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The Pusey-Barrett-Rudolph theorem excludes overlapping preparation distributions for distinct pure states in an ontological model of a quantum system satisfying preparation independence and the specified quantum predictions. It excludes neither all hidden variables nor every informational use of a quantum state. The precise overlap criterion is essential: a psi-epistemic model lets at least two distinct pure-state preparations assign positive common probability mass to the same underlying states; a psi-ontic model has mutually singular measures for each distinct pair.^[ref]

Let preparation $x$ sample $\lambda \in \Lambda$ from ${\mu }_x$. A fixed quantum measurement $M$ has nonnegative response functions ${\xi }_M(j|\lambda )$ with ${\sum }_j{\xi }_M(j|\lambda )=1$. Averaging over the preparation must reproduce the Born rule. The response is fixed by $M$ and the complete underlying state, not by additional knowledge of the preparation label. No outcome determinism or Bell local causality is assumed; the joint measurement response need not factor into local responses. For independently operated preparations, preparation independence further identifies the joint state space with the appropriate product and assignsOperationally independent devices do not logically entail this ontic factorization; it is an additional physical premise.^[ref]

The two-qubit example can be checked directly. Set $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. In computational basis order $00,01,10,11$, take the following normalized columns as an orthonormal basis:Their pairwise inner products vanish. Direct multiplication shows that the respective preparations $00,0+,+0,++$ have zero probability amplitude for respective outcomes $1,2,3,4$. This is the PBR exclusion measurement for zero and plus: the measurement does not perfectly distinguish the preparations, but each outcome rules one out.

Suppose a common submeasure $\nu$ of ${\mu }_0$ and ${\mu }_{+}$ has mass $\omega >0$; with densities one can take $\nu (\lambda )=\min \{{\mu }_0(\lambda ),{\mu }_{+}(\lambda )\}$. Preparation independence makes $\nu \otimes \nu$ a common submeasure of all four product preparations, of mass ${\omega }^2$. The zero probability for outcome $j$ under its excluded preparation and nonnegativity forceSumming and using normalization yields ${\omega }^2=0$, a contradiction. This measure formulation avoids treating a set with merely nonempty topological support as a positive-probability overlap.

For arbitrary distinct nonorthogonal pure states, write $|{\psi }_{0,1}⟩=\cos (\theta /2)|0⟩\pm \sin (\theta /2)|1⟩$, with overlap $\cos \theta$ and $0<\theta <\pi /2$. The general proof constructs an exclusion measurement on $n$ independently prepared copies when $2^{1/n}-1\le \tan (\theta /2)$. A finite $n$ always exists. The same common-submeasure contradiction then excludes positive overlap. Orthogonal preparations are already disjoint if their perfectly distinguishing measurement is reproduced. The original paper also bounds residual overlap for imperfect exclusion data; exact disjointness concerns the ideal theorem, not a finite experiment's proof of exact zero.^[ref]

The scope is consequently sharp. Bohmian mechanics is not excluded: its underlying state includes the wave function, making distinct pure preparations disjoint even though extra configuration variables remain. A psi-ontic model need not be complete with the wave function alone, and the theorem does not identify the wave function with a three-dimensional material field. Models abandoning preparation independence, or accounts not representing preparations as uncertainty over observer-independent states of this specified kind, are outside this conclusion. Epistemic uncertainty between mixed preparations is not banned. Nor does the theorem establish a collapse mechanism, choose between Bohmian and Everettian theories, or solve the measurement problem. Its achievement is a conditional impossibility theorem about a precise form of pure-state epistemic overlap.

Quantum decoherence explains the practical stability of records and suppression of observable interference; it does not alone select one outcome. The measurement problem arises when universal unitary time evolution, a complete wave function description and a unique definite outcome are all demanded. A theory must explain which commitment is changed or supplemented, not merely replace a pure composite description by a reduced one.

Model an ideal quantum measurement and its environment byTaking the partial trace over the environment gives the system–apparatus reduced density matrixEnvironmental records make $⟨E_1|E_0⟩$ very small for macroscopically distinct alternatives. If independent fragments have overlaps $d_k$, their combined overlap is ${\prod }_k{d}_k$, explaining the rapid suppression. The apparatus interaction also favours robust records, giving environment-induced pointer-basis selection rather than an arbitrary choice of basis.^[ref]

What follows is approximate agreement, for local apparatus observables, with a classical probabilistic mixture. What does not follow is that either term has ceased to exist. Even with orthogonal environmental records, the global quantum state is a pure state with quantum entanglement. For orthogonal branches $|B_0⟩,|B_1⟩$, the global observable $|B_0⟩⟨B_1|+|B_1⟩⟨B_0|$ has expectation $2\mathrm{Re}(a^{\ast }b)$ in the superposition and zero in the corresponding incoherent mixture. Its practical inaccessibility is not its mathematical disappearance. The reduced state is an improper mixed state, not proof that an unknown single branch was selected.

