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Use the Fourier transform convention $\hat{f}(s)={\int }_{\mathbb{R}}{e}^{-isx}f(x)dx$. First observe that $f(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}$ satisfies the differential equation, decay, and normalization, using the Gaussian integral. The uniqueness allowed in the question identifies it with $f$. In particular it is a Schwartz function, so the derivative and multiplication rules below are justified.

Taking the Fourier transform of the differential equation givesThe Riemann-Lebesgue lemma gives decay of $\hat{f}$. Every decaying solution $h$ of this equation is proportional to $e^{-s^2/2}$: setting $w=h^{\prime }+sh$ gives $w^{\prime }=sw$, and integration yieldsThe term with $C\ne 0$ grows like $C{e}^{s^2/2}/(2s)$ as $s\to +\infty$, so decay forces $C=0$. Thus $\hat{f}=\lambda f$ for a scalar $\lambda$. This establishes proportionality before using Fourier inversion to calculate the scalar.

On the small clockwise contour $z=ϵe^{i\theta }$, $\theta$ decreases from $\pi$ to $-\pi$. Uniformly on that arc, $e^{z-\sqrt{z}}=1+O(\sqrt{ϵ})$, and henceThe large arcs also vanish. To make this precise for the displayed geometry, take them on the left semicircle $z=\gamma +R{e}^{i\theta }$, $\pi /2\le \theta \le 3\pi /2$, split at the cut. On the principal holomorphic branch, $\mathrm{Re}\sqrt{z}\ge 0$, $|z|\ge R-\gamma$, and $\mathrm{Re}z=\gamma +R\cos \theta$. The estimation lemma givesFor the last bound, symmetry and $\sin u\ge 2u/\pi$ on $0\le u\le \pi /2$ imply that the angular integral is at most $\pi /R$. The exact choice of large arcs does not affect the limiting result. The vertical improper integral converges absolutely because $e^{-\mathrm{Re}\sqrt{\gamma +iy}}/|\gamma +iy|$ decays as $O(e^{-c\sqrt{|y|}}/|y|)$.

Taking limits in the closed contour integral now yields $2\pi iJ+2iK-2\pi i=0$, where $K={\int }_0^{\infty }{e}^{-x}\sin (\sqrt{x})dx/x$. ThusAs an optional explicit evaluation, put $x=t^2$ and define $K(a)=2{\int }_0^{\infty }{e}^{-t^2}\sin (at)dt/t$. Differentiation under the integral sign and the Gaussian integral give $K^{\prime }(a)=\sqrt{\pi }{e}^{-a^2/4}$ and $K(0)=0$. Therefore $K=K(1)=\pi \mathrm{erf}(1/2)$ and $J=\mathrm{erfc}(1/2)\simeq 0.479500$ in terms of the error function and complementary error function.

For constant $D$ and no drift, particle conservation and Fick's first law give the one-dimensional diffusion equation ${\partial }_{t}C=D{\partial }_x^{2}C$. For $C=A{t}^{-1/2}{e}^{-x^2/(4Dt)}$,so direct substitution verifies the equation for $t>0$. The Gaussian integral fixes the normalization:This is the heat kernel for a point release, with $C(x,0)=N\delta (x)$ in the distributional limit. It remains centered at zero while its width grows as $t^{1/2}$ and its peak falls as $t^{-1/2}$. To compute the second moment, set $y=x/\sqrt{4Dt}$:using $\int y^2{e}^{-y^2}dy=\sqrt{\pi }/2$ over the whole real line, obtained by integration by parts.

If the mean number of independent unbiased jumps in time $t$ is $⟨m⟩=ft$, conditioning on $m$ gives $⟨x^2(t)⟩=\Delta ⟨m⟩=f\Delta t$. Matching it to the continuum diffusion result yieldsFor fixed-magnitude steps $a$, $\Delta =a^2$. The equality of second moments also holds for Poisson jump times independent of the increments; a Gaussian continuum approximation additionally requires many steps and finite step variance.

