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The probability density is even, so its first moment vanishes. Its second moment follows by integrating the polynomial:Using the normalization from part a givesThe variance in position is consequently $a^2/7$.

Only $A_z$ depends on position. Taking the curl,The field reverses across $x=0$. Its divergence is zero, as required for a magnetic field; the vanishing field on the midplane is important for the crossing and separatrix trajectories below.

Since $p_y=0$ initially and remains constant, the motion stays in a plane of constant $y$. Setting $P=\sqrt{2m{H}_0}$ and $p=p_{z0}$ givesThe allowed interval obeys $|p+a{x}^2|\le P$. The sign of $\dot{x}$ is set by the initial direction and changes at a turning point. The specified $x_0$ fixes the starting position on the energy orbit, but the data do not specify which sign of the initial transverse velocity is chosen.

We use the equilibrium generalized Langevin equation and its generalized fluctuation-dissipation relation, with the factor $5$ in the noise autocorrelation fixing the equilibrium velocity variance. Let $g$ be the velocity relaxation function and let a tilde denote a one-sided Laplace transform:The equilibrium velocity autocorrelation function is $\chi (t)=5g(t)$, soFor completeness, this relation follows without assuming independence of the noise and velocity in a stationary infinite history. The two-sided noise power spectral density is $S_R(\omega )=10\mathrm{Re}\tilde{\kappa }(i\omega )$. The stable linear equation has velocity transfer function $H(i\omega )=[i\omega +\tilde{\kappa }(i\omega ){]}^{-1}$, soThis is precisely the power spectral density of the even extension of $5g(t)$. Stable relaxation removes initial transients with a finite second moment, giving the long-time autocorrelation function defined in the question. All the parameter choices used below have stable relaxation.

Integrating the velocity autocorrelation function twice gives the mean squared displacement from a fixed initial position:A decaying initial transient does not change its asymptotic slope. Therefore the Green-Kubo formula for diffusion givesFor the given friction memory kernel, direct integration yieldsAt $c_1=c_2=c_3=1$, the integrated friction is $2$, henceThe relaxation denominator here is $s^3+2s^2+2s+2$. Its roots have negative real parts by the cubic Routh-Hurwitz criterion, since all coefficients are positive and $2\cdot 2>2$. This justifies the decay and integrability used above. The diffusion coefficient depends on the area under the friction memory kernel, not just its value at zero.

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

Square the operator solutions, retaining the symmetrized cross term rather than treating $x^{\prime }(0)$ and $p^{\prime }(0)$ as commuting variables. The initial symmetrized expectation vanishes, soThe means remain zero. These are also the variances. Unless $\alpha =m\omega /\hslash$, the Gaussian state breathes: its position and momentum widths exchange periodically while the mean energy remains constant.

Split the time-evolution operator into $N$ factors with $ϵ=T/N$, and insert $N-1$ position completeness relations between them. Each factor is evaluated by inserting momentum states normalized by $⟨x|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipx/\hslash }$:Multiplying the $N$ kernels and integrating the intermediate positions produces the stated time-sliced path integral, including its $N/2$ normalization power. For a free particle this Gaussian short-time kernel is exact, so the convolution identity already holds at finite $N$. Oscillatory integrals are defined with a vanishing convergence regulator, consistently fixing the square-root phase.

After integrating $x_1,\dots ,x_{N-1}$, the remaining exponent contains $(x_T-x_0{)}^2/N$. The square-root factors telescope:Since $Nϵ=T$, the free-particle propagator isFor $T>0$, the square root has phase $e^{-i\pi /4}$. Its short-time limit is the position delta function in the distributional sense, consistent with unitary evolution.

For small fixed positive $E$, the stated approximate curve has maxima atIts formal minima are zero at $\kappa a=n\pi$. The peak height is the $S$-wave unitarity limit, and the peaks narrow as $E/V_0$ decreases.

To identify the bound-state threshold, let the bound-state energy be $-\epsilon$ with $\epsilon >0$. The reduced radial wave is proportional to $\sin (Kr)$ inside and $e^{-\chi r}$ outside, with $K=\sqrt{2m(V_0-\epsilon )}/\hslash$ and $\chi =\sqrt{2m\epsilon }/\hslash$. Matching gives $K\cot (Ka)=-\chi$. As a new bound state emerges at $\epsilon \downarrow 0$,Thus the limiting peak locations coincide with the appearance of successive $S$-wave bound states. At the exact threshold the zero-energy radial solution is not square-integrable; a normalizable bound state appears just beyond it. Finite-energy maxima shift slightly from the zero-energy positions.

