls reverse motion. One can prove the change exactly without approximating the wavefunction. Atthe energy eigenstate expansion and $R{\psi }_n=(-1{)}^{n+1}{\psi }_n$ give $\Psi (t_{\ast })=-R{\Psi }_0$. Differentiation shows $PR=-RP$, soThis half revival in an infinite square well is a concrete counterexample to momentum conservation. The sign from $-R$ is only a global phase.

The apparent interior identity $[H,P]=0$ cannot justify conservation here: the momentum operator does not preserve the Dirichlet boundary conditions of the Hamiltonian operator's operator domain. For sufficiently regular states, retaining the boundary terms in Ehrenfest theorem instead gives the boundary force in an infinite square well,In the strictly confined interval, $-i\hslash {\partial }_x$ with both endpoint values zero is symmetric but not itself a self-adjoint operator. The requested integral is the conventional mean momentum; it also equals the full-line momentum mean of the zero-extended state, as measured upon releasing the walls. This domain distinction explains why the formal interior commutator misses the wall force.

Take an overdamped particle with viscous drag coefficient $\gamma$, so its deterministic drift in the harmonic potential $U=k{x}^2/2$ is $\dot{x}=-kx/\gamma$. The one-dimensional Fokker-Planck equation has probability currentAt thermal equilibrium, $J=0$ and the Boltzmann distribution is $p(x)=Z^{-1}{e}^{-k{x}^2/(2k_{B}T)}$, with $p^{\prime }/p=-kx/(k_{B}T)$. Inserting this into the zero-current condition givesThe result is the Einstein diffusion relation between diffusion and mobility. It assumes a bath at one temperature, linear drag and thermal equilibrium. The same Gaussian probability density gives $⟨x^2⟩=k_{B}T/k$, providing an independent calibration of trap stiffness.

Far from a short-range potential, the particle is locally free. The imposed incoming boundary condition is a unit-amplitude plane wave travelling along $+z$, while an outgoing free spherical wave has radial dependence $e^{ikr}/r$. The scattering coefficient can depend on direction, givingFor the Schrödinger probability current, $\mathbf{j}=(\hslash /m)\mathrm{Im}({\psi }^{\ast }\nabla \psi )$, the incident flux is $j_{\mathrm{in}}=\hslash k/m$. The outgoing scattered radial flux, separated from the incident beam, is $j_{\mathrm{sc},\mathrm{r}}\sim (\hslash k/m)|f_k{|}^2/r^2$. Thus the outgoing rate through $r^{2}d\Omega$, divided by incident flux, givesThis elastic-scattering result uses the same incoming and outgoing wavenumber and the usual outgoing-wave boundary condition.

For real $x=ka$, taking the modulus squared of the amplitude gives the half-line delta-well amplitude enhancementThe supplied resonance poles have real parts $x_n\simeq n\pi (1+\lambda /2+{\lambda }^2/4)$ and negative imaginary parts $-{\gamma }_n\simeq -(\lambda n\pi /2{)}^2$ for fixed low $n$ at $\lambda \ll 1$.^[ref] Linearizing the denominator at a simple pole givesThus the sketch has narrow, tall peaks with a locally Lorentzian line shape, just above positive integer multiples of $\pi$. Their half-widths are ${\gamma }_n$, and the leading peak heights are $4/({\lambda }^2{n}^2{\pi }^2)$. In this fixed-low-mode approximation, increasing $n$ broadens and lowers the peaks. Exactly at $x=n\pi \ne 0$, the value is one. At zero it has the finite limit ${\lambda }^2/(2-\lambda {)}^2$ for $\lambda \ne 2$.