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For $\psi (x)=C{e}^{-\alpha x^2}$,The stationary Schrodinger equation becomesHence, choosing the positive root and a positive normalization constant,The state is even, so $⟨x⟩=⟨p⟩=0$. The Gaussian integrals giveand therefore $\Delta x\Delta p=\hslash /2$. These values are collected in gaussian eigenstate for a quadratic potential.

Finally, equality in the derivation of the uncertainty relation requires the centred vectors to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,In position space this sayswhose normalizable solutions areThus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the equality case of the Heisenberg uncertainty relation.

The Schrodinger equation and its complex conjugate areBecause the potential $U$ is real, its two contributions cancel when differentiating the probability density $\rho ={\psi }^{\ast }\psi$. ThereforeThe expression in square brackets is the probability current $J$, soThis probability continuity equation says that probability can leave a region only through the current across its boundary.

Suppose ${\chi }_1$ and ${\chi }_2$ are two bound-state solutions with the same energy. Their Wronskiansatisfies $W^{\prime }=0$ because the Schrodinger equation contains no first-derivative term. Hence $W$ is constant. Both wavefunctions and their derivatives decay at infinity, so $W\to 0$ there and therefore $W\equiv 0$. The two solutions are linearly dependent. Thus every one-dimensional bound-state energy is nondegenerate.

Write the expectation value asDifferentiating and retaining the possible explicit time dependence of the observable givesThe Schrodinger equation and its adjoint arewhere the Hamiltonian operator is Hermitian. Substitution yieldsThis is the proof of Ehrenfest theorem from the Schrodinger equation.

Differentiating $⟨O⟩=⟨\psi |O|\psi ⟩$ and using the Schrodinger equation and its adjoint gives the Ehrenfest theorem

The time-dependent Schrodinger equation is

For $\psi =A{e}^{-B{x}^2}$,Substitution into the Schrodinger equation and comparison of the constant and $x^2$ coefficients gives

The time-independent Schrodinger equation isIntegrating it over $(a-\epsilon ,a+\epsilon )$ and letting $\epsilon \to 0$ shows that ${\psi }^{\prime }$ has no jump because $V$ is finite. A jump in $\psi$ would create a delta term in ${\psi }^{\prime }$ and hence a derivative of a delta in ${\psi }^{\prime \prime }$, so $\psi$ is also continuous.

For a bound state $-V_0<E<0$, defineThen the proposed even wavefunction solves the equation away from the interfaces, andContinuity at $x=a$ gives $B\cos (ka)=A{e}^{-\eta a}$, while continuity of the derivative gives $-Bk\sin (ka)=-A\eta e^{-\eta a}$. Dividing yields the second required relationThus the lowest even state of the finite square well is the first-quadrant intersection, with $0<ka<\pi /2$, of the circle $k^2+{\eta }^2=2m{V}_0/{\hslash }^2$ and the curve $\eta =k\tan (ka)$.

For a neutron in the Earth's Newtonian gravitational potential ${\Phi }_E$, the Schrodinger equation isNear the surface, ${\Phi }_E(z)=gz$ up to an additive constant.

The Colella–Overhauser–Werner experiment uses crystal slabs as coherent beam splitters and mirrors in a neutron interferometer. The two paths contain horizontal segments of length $s$ separated vertically by $r$, then recombine. The gravitational potential energy difference is $m_{n}gr$. A neutron of speed $v$ spends time $s/v$ on a horizontal segment, so the gravitationally induced relative quantum phase isUsing the de Broglie wavelength $\lambda =h/(m_{n}v)$ and $\hslash =h/(2\pi )$ givesRotating the apparatus changes the vertical projection of its enclosed area from zero to $rs$. The number of full interference oscillations is thereforeHere $rs=(\sqrt{10}\mathrm{cm}{)}^2=10^{-3}{\mathrm{m}}^2$, soThe expected change is therefore $\boxed{9}$ oscillations to the nearest integer.

