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The interaction picture removes the evolution generated by $H_0$: $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$. Differentiating the Schrodinger equation gives $i\hslash {\partial }_t|{\psi }_I⟩=\lambda V_I(t)|{\psi }_I⟩$, with $V_I=e^{i{H}_{0}t/\hslash }V(t)e^{-i{H}_{0}t/\hslash }$. The integral equation, iterated once, givesDistinct-energy eigenstates are orthogonal. Their transition amplitude is therefore $-i\lambda {\hslash }^{-1}{\int }_0^t⟨\varphi |V(s)|\chi ⟩e^{i(E^{\prime }-E)s/\hslash }ds+O({\lambda }^2)$. Squaring its modulus givesThe first-order amplitude produces a second-order probability.

For the isotropic quantum harmonic oscillator, parity makes $\int x_i{x}_3|\chi {|}^2{d}^{3}x=0$ for $i=1,2$. Rotational symmetry and the given radial integral give $\int x_3^2|\chi {|}^2{d}^{3}x=\hslash /(2m\omega )$. Thus $⟨{\varphi }_i|{\hat{x}}_3|\chi ⟩={\delta }_{i3}\sqrt{\hslash /(2m\omega )}$, and the energy gap is $\hslash \omega$. The time integral is ${\int }_0^{\infty }{e}^{(-\mu +i\omega )s}ds=(\mu -i\omega {)}^{-1}$. ConsequentlyThe perturbation picks the excitation polarized along the third coordinate, rather than an arbitrary member of the degenerate first excited level.

Use the physical Coulomb bound state boundary condition: the spherically symmetric wavefunction is regular at the origin and square integrable with measure $4\pi r^{2}dr$. For $E<0$, setAt large $r$, the decaying branch behaves exponentially as $e^{-Kr}$. Substituting $\psi =f{e}^{-Kr}$ into the radial Schrodinger equation givesThe regular solution has a power series $f(r)={\sum }_{j\ge 0}{a}_j{r}^j$, with recurrenceA nonzero regular solution has $a_0\ne 0$. If the series does not terminate, its coefficients eventually have one sign after choosing an overall real phase. For any fixed $0<\delta <1/2$ and sufficiently large $j$, the ratio of consecutive magnitudes is at least $2K(1-\delta )/(j+2)$. The tail therefore grows at least as a positive constant times $e^{2K(1-\delta )r}$ divided by a power of $r$; a finite polynomial of earlier terms cannot cancel it. Multiplication by $e^{-Kr}$ leaves exponential growth, contradicting normalizability.

Thus the series must terminate at degree $N-1$, where $KN=\lambda$ and $N$ is a positive integer. Conversely, termination gives a regular polynomial times $e^{-\lambda r/N}$, which is normalizable. ThereforeThese are the $\ell =0$ members of the hydrogen spectrum. Regularity at the origin is essential: square integrability alone on the punctured radial interval would also admit a singular $1/r$ branch. Such a branch is not a physical eigenfunction of the ordinary three-dimensional Coulomb Hamiltonian; its singularity introduces an inadmissible point-source contribution. In the reduced radial variable $u=r\psi$, the physical condition is $u(0)=0$.

For a particle of mass $m$, the probability density and one-dimensional probability current areThe Schrodinger equation and its complex conjugate, for real $V$, giveThe potential terms cancel precisely because $V$ is real. Thus ${\rho }_t+j_x=0$, the continuity equation for probability. Integrating over an interval says that its probability changes by the incoming current minus the outgoing current.

The sudden approximation keeps the wavefunction unchanged at the instant of switching: a finite step of potential creates no delta-function impulse in the Schrodinger equation. Just before switching the old ground state differs from the printed real wavefunction only by an overall phase, which does not affect probabilities.

Let ${\tilde{\psi }}_0(x)=(\pi \hslash {)}^{-1/4}{e}^{-(x-d{)}^2/(2\hslash )}$, $d=qE$, be the new ground state. Its overlap with the initial state is, up to that phase,Since $x^2+(x-d{)}^2=2(x-d/2{)}^2+d^2/2$, the Gaussian integral gives $A_0=e^{-d^2/(4\hslash )}$. The Born rule and the ground-state overlap after sudden oscillator displacement thus giveThis is an overlap probability, not a claim that the original wavefunction instantaneously relaxes to the new ground state. It becomes one when $qE=0$.