Different interpretations use this result differently. In Bohmian mechanics, disjoint environmental configurations make one occupied branch an effective description of the actual apparatus. In the many-worlds interpretation, robust branches supply approximately autonomous relative records, without a unique global outcome. In the GRW model, a separate stochastic change of the wave function selects a dominant component; environmental decoherence helps explain ordinary records but is not the collapse law. These are different completions of the argument.

Nor does diagonality derive the Born rule. Interpreting the diagonal entries as experienced outcome probabilities requires a probability rule or an interpretive argument. Extremely small overlaps also leave approximations and possible quantum recoherence; a finite controlled environment can in principle be manipulated to reveal the global coherence. The appropriate verdict is that decoherence resolves much of the preferred-record-basis and effective-classicality problem while leaving the unique-outcome and probability problems dependent on further theory. It is an essential physical mechanism in many accounts, not a self-contained solution by relabelling a partial trace as collapse.

An empty wave is a nonzero component of the wave function in Bohmian mechanics containing no actual particle configuration. Its persistence is a question about ontology and explanation, not an inconsistent prediction of duplicate detected particles. In de Broglie-Bohm theory, the wave function evolves by the Schrödinger equation while the actual configuration $Q$ evolves by the guidance equation. For spinless particles without a magnetic vector potential,The physical state includes both elements, so a component of $\Psi$ need not contain another set of particles.^[ref]

After a position-recording interaction, write the universal wave function as $\Psi (q,y)=a{\psi }_0(q){\Phi }_0(y)+b{\psi }_1(q){\Phi }_1(y)$, with apparatus configurations $y$ supported in disjoint regions. If the actual $Y$ lies in the first region, ${\Phi }_1(Y)$ and its relevant local derivatives vanish in the ideal disjoint-support model. Evaluating the guidance equation at $(Q,Y)$ therefore gives the velocity field of the occupied component alone. For realistic nearly disjoint packets this is an approximation justified by decoherence. The other component is empty in configuration space, even when the system packets overlap in ordinary SPACE. This is the effective wave function in Bohmian mechanics.

The empty component cannot simply be deleted from the fundamental theory. In a coherent interferometer, separated packets can later overlap and both contribute to the actual probability current. Their relative quantum phase can change the outgoing particle's path. An empty arm is therefore capable of future dynamical significance. When an environment records the alternatives in very many degrees of freedom, restoring their configuration-space overlap is overwhelmingly difficult; that explains effective collapse without a literal change to the universal wave function.

There are two serious objections. First, if every branch has the detailed functional organization of a detector and observer, what additional work do the particles do in making only the occupied branch a world with records? A many-worlds critic can question whether the particle ontology is doing explanatory work rather than merely labelling one branch. The Bohmian response is that ordinary objects and records are constituted by actual configurations, not by every formally similar pattern in $\Psi$. Assessing that response requires an account of material constitution; counting amplitudes alone settles neither side.

Second, the universal wave function affects particles without the usual reciprocal influence of particles on it. This can look unlike a material field. A nomological interpretation of the wave function treats its role more like law than additional matter, although time-dependent effective subsystem waves make that position philosophically demanding.^[ref]

Neither objection licenses assigning ordinary particle energy separately to every empty component as though it contained a second classical apparatus. Such an assignment would require further assumptions. The definite mathematical theory reproduces standard equilibrium statistics while retaining empty components. The problem is the cost and intelligibility of that ontology, especially relative to alternatives, not an unexplained failure of its specified guidance equation.

For the oppositely travelling components, write $u=c_1{e}^{ikx}$ and $v=c_2{e}^{-ikx}$ after removing the common time phase. Then ${\partial }_x\psi =ik(u-v)e^{-i\omega t}$. In $\mathrm{Im}({\psi }^{\ast }{\partial }_x\psi )$ the interference terms contribute no imaginary part, soA standing-wave interference pattern can therefore change density without adding a cross-term to the net current.

For barrier-top scattering, the stationary equation inside the barrier is $-{\hslash }^2{\psi }_2^{\prime \prime }/(2m)=0$. Thus its spatial part is linear: ${\psi }_2=(bx+c)e^{-i\omega t}$. Take incident amplitude one, $k=\sqrt{2mE}/\hslash$ and $\omega =E/\hslash$. Continuity of the spatial wave function and its derivative at the two finite steps givesSolving these four equations yieldsThe equal exterior velocities mean flux ratios are squared amplitude ratios:This is the threshold limit of finite-barrier scattering, not exponential decay with a nonzero interior decay constant.

The interior constants areSince $\mathrm{Im}[(bx+c{)}^{\ast }b]=\mathrm{Im}(c^{\ast }b)=k/(1+k^2{a}^2)$,It is positive and independent of position, matching the transmitted flux. Although the classical kinetic energy inside is zero, the complex linear quantum wave function carries conserved probability current; its phase varies with position.

identical particles have the same intrinsic properties, including mass, charge and spin. Their labels are bookkeeping devices rather than measurable identities. When wave functions overlap, quantum mechanics supplies no observable trajectory that permanently identifies “particle 1” independently of “particle 2”. A many-particle description must therefore treat exchange consistently. The Hamiltonian and physical observables are invariant under relabelling identical particles.