The Gamma function identity $\Gamma (1/2)=\sqrt{\pi }$ gives the Bromwich integral of an inverse square root directly:Equivalently, put $x=u^2$ in the real integral. It becomes $2{\int }_0^{\infty }{e}^{-u^2}du=\sqrt{\pi }$ by the Gaussian integral.

For $t>0$, $B_t$ has the normal distribution $N(0,t)$. Evaluate its Gaussian integral:At $t=0$ the defining random variable is one, consistent with the same formula. Thus

Put $c=\beta /2-\alpha$. The change of measure and part (a)(iii) yieldFor a centered normal random variable $G$ of variance $v>0$, a Gaussian integral gives the Gaussian quadratic exponential momentHere $c\le \beta /2$ and $\beta v_t<1/2$ for finite positive $t$, so the required positivity holds, including when $c>0$. ThusExpanding the two exponentials in terms of hyperbolic functions givesAt $t=0$ this is one. As $\beta \downarrow 0$, the denominator tends to $1+2\alpha t$, recovering part (a)(i). At $\alpha =0$ it gives the Laplace transform of the integrated square of Brownian motion. The joint terminal-and-integral transform is the Joint Laplace transform of Brownian square and its time integral.

The normalized initial Gaussian wave packet is real and even, so the integrands for $⟨x^{\prime }{⟩}_0$ and $⟨p^{\prime }{⟩}_0$ are odd. Its second position moment is the elementary Gaussian integral $⟨x^{\prime 2}{⟩}_0=1/(2\alpha )$. Since ${\partial }_x\Psi =-\alpha x\Psi$ and ${\partial }_x^2\Psi =({\alpha }^2{x}^2-\alpha )\Psi$,Moreover $⟨x^{\prime }{p}^{\prime }{⟩}_0=i\hslash \alpha ⟨x^{\prime 2}{⟩}_0=i\hslash /2$ and $⟨p^{\prime }{x}^{\prime }{⟩}_0=-i\hslash /2$. Thus

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

Use the Dirac delta function to set $p^{\prime }=p+\alpha T$, and write $X=x-x^{\prime }$. The remaining momentum-dependent quantum phase isShifting the integration variable by $\alpha T/2$ reduces the integral to the free Gaussian integral. ThereforeThe force is $-\alpha$, not $+\alpha$. Both the linear endpoint term and the cubic-time term disappear in the zero-force limit. The prefactor remains the free one.

Specify the Fourier transform before performing the Gaussian integral:At $u=0$ the functional is diagonal in both component and momentum, with inverse kernels $q^2+t+r$, $q^2+t-r$, and $q^2+t$. The condition $0<r<t$ makes all three positive. The covariance of a Gaussian distribution with statistical weight $e^{-F/(k_{B}T)}$ is $k_{B}T$ times the inverse kernel, henceIf $F$ is already the dimensionless free energy divided by $k_{B}T$, the numerators are one instead. Translational invariance forces opposite momenta to pair. There is no complex conjugate on the second field, so the delta function is at $\mathbf{p}=-\mathbf{q}$, not at $\mathbf{p}=\mathbf{q}$. A correlation with $m_j(\mathbf{p}{)}^{\ast }$ would have the latter convention. Rotational symmetry in real space leaves dependence on $q^2$ alone once the momentum-conservation delta function is factored out. The correlation lengths are ${\xi }_1=(t+r{)}^{-1/2}$ and ${\xi }_2=(t-r{)}^{-1/2}$; the second component becomes soft as $r\uparrow t$. The skipped letter in the source is preserved: this solution is part (e), with no inferred part (d).