Inside the infinite square well, the time-independent Schrodinger equation is $-({\hslash }^2/(2m)){\psi }^{\prime \prime }=E\psi$. The infinite exterior potential imposes $\psi (0)=\psi (a)=0$. For $E>0$, the first condition leaves a sine and the second requires $ka=n\pi$. A zero or negative energy gives only the zero solution under these conditions. The normalized energy eigenstates and eigenvalues areThe wavefunctions vanish outside the well, and a constant phase is immaterial. Normalization uses ${\int }_0^a{\sin }^2(n\pi x/a)dx=a/2$.

The wavefunction is a plane wave and a simultaneous formal eigenfunction of the momentum operator and the time-translation energy operator:Its momentum and energy are therefore definite, while its probability density is spatially uniform:For a particle of mass $m$, the probability current is $j=(\hslash /m)\mathrm{Im}(\overline{\varphi }{\partial }_x\varphi )=\hslash k{A}^2/m$, directed according to the sign of $k$.

On the whole real line, a nonzero constant-amplitude plane wave is not square-integrable and cannot be normalised to total probability one. It represents an idealised momentum eigenstate, interpreted through generalized eigenfunctions or as a limit of wave packets. Also, arbitrary $k,E$ do not automatically solve the Schrodinger equation for a specified potential energy: for a free particle, $E={\hslash }^2{k}^2/(2m)$, or this value plus the constant potential energy for a constant-potential region. If $A=0$, the zero function is not a physical quantum state.

A Hermitian operator $A$ satisfies $⟨\phi ,A\psi ⟩=⟨A\phi ,\psi ⟩$ on its operator domain. HereTwo integration by parts operations, together with the boundary conditions in the Hamiltonian domain, show that the kinetic term is Hermitian; multiplication by the real function $V$ is Hermitian as well. Thus $H$ is Hermitian.

For a time-independent operator $A$, the Schrodinger equation and its adjoint give the Ehrenfest theoremUsing the canonical commutation relation $[x,p]=i\hslash$,Consequently

The starting point of a conforming finite element method is a weak formulation in a Hilbert space $X$. For example, a homogeneous Dirichlet boundary condition on a bounded Lipschitz domain gives $X=H_0^1(\Omega )$. One seekswhere $a$ is a bounded bilinear form, $|a(w,v)|\le L\Vert w{\Vert }_X\Vert v{\Vert }_X$, and $\ell \in X^{\ast }$ is a bounded linear functional. A conforming finite element space $X_h\subset X$ is finite dimensional and is typically built from functions that are polynomial on each element of a finite element mesh. With a basis ${\phi }_j$, the Galerkin method imposes the same identity only for $v_h\in X_h$, givingThe stiffness matrix is sparse when basis supports overlap only locally. Local element integrals assemble its entries; the mass matrix represents an $L^2$ term. This separates the choice of trial functions from the variational principle determining their coefficients.

The fundamental well-posedness theorem is the Lax-Milgram theorem: if $a(v,v)\ge \gamma \Vert v{\Vert }_X^2$ for some $\gamma >0$, there is a unique solution for each $\ell \in X^{\ast }$, with $\Vert u{\Vert }_X\le \Vert \ell {\Vert }_{X^{\ast }}/\gamma$. The same theorem on $X_h$ gives a unique discrete solution. The constants must be independent of the mesh for uniform stability of a numerical method. Boundary conditions are built into the trial space when they are essential constraints; flux conditions arise through boundary terms in integration by parts and are natural constraints in the weak formulation.

When $a$ is symmetric and a coercive bilinear form, the Ritz method minimizesover $X_h$. Its first variation is exactly the Galerkin method, so Ritz minimization and Galerkin testing coincide for symmetric coercive problems. The energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ turns the approximation into an orthogonal projection. Indeed, subtracting the exact and discrete weak equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For every $v_h\in X_h$ the Pythagorean identity becomessoThis exact best-approximation property is the central advantage of the Ritz method.

Symmetry is unnecessary for the Galerkin method. For any bounded, coercive form, the Céa lemma givesTo prove it, put $e=u-u_h$. Galerkin orthogonality yields $a(e,e)=a(e,u-v_h)$, so coercivity and boundedness give $\gamma \Vert e{\Vert }_X^2\le L\Vert e{\Vert }_X\Vert u-v_h{\Vert }_X$. Divide when $e\ne 0$ and take the infimum. Thus approximation capability plus uniform coercivity gives convergence. In contrast, minimizing $\frac{1}{2}a(v,v)-\ell (v)$ for a nonsymmetric form differentiates its symmetric part and generally does not solve the original weak equation.