The formal minima require care because the assumption used in part d fails there. Retaining $ka$ in the exact matching condition gives the spherical square-well scattering lengthaway from a threshold. At $\kappa a=n\pi$, $a_s=a$, so the true low-energy S-wave scattering contribution is $4\pi a^2$, not zero. Genuine leading-order zeros of that contribution instead satisfy $\tan q=q$, for example $q\simeq 4.493409$ and $7.725252$. The original sketch compares the exact finite-energy $S$ wave with the requested approximation, making this nonuniformity visible. Dominance by the $S$ wave must be checked separately near a higher-partial-wave threshold. The scattering-length divergence and its bound-state interpretation are also derived in institutional lecture notes.^[ref]

Multiply each time-sliced path weight by $\exp [-{\sum }_{j}V(x_j)\Delta t]$. A nonnegative finite $V$ is a local killing rate: the Feynman-Kac weight attenuates paths in proportion to the time spent in regions where $V>0$. Let that rate tend to infinity outside the allowed interval. Paths that exit are removed, leaving Brownian motion killed at its first boundary hit.

Consequently the required kernel is the Gaussian path measure restricted to surviving paths, equivalentlywith the infinite-barrier interpretation stated above. It solves ${\partial }_{t}K=D{\partial }_x^{2}K$ in the interior with absorbing Dirichlet boundary conditions at $\pm L$. A reflecting boundary condition for a diffusion would describe a different process.

Strictly, this is a transition probability density in the final position, jointly with survival; a probability for an interval of final positions includes an integral over $x_f$. Its integral over the whole allowed interval is at most one. It is not the normalized endpoint density conditional on survival, which would require division by the survival probability. In these diffusion units $V$ and the eigenvalues of $H$ have units of inverse time.

The normalized Dirichlet eigenfunctions on $(-L,L)$ and their decay rates areTheir norms are one because the sine-squared integral is $L$. At the initial position zero, ${\phi }_n(0)=L^{-1/2}\sin (n\pi /2)$, so all even $n$ disappear. For $n=2j+1$, the two signs from ${\phi }_n(x)$ and ${\phi }_n(0)$ cancel, yieldingThis holds for $|x|<L$ and is zero outside. The kernel is even in $x$, vanishes at both boundaries and tends weakly to the initial delta. The eigenvalues are decay rates rather than quantum energies unless one makes an additional unit convention.

The survival probability from an absorbing transition density follows by integrating over final position. Sincethe exact survival probability for $t>0$ isThe slowest Dirichlet mode dominates at large time:The first relative correction is $-\frac{1}{3}{e}^{-8u}$. The approximation is a late-time result; extrapolating its prefactor to $t=0$ would incorrectly give a probability larger than one. The full alternating series tends to one there.

Complete the square in the potential energy. With $x_F=F/(M{\omega }^2)$,The quantum harmonic oscillator is translated without changing its level spacing. If ${\phi }_n(x)$ is its normalized unforced energy eigenstate, thenEquivalently, $|n;F⟩=e^{-i{x}_{F}p/\hslash }|n⟩$. The ground state position is centered at $x_F$, and its position uncertainty remains $\sqrt{\hslash /(2M\omega )}$.

Put $\alpha =\Delta \tau /\hslash$ and $\beta =E{x}_0\tau /\hslash$. The Pauli matrix multiplication law gives the pulse-and-precession spin one-half mapBecause $C^2+|\mathbf{v}{|}^2=1$, a suitable angle and unit vector for the rotation areIf the denominator vanishes, $U=\pm I$ and the axis is arbitrary. The Bloch vector $\mathbf{r}=⟨\mathbf{\sigma }⟩$ rotates through $\theta$ about $\mathbf{n}$ per period; after $N$ periods the angle is $N\theta$. The eigenphases equivalently give quasienergies $\pm \hslash \theta /(2\tau )$ modulo $2\pi \hslash /\tau$.

In the two-orbital subspace the position operator is $X=x_0{\sigma }_x$. The Heisenberg equation of motion between pulses gives $P/m=(i/\hslash )[H_0,X]=(2\Delta x_0/\hslash ){\sigma }_y$, henceThus two components of the Bloch vector directly describe the mean position and momentum; its third gives the orbital population imbalance. The electric kick rotates about the position axis, leaving $⟨X⟩$ unchanged at the kick, while free evolution mixes the position and momentum components. Starting in $|1s⟩$, one complete period giveswhich exhibits the induced motion explicitly. These are projected two-orbital observables.

