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In the Schrödinger picture, a time-independent Hamiltonian gives $|{\psi }_S(t)⟩=U(t)|\psi (0)⟩$, where $U=e^{-iHt/\hslash }$. The Heisenberg picture transfers this evolution to each observable:Both pictures give the same probabilities and expectation values. Differentiating yields the Heisenberg equation of motionThe last term is an explicit change of the measured operator. If it is absent and $[H,A]=0$, the observable is conserved.

The corresponding classical equation is $dA/dt=\{A,H{\}}_{\mathrm{PB}}+\partial A/\partial t$. The correspondence principle relates the Poisson bracket to $[A,H]/(i\hslash )$, with suitable operator ordering and a classical limit. For $H=p^2/(2m)+V(x)$, the canonical commutation relation givesTaking means gives Ehrenfest theorem. Its second equation contains $-⟨V^{\prime }(x)⟩$, which need not equal $-V^{\prime }(⟨x⟩)$. A narrow wave packet in a slowly varying force field can follow the classical equations approximately; a quadratic potential makes the equality exact for the means.

For a quantum harmonic oscillator, direct solution gives $x_H(t)=x_H(0)\cos \omega t+p_H(0)\sin \omega t/(m\omega )$ and $p_H(t)=p_H(0)\cos \omega t-m\omega x_H(0)\sin \omega t$. These have the classical form as operator identities. A coherent state supplies a localized center moving on the classical orbit, whereas an energy eigenstate has stationary density and $⟨x⟩=0$, even at large excitation. Thus large quantum numbers alone do not create a localized classical trajectory. Classical behavior also concerns state preparation, relative fluctuations and the resolution of the measurements.

The total angular momentum operator is $\mathbf{J}=\mathbf{L}+\mathbf{S}$. Since orbital and spin operators act on different factors of the tensor product, they commute, and expanding $J^2$ givesA coupled angular momentum eigenstate $|l,s;j,m_j⟩$ is a simultaneous eigenstate of these three squared operators, with eigenvalues ${\hslash }^{2}j(j+1)$, ${\hslash }^{2}l(l+1)$ and ${\hslash }^{2}s(s+1)$. Thus the angular factor in the spin-orbit interaction is the numberThe other factor, $r^{-1}dV/dr$, is radial. It commutes with the angular operators, and a central-potential energy eigenstate separates into a radial wavefunction and an angular spinor. Taking its radial expectation value therefore givesThis spin-orbit energy shift in a central potential is independent of $m_j$, so rotational symmetry preserves the $2j+1$ magnetic degeneracy within a fixed $j$ multiplet. It is a first-order fine structure correction; the given coefficient already includes the relativistic factor needed for the electron's spin-orbit coupling.

Write $a=a_{-}$ and $a^{†}=a_{+}$. Inverting the annihilation operator and creation operator definitions givesThe ladder operators take $|n⟩$ to orthogonal quantum harmonic oscillator energy eigenstates $|n-1⟩$ and $|n+1⟩$. Thus $⟨n|a|n⟩=⟨n|a^{†}|n⟩=0$, andFor $n=0$, the lowering term is simply zero. The conclusion concerns an individual energy eigenstate; interference between different eigenstates can produce nonzero expectation values.

Each energy eigenstate picks up its own phase under unitary time evolution. The relative phase therefore changes at $(E_1-E_0)/\hslash =\omega$. After removing an irrelevant global phase, the state isThe diagonal position matrix elements vanish, but the annihilation operator and creation operator giveThe two off-diagonal cross terms in the expectation value yieldThus the angular frequency is $\omega$, and the magnitude of the oscillation, meaning its amplitude, isThe peak-to-peak displacement is $2A$. The phase $\varphi$ fixes where the oscillation starts but does not change $A$. A pure $|0⟩$ or $|1⟩$ preparation has no mean oscillation, whereas an equal-weight quantum superposition maximizes its amplitude within this two-state family. The motion is produced by interference between energy eigenstates, rather than by a changing occupation of either one.

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 normalized Hartree product state and its equivalent second quantization form areThe product is already symmetric under exchanging bosons; no additional $1/\sqrt{N!}$ belongs in the coordinate wavefunction. The factor in the second quantization expression instead normalizes repeated creation operators acting on the vacuum state. This is a fixed-number two-mode condensate, with every boson in the same normalized orbital.

