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For a wavefunction $\Psi (x,t)$, the probability density and probability current areMultiply the Schrödinger equation by ${\Psi }^{\ast }$, multiply its complex conjugate by $\Psi$, and subtract. The real potential energy terms cancel, leavingwhich is the probability continuity equation ${\rho }_t+j_x=0$.

For a stationary state, ${\rho }_t=0$, so $j_x=0$ and $j$ is constant in space. A normalizable wavefunction has vanishing current at spatial infinity, forcing this constant to be zero. By contrast the nonnormalizable plane wavehas stationary density $|A{|}^2$ and nonzero currentwhen $k\ne 0$.

A stationary state has the formFor a time-independent quantum observable $Q$, the global phases cancel in the expectation value:It is therefore independent of time.

For the infinite square well,The possible energy measurement results arewith respective Born rule probabilities $|c_1{|}^2$ and $|c_2{|}^2$. Normalization gives $|c_1{|}^2+|c_2{|}^2=1$.

At time $t$,The required matrix elements areConsequentlywhere the angular frequency isFinally, $2|c_1{c}_2|\le |c_1{|}^2+|c_2{|}^2=1$, so

For a one-dimensional wavefunction,The probability continuity equation isFor a stationary state $\psi (x,t)=\varphi (x)e^{-iEt/\hslash }$, the probability density is independent of time, so ${\partial }_{x}j=0$ and the probability current is constant in space.

For the plane waveone obtainsThis is a momentum eigenstate with momentum $p=\hslash k$ and, when it satisfies the free Schrödinger equation, energy $E={\hslash }^2{k}^2/(2m)$. Its constant density and current describe a spatially uniform beam carrying probability in the sign of $k$. It is not a normalizable wavefunction, so it represents an idealized state rather than a localized particle.

The probability density and probability current areFor a stationary state, $\psi (x,t)=\chi (x)e^{-iEt/\hslash }$, so $\rho =|\chi {|}^2$ is independent of time. The continuity equation then gives $\partial J/\partial x=0$. Hence

For a plane wave $A{e}^{i(kx-Et/\hslash )}$, the probability current isThe incident amplitude is one and the transmitted amplitude is $T$, soThis is the transmission probability for quantum tunnelling. In the stationary state, $\partial \rho /\partial t=0$, so the continuity equation makes the net current independent of position. The reflected current therefore supplies the remainder:

Only odd stationary states contribute, and for $j=2n+1$ their squared coefficients are $128/[{\pi }^2(j^2-16{)}^2]$. Since $E_{av}/E_1=16$, substitution givesso $B=8$.

$\rho =|\psi {|}^2$ and $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }_x)$. Schrödinger’s equation and its conjugate give ${\rho }_t+j_x=0$; a stationary state has $j_x=0$. Continuity of $\psi ,\psi \prime$ at $0,a$ and matching plane waves giveso $F=1/3$ and $G=a\sqrt{3m}/\hslash$. Transmission is minimised when $G\sqrt{U_0}=\pi /2+n\pi$, i.e. $U_0=(\pi /2+n\pi {)}^2/G^2$.

The wave function vanishes in regions I and III. In region II it is sinusoidal and satisfies $\psi (0)=\psi (L)=0$. The time-independent equation there isA nonzero stationary state therefore has standing-wave nodes at the two walls; it cannot have a nonzero constant ground-state wave function.

The Time-dependent Schrödinger equation isA stationary state of energy $E$ has a wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, whereIts time dependence is only a global phase, and its probability density is independent of time.

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 Time-dependent Schrödinger equation isA stationary state of energy $E$ has a wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$, whereIts time dependence is only a global phase, and its probability density is independent of time.

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.

A pure stationary state of a time-independent Hamiltonian operator is $\Psi (\mathbf{r},t)=\psi (\mathbf{r})e^{-iEt/\hslash }$. Its probability density and all expectations of time-independent observables are constant. This does not require zero probability current.

For real scalar potential energy and no vector potential, substitute $i\hslash \dot{\Psi }=-({\hslash }^2/2m){\nabla }^2\Psi +V\Psi$ and its complex conjugate into $\dot{\rho }={\Psi }^{\ast }\dot{\Psi }+{\dot{\Psi }}^{\ast }\Psi$. The potential terms cancel:Thus the probability current is $\mathbf{j}=(\hslash /m)\mathrm{Im}({\Psi }^{\ast }\nabla \Psi )$. In a stationary state $\nabla \cdot \mathbf{j}=0$: flow may circulate without changing the density.

Here only the azimuthal phase contributes an imaginary logarithmic derivative. Since $\nabla \varphi =\hat{\mathbf{\varphi }}/(r\sin \theta )$,There is no radial or polar flow; the azimuthal current is independent of $\varphi$, so its divergence vanishes. The apparent coordinate singularity at $\sin \theta =0$ cancels in the final expression.

The z-component of orbital angular momentum follows from the moment of the mass current:Indeed $A^2(24a_0^5)(4/3)(2\pi )=1$. This agrees with $L_z=-i\hslash {\partial }_{\varphi }$ acting on the wavefunction.

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.

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.

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.

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.

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 issue is the square-integrable condition on the whole real line, not just decay at one end. Because the quantum harmonic oscillator potential is even, the even and odd parts of a square-integrable wavefunction are themselves square integrable and satisfy the same equation. It is therefore enough to examine each parity sequence separately.

For a nonterminating sequence, put $b_j=a_{2j+p}$, where $p=0$ or $1$. Its recurrence relation givesSince $ϵ$ is real for an energy eigenvalue, the ratio is eventually positive. After multiplying by one constant phase, the tail coefficients are all positive. Choose $0<\delta <1/2$. For sufficiently large $j$, the ratio is at least $(1-\delta )/(j+1)$, so for some $K>0$ the tail coefficients obey $b_j\ge K(1-\delta {)}^j/j!$. Consequently, for $\xi >0$, the magnitude of $f$ eventually exceeds a positive multiple of ${\xi }^p{e}^{(1-\delta ){\xi }^2}$ after subtraction of a fixed polynomial. Multiplication by $e^{-{\xi }^2/2}$ leaves exponential growth like $e^{(1/2-\delta ){\xi }^2}$, incompatible with the square-integrable condition. Thus each nonzero parity series must terminate. The parity argument also excludes cancellation between the two nonterminating sequences at both infinities. This proves Normalization forces Hermite-series termination.

If the terminating polynomial has degree $n$, its top nonzero coefficient requires $2n+1-ϵ=0$. Conversely, when $ϵ=2n+1$, choosing the seed of parity $n$ nonzero makes the recurrence relation terminate exactly at degree $n$. The opposite seed must be zero, since that sequence does not terminate. ThereforeThe resulting polynomials are proportional to the physicists' Hermite polynomials, and ${\psi }_n\propto H_n(\xi )e^{-{\xi }^2/2}$ is normalizable because any polynomial times this Gaussian function is square integrable. The two parity seeds describe the full solution space of the second-order differential equation; the preceding argument therefore exhausts the normalizable stationary states. Each eigenspace has dimension one. General time-dependent quantum states can of course be superpositions of these energy eigenstates.