Keep the same measure normalization as in the free generating functional and defineHere $W$ is the source-dependent partition function, despite the common alternative convention in which that letter denotes its logarithm. A functional derivative of $e^{\int Jx}$ brings down one factor $x(\tau )$ at the differentiated point. Expanding the interaction exponential and replacing each such factor by $\delta /\delta J(\tau )$ therefore givesCompleting the square in the free Gaussian integral gives $W_0[J]=W_0[0]\exp (\frac{1}{2}\int J{G}^{(0)}J)$ because $G^{(0)}=D^{-1}$ with periodic boundary conditions. Restoring $\hslash$ replaces the upper limit by $\hslash \beta$ and the interaction coefficients by $\lambda /(4!\hslash )$ and $\mu /(6!\hslash )$, exactly as in the source.

For $N$ indistinguishable classical particles the canonical partition function is an integral over phase space,The factor $1/N!$ removes overcounting under particle permutations, and $h^{3N}$ makes the classical phase-space measure dimensionless. The pair potential depends only on positions, so all momentum integrations separate. The three-dimensional Gaussian integral for each particle givesThe grand canonical partition function is $\Xi ={\sum }_{N=0}^{\infty }{e}^{\beta \mu N}{Z}_N$, where chemical potential $\mu$ controls the mean number of particles. ThereforeFor $N=0$, the empty products and configurational integral equal one. The factor $1/2$ in the interaction prevents double counting of unordered particle pairs.

Normalize the probability density function on $x\ge 0$:For $y=\sqrt{x}\ge 0$, the change of variables is $x=y^2$, $dx=2ydy$. Conservation of probability gives the transformed probability density functionIt is a Rayleigh distribution with scale $\sigma =1/\sqrt{2\alpha }$. Its derivative is $g^{\prime }(y)=2\alpha e^{-\alpha y^2}(1-2\alpha y^2)$, so it rises from zero, has one maximum, and then decays:The expected value follows by putting $z=\sqrt{\alpha }y$:Here ${\int }_0^{\infty }{z}^2{e}^{-z^2}dz=\sqrt{\pi }/4$, obtained from the Gaussian integral by integration by parts. In particular $⟨\sqrt{x}⟩\ne \sqrt{⟨x⟩}$, and the mode changes under a nonlinear variable transformation because a density includes its Jacobian determinant.

A phase-space path integral represents a quantum transition kernel by integrating both coordinates and conjugate momenta with action $\int (p\dot{q}-H)dt$. It follows from inserting position and momentum completeness relations between short-time evolution operators. For a quadratic momentum dependence the momentum variables can be integrated by a Gaussian integral.

The quantum harmonic oscillator describes a particle of mass $m>0$ in a quadratic potential energy with angular frequency $\omega >0$. In the position representation its Quantum Hamiltonian isThe model is useful far beyond a literal spring: a Taylor series about a stable equilibrium with positive potential curvature has this quadratic term as its leading nonconstant approximation. The pure quartic oscillator has zero curvature at its minimum, so that particular approximation does not apply. ^[ref].

In units with $\hslash =1$, creation and annihilation operators giveHere $N$ is the number operator. Restoring the reduced Planck constant, the energy eigenvalues are $E_n=\hslash \omega (n+1/2)$ for $n=0,1,2,\dots$: neighboring levels are separated by $\hslash \omega$, and the ground-state energy is the zero-point energy $\hslash \omega /2$.

With length scale $\ell =\sqrt{\hslash /(m\omega )}$ and dimensionless coordinate $u=x/\ell$, the ground state is the normalized Gaussian wave packetFor $\hslash =m=\omega =1$, this is precisely the $a=1$ trial wavefunction in the quantum variational principle. The balance of kinetic energy and potential energy therefore recovers the exact ground-state energy using only Gaussian integrals.

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

The quantum rotor with angle period $2\pi$ has both the spectral kernel $(2\pi {)}^{-1}{\sum }_n{e}^{in\Delta -i\hslash n^{2}T/(2I)}$ and the displayed winding representation. The Poisson summation formula converts one into the other by a Gaussian integral. Each winding sector has a constant-velocity classical path and action $I(\Delta +2\pi m{)}^2/(2T)$. Quadratic fluctuations give the same prefactor in every sector, so this is exact. Real time is understood by $T\to T-i0$, which fixes the square-root branch and convergence prescription.