The patched conic approximation models different portions of a spacecraft trajectory by separate two-body Kepler orbits, matching position and velocity where one attracting body gives way to another. For a surface departure from a nonrotating planet, conservation of energy gives $u^2=v_e^2+w^2$, where $u$ is surface launch speed, $v_e$ is the planet's surface escape speed, and $w$ is the planet-relative hyperbolic excess speed. Adding that excess velocity vector to the planet's orbital velocity supplies the heliocentric initial condition. It neglects the simultaneous full gravitational action of both bodies near the matching region.

A modern teaching experiment slows gravity-driven motion enough to resolve it with a photogate. PASCO scientific's Inclined Plane Dynamics Set, ME-6966, Instruction Manual 012-10874A describes “Acceleration Down an Incline” on printed pages 15–16. Its apparatus is a GOcar on a PAStrack with an adjustable incline support, a bracket-mounted photogate, a cart-mounted picket fence, and an interface with recording software. For the interface method, the fence's $1\mathrm{cm}$ pattern is aligned to interrupt the beam. Recording starts before release from rest; the cart is caught at the bottom bumper. The procedure specifies three runs per angle, starting at $14^{\circ }$ and reducing by $2^{\circ }$, with the release position fixed. The readout is velocity against time; its slope estimates acceleration. Mean accelerations are plotted against $\sin \theta$, testing the ideal slope $g$. An added-mass comparison is also requested. These are protocol settings, not observed results.^[ref]

The cart undergoes rolling without slipping at its wheels, whereas the worked block undergoes contact slip. Even ideal wheel moment of inertia changes the cart prediction. For total cart mass $M$, wheel radii $r_i$ and axial moments of inertia $I_i$, rigid wheels with fixed axles and no losses giveTo derive it, combine translational kinetic energy $M{v}^2/2$ with wheel rotational kinetic energy ${\sum }_i{I}_i(v/r_i{)}^2/2$, then equate their rate of change to the release of gravitational potential energy $Mg\sin \theta v$. This is the multiwheel counterpart of rolling acceleration with rotational inertia. Neglecting wheel moment of inertia yields $a\simeq g\sin \theta$; treating a cart as a solid rolling cylinder would give a different correction.

Useful additional controls are repeated releases without a push, checking the fence alignment and spacing, and checking that the recorded velocity trace is straight over the fitted interval. A gently moving cart on a level track can expose losses or unintended tilt. These checks help distinguish an angle or timing error from neglected wheel moment of inertia, bearing friction, or drag force. The three-run spread measures repeatability but cannot by itself reveal a shared calibration error. The inspected protocol makes the gravity component measurable; it does not measure the sliding block's coefficient of kinetic friction.

A generalized eigenstate of the position operator is labeled by its position eigenvalue. In a continuous position coordinate the normalization is distributional: $⟨q|q^{\prime }⟩=\delta (q-q^{\prime })$ and $\int dq|q⟩⟨q|=1$. Individual position eigenstates are not normalizable Hilbert-space vectors. The Dirac delta distribution normalization makes $⟨q_f|U|q_i⟩$ the integral kernel of a time-evolution operator, such as the free-particle propagator.

The position uncertainty of a quantum state is the standard deviation of its measured position, provided the second moment is finite. It is computed from the position operator with the Born rule. An oscillator coherent state has $\Delta x=\sqrt{\hslash /(2m\omega )}$ independently of its mean position; translating the state changes the mean without changing this variance.

Preparation uncertainty describes the outcome spread intrinsic to a specified quantum state preparation. Operationally, prepare many copies of the same quantum state, measure position on one group and momentum on another, and estimate the two variances. Ideal measurements still have the spreads predicted by the Born rule. No earlier position measurement need occur in the momentum group. The Heisenberg uncertainty relation constrains these two distributions belonging to the same preparation.

Measurement backaction concerns how a particular measurement in quantum mechanics changes a quantum state. Instrument resolution concerns how its reported outcomes relate to the intended observable. Those questions require a model of the instrument and its interaction with the system. Substituting an instrument error and an induced change for $\Delta x$ and $\Delta p$ in the Heisenberg uncertainty relation does not follow from its proof. The practical value of preparation uncertainty is that it bounds achievable input spreads before any such instrument model is chosen.

Motion after launch under a specified gravitational field, with negligible drag force and no thrust. In a constant downward gravitational acceleration $g$, horizontal velocity is constant and vertical position obeys $y(t)=y_0+v_{y0}t-g{t}^2/2$. Eliminating time gives a parabola when horizontal velocity is nonzero. Over larger distances the changing direction and magnitude of gravity require an orbital model instead.