A concrete symmetric example is the one-dimensional reaction-diffusion problem from Question 5. Continuous piecewise-linear hat functions give the diffusion stiffness matrix $S$ with diagonal $2/h$ and neighbouring entries $-1/h$, and the mass matrix $M$ with diagonal $2h/3$ and neighbouring entries $h/6$. Therefore $K=S+M$ and $F_i=i{h}^2$. Its positive-definite matrix is the coordinate form of the continuous energy. The exact smooth solution $u(x)=x-\sinh (x)/\sinh 1$ makes this an explicit test of the method, not just an abstract existence result.

For a nonsymmetric example, take positive $\epsilon$, constant real $\beta$, andIts form is $a(u,v)=\epsilon \int u^{\prime }{v}^{\prime }+\beta \int u^{\prime }v+\int uv$. Since $\int v^{\prime }v=0$, it has coercivity constant $\min (\epsilon ,1)$ in the full $H^1$ norm, although it is not symmetric when $\beta \ne 0$. The Galerkin method remains well posed. On the uniform hat basis, $K=\epsilon S+\beta C+M$, where $C_{i,i+1}=1/2$, $C_{i,i-1}=-1/2$ and $C_{ii}=0$. The skew-symmetric matrix $C$ cancels from $c^{T}Kc$, proving nonsingularity. A Ritz energy would omit precisely this advection contribution. Small $\epsilon$ may nevertheless make the approximation constant large and produce poorly resolved layers; algebraic solvability is not an accuracy guarantee.

Quantitative convergence uses a finite element interpolation estimate. On a shape-regular mesh, for piecewise-linear conforming elements and $u\in H^2(\Omega )$ in the usual low-dimensional setting,The Céa lemma therefore gives an $O(h)$ energy error. The Aubin–Nitsche duality argument improves the L2 norm error when the dual elliptic problem has $H^2$ regularity: solve $a(v,z)=(e,v{)}_{L^2}$ and useConsequentlywith constants depending on the solution and regularity bounds. The second estimate is conditional on dual regularity; corners or insufficient data regularity can reduce the rate.

The same Galerkin method handles evolution equations. For the homogeneous heat equation, the semidiscrete finite element heat equation is $M\dot{c}+Sc=0$. Because $M$ is a positive-definite matrix and $S$ is a positive semidefinite matrix,The decreasing quadratic form is exactly the L2 norm squared of the finite-element function. This separates spatial stability of a numerical method from the subsequent time integrator, just as in the Schrodinger equation example. Conformity, coercivity and approximation estimates together explain convergence.

Choose any basis of two independent local solutions and collect their coefficients into a column $c_n$ immediately to the right of the $n$th cell boundary. The Floquet matrix is the one-period matrix $F(E)$ defined byEquivalently, it advances the Cauchy data $(\psi ,{\psi }^{\prime }{)}^T$ by one period. The Wronskian of two solutions of the stationary Schrodinger equation is constant between the delta functions, and the derivative jump at each delta function also has determinant one. ConsequentlyFor a real potential the Floquet matrix in the real Cauchy-data basis is real, so its trace is real in every basis. Its Floquet multipliers ${\rho }_{\pm }$ satisfyBy Bloch theorem, an energy is in the interior of an allowed energy band when ${\rho }_{\pm }=e^{\pm iqa}$ are distinct unit-modulus numbers. Since $\mathrm{tr}F$ is real, this is equivalent to

For the attractive delta-comb Kronig-Penney model, integrating the Schrodinger equation across a lattice point givesAt negative energy write $E=-{\hslash }^2{\mu }^2/(2m)$ and use the local basis $e^{\pm \mu (x-na)}$. Free propagation through one cell followed by the derivative jump givesThusThe band condition is thereforeUsing $(\cosh u-1)/\sinh u=\tanh (u/2)$ and $(\cosh u+1)/\sinh u=\coth (u/2)$ gives the requested inequality

At positive energy write $E={\hslash }^2{k}^2/(2m)$ and use $e^{\pm ik(x-na)}$. The corresponding matrix issoHence the positive-energy bands are exactly the values of $k>0$ for whichFinally, the zero-energy limit has $\frac{1}{2}\mathrm{tr}F=1-\lambda a$. If $\lambda a>2$, then $|1-\lambda a|>1$, and therefore zero energy lies in a forbidden gap.

Inside the infinite square well, the stationary Schrodinger equation for unit mass isIts normalized eigenstates and energies areThe general normalized time-dependent solution is

Immediately after removing the barrier, the normalized wavefunction isIts Fourier sine series coefficients in the full-well basis areThey areThe Born rule therefore gives

For a wavefunction $\Psi (x,t)$, the probability density and probability current areMultiply the Schrodinger 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 $k\ne 0$.