For $\alpha >0$, polar coordinates reduce the integral to $2\pi {\int }_0^{\infty }r{e}^{-(\alpha -i\lambda )r^2}dr$. The result is $\pi /(\alpha -i\lambda )$. Its damped limit for real $\lambda >0$ is $i\pi /\lambda$, which fixes the sign of the oscillatory complex Gaussian integral. The Fresnel integral convention can be compared with NIST definitions.

The Schrödinger equation governs the evolution of a quantum state between instances of measurement in quantum mechanics. For a particle of mass $m>0$ moving in a prescribed real potential energy $V(\mathbf{r},t)$, its position wavefunction satisfiesHere $h$ is the Planck constant, ${\nabla }^2$ is the Laplace operator, and the Hamiltonian operator combines kinetic energy with potential energy. Specify an initial wavefunction, an operator domain and any boundary conditions: the differential expression alone does not completely specify a physical model. The Born rule then turns the evolving probability amplitude into position probabilities through $\rho =|\psi {|}^2$ and integration over a region.

This scalar Schrödinger operator describes nonrelativistic quantum mechanics, with relevant momenta much smaller than $mc$, where $c$ is the speed of light. It omits the constant rest energy $m{c}^2$, assumes fixed particle number, and suppresses active spin dynamics and coupling to a magnetic vector potential. A free electron with a fixed, uncoupled spin state can still use this scalar model; the Pauli equation supplies the additional spin components and magnetic coupling when needed. A prescribed real $V$ describes a closed evolution model or an approximation in which the source of that potential energy is treated as external. Dissipation, particle creation and a general measurement in quantum mechanics need further modelling.

For a time-independent self-adjoint Hamiltonian operator, the spectral theorem defines unitary time evolution byInitial states in the operator domain satisfy the differential equation in the Hilbert space sense; the unitary operator also evolves all other states in that Hilbert space. Time-dependent Hamiltonian operators need suitable regularity and domain hypotheses; their evolution is not generally the exponential of $-itH(t)/\hslash$.

The local reason for conservation is especially tangible. Multiply the Schrödinger equation by the complex conjugate ${\psi }^{\ast }$, subtract its conjugate multiplied by $\psi$, and use the fact that the real potential energy terms cancel. For smooth wavefunctions this gives the probability continuity equationThe divergence theorem identifies the change in probability inside a region with minus the outward probability current. Decay at infinity, or appropriate probability-conserving boundary conditions, preserves wavefunction normalization. Conservation of total probability does not require a stationary probability density.

Its usefulness comes from being a linear partial differential equation: quantum superpositions of solutions are solutions, while different quantum phases can produce interference. For a fixed Hamiltonian operator, the time-independent Schrödinger equation finds the energy eigenstates; each then acquires just its own time-dependent quantum phase. One eigenvalue problem therefore organizes both allowed energies and the evolution of arbitrary quantum superpositions, with a continuous spectrum included when required. This is the attractive bridge between a spatial wave problem and a clock: energy differences determine the rate at which relative quantum phases change. It makes atomic spectral lines, quantum tunnelling and quantum scattering calculable from a specified model.