For particle positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-particle Schrodinger equation isWrite $|a{⟩}_i$ for the localized wave packet ${\psi }_{ai}$ and $d_{ab}=|x_{a1}-x_{b2}|$. Neglecting packet spreading and branch overlap, the initial product state evolves branchwise asup to phases generated independently on the two particles. These branch-dependent phases generally cannot be separated into one phase depending only on $a$ and one depending only on $b$, so the Newtonian gravitational potential energy creates gravitationally induced entanglement.

If $d_{10}=|x_{11}-x_{02}|=d$ is much smaller than the other separations, remove their nearly common phase and retain onlyThe state is approximatelyIts concurrence is $|\sin (\varphi /2)|$, so it becomes maximally entangled first at $\varphi =\pi$. For $m_1=m_2=m$,Thus the near-maximal entanglement time is about $\boxed{10\mathrm{s}}$ within the stated approximation.

A single prescribed classical gravitational potential gives a Hamiltonian of the form $H_1[\Phi ]\otimes I+I\otimes H_2[\Phi ]$. Its evolution factorizes as $U_1\otimes U_2$ and preserves every initial product state, so it cannot generate this entanglement. A semiclassical mean field sourced only by expectation values likewise gives each particle a local one-body potential and does not provide a quantum mediator carrying branch correlations.

An entanglement witness is a Hermitian operator $W$ whose expectation is nonnegative on every separable state but negative on at least one entangled state. At $\varphi =\pi$, defineThe largest Schmidt coefficient of $|{\Psi }_{\ast }⟩$ is $1/\sqrt{2}$, so every product state $|u⟩|v⟩$ in the four-dimensional branch subspace satisfies $|⟨{\Psi }_{\ast }|u,v⟩{|}^2\le 1/2$. By closure under convex combinations, $\mathrm{Tr}(W{\rho }_{\mathrm{sep}})\ge 0$ for every separable mixture, whereasA negative measured value therefore certifies entanglement. Under the assumptions that the masses began unentangled and interacted only through gravity, such certification would show that the mediator can transmit quantum coherence; it would be evidence against a purely classical gravitational channel and for the quantum nature of gravity.

Write $H=-{\partial }_x^2+V(x)$. The real potential and periodic boundary conditions make $H$ a self-adjoint operator. Since $u_t=-iHu$,because the expectation of a self-adjoint operator is real. Equivalently, integrating the kinetic term by parts leaves $\int |u_x{|}^{2}dx$ and the periodic boundary term cancels. Thus the continuous Schrodinger equation preserves its $L^2$ norm.

Suppose a semidiscrete linear PDE is $u_t=(A+B)u$. Direct evaluation of $e^{h(A+B)}$ may be expensive, while $e^{hA}$ and $e^{hB}$ can be cheap, parallelizable, or exactly structure-preserving. Splitting is especially effective when $A$ and $B$ represent different spatial directions, kinetic and potential energy, diffusion and reaction, or linear and nonlinear pieces.

The Lie-Trotter splitting commutator error follows from multiplying the exponential series:It therefore has local error $O(h^2)$ and global order one. Reversing the factors changes the sign of the leading commutator. The symmetric average in composition form is Strang splitting,whose symmetry $S_{-h}=S_h^{-1}$ removes even powers from the modified generator. Its local error is $O(h^3)$, involving nested commutators such as $[A,[A,B]]$ and $[B,[A,B]]$, and its global order is two.

For multidimensional diffusion, directional splitting replaces one large elliptic solve by successive one-dimensional tridiagonal solves; alternating-direction implicit methods are standard examples. For the Schrodinger equation, take $A=i{\partial }_x^2$ and $B=-iV(x)$. The kinetic subflow is diagonal in Fourier space, the potential subflow is pointwise multiplication, and Strang splitting is both second order and exactly norm-preserving.

A symmetric method has even global order. If its leading modified-flow defect is $h^{2p+1}E$, a symmetric composition $S_{a_{1}h}\cdots S_{a_{s}h}$ raises the order when $\sum a_i=1$ and $\sum a_i^{2p+1}=0$, together with the required higher commutator conditions. In particular, the higher-order composition of a symmetric splittingis fourth order because $2\gamma +\delta =1$ and $2{\gamma }^3+{\delta }^3=0$. Its negative middle step is harmless for reversible unitary problems but can be ill-posed or strongly unstable for parabolic semigroups. Higher-order parabolic splittings therefore use more specialized complex coefficients, commutator corrections, or extrapolation.