Set $c=\cos \theta$ and $s=\sin \theta$. Its binomial distribution of occupations givesThe expectation value of the Hamiltonian operator is thereforeKeeping $N(N-1)$ rather than replacing it by $N^2$ retains the exact pair count of the trial quantum state. Equivalently, with $ϵ=w-\mu$, its interaction contribution is $wN(N-1)(c^4+s^4)$ and its one-particle contribution is $ϵN-tN\sin (2\theta )\cos \chi$.

Choose the nonredundant parameter range $0\le \theta \le \pi /2$, with $\chi$ modulo $2\pi$. For $t>0$, minimizing over the relative phase gives $\chi =0$. Putting $x=\sin (2\theta )\in [0,1]$, the remaining nonconstant energy is $-tNx-\frac{1}{2}wN(N-1)x^2$, which decreases with $x$ for the repulsive coupling $w>0$. HenceThe optimal orbital is the symmetric ${\varphi }_0=({\phi }_L+{\phi }_R)/\sqrt{2}$. Both the quantum tunnelling term and the repulsive interaction favor this balanced occupation within the Hartree product state family. This result uses a variational method: for strong repulsion relative to quantum tunnelling, the true many-body ground state can have smaller occupation fluctuations than this fixed-number two-mode condensate.

The Hartree-Fock method approximates the ground state wavefunction by a single Slater determinant, minimizing the expectation value of the Hamiltonian over its occupied orbitals. It retains fermionic exchange symmetry exactly within that determinant but omits electron correlation beyond a single Slater determinant.

For a normalized orbital $\int d^{2}r|\varphi {|}^2=1$, the symmetric Hartree product state and its equivalent Fock space form areThere are exactly $N(N-1)/2$ interacting pairs, so the expectation value of the grand Hamiltonian is the functionalAt fixed $N$ the term $-\mu N$ is constant. Enforce wavefunction normalization separately by varying ${\mathcal{E}}_{\mu }-N\eta (\int |\varphi {|}^2-1)$ with respect to ${\varphi }^{\ast }$. The resulting stationary Gross–Pitaevskii equation for the normalized orbital isThe Lagrange multiplier $\eta$ enforces the norm; fixed-$N$ variation cannot identify it with the independently inserted chemical potential merely by differentiating the constant $-\mu N$.

For the standard large-$N$ convention put $\Psi =\sqrt{N}\varphi$, so $n=|\Psi {|}^2$ and $\int n=N$, and replace $N-1$ by $N$. Varying the resulting grand energy functionalwith unrestricted $\Psi$ givesEquivalently, at fixed $N$ this same $\mu$ is chosen as the normalization Lagrange multiplier. This is the chemical potential convention used below. The contact interaction coupling $g$ has dimensions of energy times area in two dimensions.

For a nonzero quantum state $\psi$ and a self-adjoint operator $A$, the expectation value and standard deviation of its measurement areThe squared deviation is also called the variance of the observable. All displayed products are understood on quantum states for which the relevant moments exist. Self-adjointness gives $\overline{⟨\psi ,A\psi ⟩}=⟨A\psi ,\psi ⟩=⟨\psi ,A\psi ⟩$, so $a$ is real. Expanding the squared norm gives the variance expression. A norm is nonnegative and is zero exactly for the zero vector, hence ${\Delta }_{\psi }(A)=0$ precisely when $A\psi =a\psi$. Thus zero dispersion is equivalent to being an eigenstate.

For bounded $H$, the operator exponential is the norm-convergent power seriesFor a general unbounded self-adjoint Hamiltonian operator, the spectral theorem defines the same notation byThis definition does not require the power series to converge on every quantum state. complex conjugation of the spectral multiplier gives $U_t^{\ast }=U_{-t}$, and multiplication of the phases gives $U_{-t}{U}_t=I$. Thus $U_t$ is unitary, preserves the quantum state norm, and has $U_0=I$.

For initial data in the domain of $H$, differentiating the spectral expression yields ${\partial }_t{U}_t{\psi }_0=-(i/\hslash )H{U}_t{\psi }_0$. Consequentlysolves the Schrödinger equation with the prescribed initial quantum state. Uniqueness follows by differentiating $U_{-t}{\psi }_t$: the two terms cancel, leaving the constant ${\psi }_0$.

In the two-dimensional energy basis, expand the initial quantum state as ${\psi }_0=c_1{\Psi }_1+c_2{\Psi }_2$. Each energy eigenvector receives its own phase under the operator exponential, givingThis is the general solution because every initial vector has a unique basis expansion.