For an explicit example without confinement, take a free particle on the whole line, $V=0$, initially in a Gaussian wave packet with position standard deviation ${\sigma }_0$ and mean momentum $\hslash k_0$:Unlike a single plane wave, this wavefunction satisfies wavefunction normalization. A Fourier transform writes it as a quantum superposition of plane waves. Their dispersion relation is $\omega (k)=\hslash k^2/(2m)$, so each Fourier component gains the factor $e^{-i\hslash k^{2}t/(2m)}$. Completing the square in the resulting Gaussian integral giveswith the square root chosen continuously from $t=0$. Its position probability density and standard deviation areThe centre moves with the group velocity ${\omega }^{\prime }(k_0)=v_0$ and the wave packet spreads because different wavenumbers have different group velocities. This is unitary wave dispersion, without a random force or loss of probability. Independently, the initial momentum standard deviation is $\Delta p=\hslash /(2{\sigma }_0)$ and the initial position-momentum covariance vanishes. Free evolution gives $x(t)=x(0)+p(0)t/m$ in the Heisenberg picture, henceThis independently checks the spreading law. The Gaussian integral of $\rho$ is one at every time; the decreasing peak is exactly offset by increasing width.

As hypothetical preparation parameters, choose a free electron with ${\sigma }_0=1\mathrm{nm}$ and $k_0=2{\mathrm{nm}}^{-1}$. Using $\hslash \simeq 1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$ and $m\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ givesThe rounded electron mass uses the 2022 adjustment.^[ref] At $t=t_s$, the centre is $4\mathrm{nm}$ and ${\sigma }_t=\sqrt{2}\mathrm{nm}$; at $t=2t_s$, they are $8\mathrm{nm}$ and $\sqrt{5}\mathrm{nm}$. The mean kinetic energy staysIt includes both mean motion and the momentum spread. The mean speed is about $7.72\times 10^{-4}c$ and the momentum spread corresponds to $1.93\times 10^{-4}c$, making the nonrelativistic model accurate for the appreciably occupied part of this ideal Gaussian distribution. Its mathematically unbounded tails are an idealization.

The complex width also explains where the probability goes. Write $\psi =\sqrt{\rho }{e}^{i\theta }$, where $\theta$ is the dimensionless quantum phase. Differentiating the exact solution givesRelative to the moving centre, the probability current $j_{\mathrm{rel}}=j-v_0\rho$ points outward on both sides for $t>0$. This is a flux statement, not an additional assumption about individual particle trajectories. At $t=t_s$ and $x-v_{0}t=\pm {\sigma }_0$, the relative flow speeds $j_{\mathrm{rel}}/\rho$ are $\pm {\sigma }_0/(2t_s)$, about $\pm 2.8942\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$. The probability within a moving interval of half-width ${\sigma }_0$ falls from about $0.6827$ initially to $0.5205$ at $t_s$, exactly because of this outward probability current.

Historically, the 1926 Quantisierung als Eigenwertproblem (Erste Mitteilung) by Erwin Schrödinger began with the stationary hydrogen atom problem. It distinguishes discrete negative energies from a positive continuous spectrum; it is not yet the general evolution equation displayed above.^[ref] The opening of quantum mechanics discusses its original vibrating-string analogy.

In the 12 December 1933 lecture The fundamental idea of wave mechanics by Erwin Schrödinger, Erwin Schrödinger explains the move from independent rays to waves through diffraction, then applies that analogy to atomic physics. His closing pages explicitly acknowledge the difficulty of reconciling wave language with observed particle tracks. This retrospective physical argument motivates the model; it does not derive the modern Born rule from classical mechanics.^[ref]

The dates separate proposal from measurement: the de Broglie wavelength belongs to the 1924 thesis, the Schrödinger equation to 1926, and the Davisson–Germer experiment's electron diffraction beams to 1927.^[ref] That linked experiment describes the prepared nickel crystal, evacuated electron gun, angular collector scans, retarding-potential selection and heating controls. Its measured collector electric current tests wave-dependent scattering probabilities; it neither measures a wavefunction directly nor establishes this equation uniquely.

For $\mathrm{Re}A>0$ and $\mathrm{Re}v>0$, rescaling the defining gamma function integral proves this representation, with compatible complex-power branches. It turns denominator powers into exponential factors, making Gaussian integrals possible. Combining two parameters by their sum and ratio gives the usual Feynman parameter identity. Outside its convergence domain a further continuation or pole prescription is needed.