For two solutions of the Schrodinger equation, their Wronskian is constant because the equation has no first-derivative term. The determinant of the Floquet matrix is the ratio of the Wronskian at $x+a$ to that at $x$, hence

Writing $D_A=\nabla +ieA/\hslash$, the current isThe two covariant factors acquire opposite phases under a gauge transformation, so $J$ is invariant.

Multiply the Schrodinger equation by ${\psi }^{\ast }$, subtract its complex conjugate multiplied by $\psi$, and useThe real scalar-potential terms cancel, leaving

For positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-particle Schrodinger equation isNeglecting packet spreading and writing $|a{⟩}_i$ for ${\psi }_{ai}$, branchwise evolution givesup to local kinetic phases. This branch-dependent Newtonian gravitational potential energy produces gravitationally induced entanglement.

Under the stated distance approximation, only the three branches $a=b$ acquire an appreciable common phaseThe coefficient matrix iswhere $J$ is the $3\times 3$ all-ones matrix. The reduced density matrix isIt equals $I/3$ when $\cos \varphi =-1/2$, so the first maximally entangled state occurs at $\varphi =2\pi /3$. Therefore

The state does not remain entangled for every $t>0$. Whenever $\varphi =2\pi k$, all branch phases again agree and the state returns to its initial product state. The revival period is

PutAway from $x=x^{\prime }$, this satisfies $(d^2/d{x}^2+k^2)G_k=0$. Its first derivative has jump $2ik$, so, distributionally,Applying $d^2/d{x}^2+k^2$ to the proposed integral equation therefore givesSince $k^2=2mE/{\hslash }^2$, this is precisely the stated Schrodinger equation. The term $e^{ikx}$ is the incident wave and the choice $e^{ik|x-x^{\prime }|}$ gives outgoing waves on both sides of each source point.

Direct differentiation verifies the free-particle Schrodinger equation, so $V=0$. At $t=0$, normalisation of $A{e}^{-x^2/2}$ gives $A={\pi }^{-1/4}$. Symmetry gives $⟨x⟩=0$, while Gaussian integration givesIts growth shows wave-packet spreading and therefore a nonstationary state.

For equal masses with positions ${\mathbf{r}}_1,{\mathbf{r}}_2$, the two-body Schrodinger equation isNeglecting wave-packet spreading and overlap, write $|a{⟩}_i$ for the packet ${\psi }_{ai}$. The initial state and its branchwise gravitational evolution areup to local kinetic phases. The branch dependence of the Newtonian gravitational potential energy is the source of gravitationally induced entanglement.

Under the approximation stated in the question, only the matching pairs $a=b$ acquire an appreciable common phaseThe coefficient matrix of the resulting bipartite state iswhere $J$ is the all-ones matrix. ThereforeA maximally entangled state would require ${\rho }_1=I/n$. This demandsThat equation has no solution for $n>4$. Thus the requested large-$n$ conclusion does not follow from the assumptions printed in the paper: within the stated approximation, the system never becomes maximally entangled in the large-$n$ limit. For completeness, maximal entanglement is possible for $n=3$ at $\varphi =2\pi /3$ and for $n=4$ at $\varphi =\pi$; for $n=2$ it occurs at $\varphi =\pi /2$ or $3\pi /2$. The $n=4$ time would bebut it is not a large-$n$ answer.

The system also does not remain entangled for every $t>0$. Whenever $\varphi =2\pi k$, the phase matrix again factorizes and the state returns to its initial product state. The revival period isFor $n>2$ it is entangled at all intervening times; for $n=2$ there is the additional product-state revival at $\varphi =\pi$.

For a wavefunction satisfying the Schrodinger equation with a real potential,This continuity equation expresses local conservation of probability.