In that orthonormal basis the other observable has the real symmetric matrixIts conjugate transpose equals itself, so it is self-adjoint. Also $A^2=I$ and $\mathrm{tr}A=0$, giving eigenvalues $+1$ and $-1$. Direct substitution gives the normalized orthogonal eigenvectorsAn irrelevant global phase may multiply either vector. With the chosen lower eigenstate initially,The Born rule says that the probability for the higher eigenstate is $|⟨v_{+},{\psi }_t⟩{|}^2$. Its amplitude is $\sin \alpha \cos \alpha (e^{-it{E}_2/\hslash }-e^{-it{E}_1/\hslash })$. Using $|e^{-iu}-e^{-iv}{|}^2=4{\sin }^2((u-v)/2)$ givesThis is the transition probability for a rotated two-level observable. Its time dependence comes from the relative phase between two energy eigenstates. If their energies coincide, or if the measured observable has the energy eigenvectors as its eigenvectors, this transition probability is zero at all times.

Write $b=(\ell +1)a$, $R(r)=r^{\ell }{e}^{-r/b}$ and $Y(\theta ,\varphi )={\sin }^{\ell }\theta e^{i\ell \varphi }$, for nonnegative integer $\ell$. The angular Laplacian satisfies ${\Delta }_{S^2}Y=-\ell (\ell +1)Y$. Direct radial differentiation givesBecause $(\ell +1)/b=1/a$, the $1/r$ contribution cancels the Coulomb potential in the Hamiltonian. ThereforeThe volume element contributes $r^2\sin \theta$. The squared state factors into radial, polar and uniform azimuthal densities, so the angular normalization cancels in every radial expectation. With $I_j={\int }_0^{\infty }{r}^j{e}^{-2r/b}dr$,The supplied integration-by-parts recurrence gives $I_{j+1}=(j+1)b{I}_j/2$. HenceIn particular $\mathrm{Var}(r)/{\mu }^2=(2\ell +3{)}^{-1}$. Put $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality gives

For the polar variable set $J_j={\int }_0^{\pi }{\sin }^j\theta d\theta$. The polar density is proportional to ${\sin }^{2\ell +1}\theta$, soAn exact expression for the requested tail, as well as its useful estimate, isThe inequality is Markov's inequality applied to ${\cos }^2\theta$. If a finite expression is desired, expand $(1-z^2{)}^{\ell }$: the numerator is ${\sum }_{j=0}^{\ell }(-1{)}^j(\genfrac{}{}{0ex}{}{\ell }{j})(1-h^{2j+1})/(2j+1)$ and the denominator is the same sum with $h=0$.

By the union bound, with probability at least $1-2h$ both $|r/\mu -1|\le h$ and $|\cos \theta |\le h$. The height above the equatorial plane is then at most $(1+h)\mu h$. The cylindrical radius differs from $\mu$ by at most $\mu h+(1+h)\mu h^2$. Thus the distance from the circle of radius $\mu$ in the equatorial plane is $O(\mu h)$ with probability tending to one. This is relative concentration around a growing circle; it does not assert an absolutely shrinking thickness. The probability concentration of a circular Coulomb eigenstate is a precise version of the large-angular-momentum orbit picture.

For a normalized state and a self-adjoint operator, the expectation value and dispersion areUse the physics convention that the inner product is conjugate-linear in its first argument, and assume the displayed products and second moments are defined. Self-adjointness gives $\overline{⟨\psi |A\psi ⟩}=⟨A\psi |\psi ⟩=⟨\psi |A\psi ⟩$, so the expectation is real. MoreoverIt vanishes exactly when $\psi$ is an eigenstate, with eigenvalue its expectation.

The product rule gives $XPf=-i\hslash x{f}^{\prime }$ and $PXf=-i\hslash (f+x{f}^{\prime })$, hence $\boxed{[X,P]f=i\hslash f}$ on their common differentiable domain. For general operators, put $z=⟨A\psi |B\psi ⟩$. ThenThe Cauchy-Schwarz inequality yieldsApply this to centered position and momentum operators, whose commutator is unchanged. Their squared norms are the variances, givingEquality requires both proportionality in Cauchy-Schwarz and a purely imaginary inner product. Thus for real means $x_0,p_0$,with real $\gamma$. Solving this first-order equation givesNormalizability on the real line forces $\gamma >0$. Writing ${\sigma }^2=\hslash /(2\gamma )$, all normalized equality states are thereforeThey have $\Delta X=\sigma$ and $\Delta P=\hslash /(2\sigma )$. A nonconstant quadratic phase would increase the product, so the full family here has a real positive Gaussian width and only a linear spatial phase.