Write $T=\tanh (ax)$ and $S={\mathrm{sech}}^2(ax)$. Usingtwo differentiations ofgiveThus it solves the time-independent Schrodinger equation in the stated Pöschl-Teller potential, with

As $x\to -\infty$ and $x\to +\infty$, respectively,There is no left-moving term, so this is a reflectionless potential:The incident and transmitted amplitudes have equal modulus becauseHence

Now set $k=i\lambda$ with $\lambda >0$. ThenAt $+\infty$ this decays for every positive $\lambda$. At $-\infty$, however, the factor tends to $\lambda -a$ while $e^{-\lambda x}$ grows. Normalizability therefore requires $\lambda =a$. In that casewhich decays at both ends. The bound-state energy is

For a real short-range spherically symmetric potential, the stationary wavefunction must solve the Schrodinger equation, be regular at the origin whenever the potential is finite there, and satisfy the scattering boundary conditionHere $e^{ikz}$ is the incident plane wave, the second term is an outgoing spherical wave, and its coefficient $f(\theta )$ is the scattering amplitude.

Comparing the stated asymptotic expression with the partial-wave expansion of a scattering amplitude giveswhere $P_l$ is a Legendre polynomial. The outgoing-to-incoming coefficient in channel $l$ isPartial-wave unitarity requiresor equivalentlyThe real scattering phase shift ${\delta }_l$ is defined modulo $\pi$ by

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 Schrodinger 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 equation can be writtenFor the squared L2 norm, periodic integration by parts and the reality of $V$ giveThus the Schrodinger equation generates a unitary flow and $\Vert u(\cdot ,t){\Vert }_2$ is constant.

Inside the infinite square well, the time-independent Schrodinger equation has Dirichlet boundary conditions at $0$ and $a$. Its normalized energy eigenstates and energy eigenvalues are

Substitute $g=r^a{e}^{-br}$ into the radial Schrodinger equation and equate the coefficients of $r^{-2}$, $r^{-1}$, and the constant term. Regularity selectsThe corresponding bound-state energy is

Let $q=({\mathbf{q}}_1,\dots ,{\mathbf{q}}_N)$ denote the particle positions. In the Ghirardi--Rimini--Weber model, the wavefunction obeys the ordinary Schrodinger equationbetween random collapses. Each particle $i$ has an independent Poisson process of collapse times with rate $\lambda$. At such a time the state jumps according towhere the random centre has probability density $\Vert L_i(\mathbf{X})\psi {\Vert }^2$. The normalization is arranged so that $\int d^{3}X{L}_i(\mathbf{X}{)}^2=I$.

The original GRW scales are approximatelyAn isolated microscopic particle is therefore exceedingly unlikely to collapse during a laboratory experiment. A macroscopic pointer containing about $N\sim 10^{23}$ relevant particles has total collapse rate $N\lambda \sim 10^7{\mathrm{s}}^{-1}$ and collapse time about $10^{-7}\mathrm{s}$. After a measurement interaction correlates different microscopic outcomes with pointer positions separated by much more than $r_C$, one constituent's localization suppresses all incompatible pointer branches. This GRW amplification mechanism produces one definite macroscopic outcome with probabilities given by the Born rule, while leaving ordinary microscopic unitary time evolution almost unchanged.

For one spatial coordinate, average over the random centre of one collapse. The resulting density operator has position-space kernelIts diagonal is unchanged, so $⟨x⟩$ and $⟨x^2⟩$ are unchanged. The first derivative of the Gaussian factor vanishes at $x=x^{\prime }$, so $⟨p⟩$ is also unchanged. Its second derivative does not vanish, and the position representation of the momentum operator givesIn three dimensions the increase in total ${\mathbf{p}}^2$ is $3{\hslash }^2/(2r_C^2)$, corresponding to kinetic-energy increase $3{\hslash }^2/(4m{r}_C^2)$ per collapse. These are ensemble statements: conditioning on one specified collapse centre can shift the position moments. Repetition at rate $\lambda$ predicts GRW spontaneous heating, so precision searches for anomalous bulk heating, spontaneous radiation, momentum diffusion, and loss of matter-wave interference test the model.

Write $H={\partial }_x^2-V(x)$. The real potential and periodic boundary condition make $H$ a self-adjoint operator on the periodic domain. Since $u_t=-iHu$,because the inner product $⟨u,Hu⟩$ is real. Thus the Schrodinger equation preserves the $L^2$ norm.