For a normalized wavefunction, the expectation value of a function of position isprovided the integral exists. For an unnormalized state, divide this expression by $\int |\psi {|}^{2}dx$. Put $b=m\omega /\hslash >0$. The ground state has probability density $\sqrt{b/\pi }{e}^{-b{x}^2}$, which is even. ThusThe zero mean position ${\mathbb{E}}_{{\psi }_0}(x)=0$ expresses symmetry, whereas this positive expectation value measures the mean distance from the centre.

A normalized quantum state satisfies $⟨\psi ,\psi ⟩=1$. Its expectation value is $\boxed{{\mathbb{E}}_{\psi }(A)=⟨\psi ,A\psi ⟩}$. The conjugate symmetry of the inner product and self-adjointness giveso the expectation value is real. If $A\varphi =\lambda \varphi$ with $\varphi \ne 0$, normalize $\varphi$ by dividing by its norm. ThenEvery eigenvalue of a self-adjoint operator is real.

Centre the position operator and momentum operator by their real expectation values:Scalar multiples of the identity operator commute with every linear operator, so the canonical commutation relation givesThe second moments of the centred linear operators are the variances, ${\mathbb{E}}_{\psi }(A^2)={\Delta }_{\psi }(X{)}^2$ and ${\mathbb{E}}_{\psi }(B^2)={\Delta }_{\psi }(P{)}^2$. Applying the preceding inequality and taking nonnegative square roots proves the Heisenberg uncertainty relation:Subtracting the means is essential: the uncentred second moments alone do not equal the variances. The equality case of the Heisenberg uncertainty relation requires the centred vectors to be imaginary multiples, as in the preceding equality conditions.

If $a_{-}\psi \ne 0$, it has energy $E-\hslash \omega$, distinct from the energy of $\psi$. Eigenstates of a self-adjoint operator with distinct eigenvalues are orthogonal. Explicitly,so $⟨\psi ,a_{-}\psi ⟩=0$. If $a_{-}\psi =0$, the same conclusion is immediate.

Dividing by $\Vert \psi {\Vert }^2$ and using the definition of $a_{-}$ givesBoth expectation values are real because $P$ and $X$ are self-adjoint operators. Its real and imaginary parts therefore vanish separately:An oscillator energy eigenstate is centered in both position and momentum, despite having nonzero quantum uncertainties.

The radial probability density includes the spherical-shell volume factor:Consequently the radial expectation value isKeeping $m$ and $K$ fixed, the hydrogenic ground-state radius scaling is therefore $⟨r⟩\propto 1/Z$: the $Z=2$ state has half the mean radius of the $Z=1$ state. Equivalently, with $a_{\ast }={\hslash }^2/(mK)$, the mean radii are $3a_{\ast }/2$ and $3a_{\ast }/4$. For the physical electron-nucleus Coulomb potential, $a_{\ast }$ is the Bohr radius of the fixed-nucleus hydrogen atom model.

The exact $Z=2$ comparison in this one-electron model describes the hydrogen-like ion ${\mathrm{He}}^{+}$. A neutral helium atom has two electrons; its exact size also depends on their mutual Coulomb interaction and is not determined by this one-electron calculation. The stated half-radius conclusion is the prediction of the model supplied in the paper.

Write $k=\pi /a$, $A=\sqrt{2/a}$ and ${\Psi }_0=A{e}^{ikx}\sin kx$. Its derivative isIntegration by parts with the Dirichlet boundary conditions gives the energy expectationIt is constant in time, since the time-independent Hamiltonian operator has fixed energy measurement probabilities: $⟨H{⟩}_t={\sum }_n{E}_n|c_n{|}^2$.

For the midpoint parity operator, $R{\Psi }_0=-A{e}^{-ikx}\sin kx$. Hence, again putting $u=kx$,Indeed the imaginary part integrates to zero by reflection and ${\int }_0^{\pi }\cos 2u{\sin }^{2}udu=-\pi /4$. This expectation value is also constant in time, because the bounded self-adjoint operator $R$ commutes with $H$ and hence with its time-evolution operator. Equivalently, $⟨R{⟩}_t={\sum }_n(-1{)}^{n+1}|c_n{|}^2$.

For the momentum operator, direct evaluation givesThe real sine-cosine integral vanishes. This expectation value is not constant in time: the infinite square well walls reverse motion. One can prove the change exactly without approximating the wavefunction. Atthe energy eigenstate expansion and $R{\psi }_n=(-1{)}^{n+1}{\psi }_n$ give $\Psi (t_{\ast })=-R{\Psi }_0$. Differentiation shows $PR=-RP$, soThis half revival in an infinite square well is a concrete counterexample to momentum conservation. The sign from $-R$ is only a global phase.