For two particles, the exchange operator obeys ${\hat{P}}_{12}^2=\hat{I}$. The usual three-dimensional symmetrization postulate selects states symmetric under exchange for bosons and antisymmetric for fermions. In relativistic quantum theory, the Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions under its locality and positivity assumptions; the connection is not derived merely from ${\hat{P}}_{12}^2=\hat{I}$ in elementary nonrelativistic mechanics. For orthonormal one-particle states $u$ and $v$, the two-particle spatial combinations areThe sign concerns the complete state, including spin and other internal variables, not necessarily the spatial factor alone.

The antisymmetric state vanishes if both fermions occupy the same one-particle state: this is the Pauli exclusion principle. For $N$ fermions in orthonormal spin orbitals ${\chi }_j$, the normalized Slater determinant $\Psi =(N!{)}^{-1/2}\det [{\chi }_j(i)]$ supplies the required exchange sign. Bosonic occupation has no corresponding upper limit, and symmetrized products describe many bosons. Fermionic occupation is zero or one per complete one-particle state; two electrons can still occupy the same spatial atomic orbital with opposite spin.

Spin and spatial symmetry are linked. A two-electron spin singlet state is antisymmetric, so its spatial factor must be symmetric. A symmetric triplet state instead requires an antisymmetric spatial factor. Exchange then changes pair correlations and the energies available when interactions depend on particle separation, even without introducing a new classical “exchange force”. It contributes to atomic shell structure, bonding and magnetic behaviour.

In an occupation-number description, bosonic creation and annihilation operators satisfy $[{\hat{a}}_i,{\hat{a}}_j^{†}]={\delta }_{ij}$, whereas fermionic operators satisfy $\{{\hat{c}}_i,{\hat{c}}_j^{†}\}={\delta }_{ij}$ and anticommute among themselves. The latter imply $({\hat{c}}_i^{†}{)}^2=0$. This formalism implements the same exchange constraints without assigning permanent particle labels.

For noninteracting particles in thermal equilibrium, the mean occupation of a state of energy $ϵ$ isFermi-Dirac statistics produces a filled Fermi sea at zero temperature and degeneracy pressure; Bose-Einstein statistics permits macroscopic Ground state occupation when the density of available states and thermodynamic conditions allow condensation. The different statistics therefore have measurable collective consequences, not just different notation.^[ref]

When phase-space occupation is dilute, both distributions reduce to the Maxwell–Boltzmann statistics limit. Likewise, well-separated localized wave packets have negligible overlap and small exchange corrections, so effectively labelled classical trajectories can become a useful approximation. In two spatial dimensions more general exchange statistics are possible; the standard boson/fermion account here concerns ordinary three-dimensional particles and their total states.

In the time-independent Schrödinger equation, $-{\hslash }^2{\psi }^{\prime \prime }/(2m)$ is the kinetic-energy operator acting on the state, $V\psi$ its potential-energy contribution and $E\psi$ the energy-eigenvalue term. The complex wave function is a probability amplitude: the Born rule assigns density $|\psi {|}^2$ for a normalized single-particle state. A plane-wave beam uses an incident-amplitude or flux normalization instead of a unit-integral wavefunction.

For a free plane wave, ${\psi }^{\prime \prime }=-k^2\psi$, so $\boxed{E={\hslash }^2{k}^2/(2m)}$ and its time factor obeys $\hslash \omega =E$. Above the upward potential step, the bounded subthreshold solution has decay constantFor the PDF's unit incident amplitude, write $\psi =e^{ikx}+R{e}^{-ikx}$ on the left and $\psi =T{e}^{-\alpha x}$ on the right, omitting their common time phase. Continuity of both $\psi$ and ${\psi }^{\prime }$ at this finite step givesConsequentlyThe evanescent amplitude versus transmitted current distinction matters: the second quantity is a squared amplitude, not a transmitted-flux probability. The decaying tail has zero probability current, despite nonzero density and the possible value $|T{|}^2>1$.

Using $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$ and the electron mass and elementary charge specified above, the energy difference1eV gives $\alpha =5.12\mathrm{n}{\mathrm{m}}^{-1}$. Thus at the stated positionThis is relative to the unit incident density in the source figure, not an absolute normalized probability per metre. ^[ref]

Here $V(x)$ is potential energy, $E$ is the stationary state's total energy, and $\psi (x)$ is its spatial probability amplitude. The wave function has time factor $e^{-iEt/\hslash }$; $|\psi {|}^2$ is a position probability density for a normalized state. Stationary plane-wave scattering solutions are idealizations described by probability-current ratios, or components of a normalizable wave packet.

Above an upward step, a classical particle is fully transmitted with reduced speed. Quantum mechanically the abrupt wavelength change generally produces both reflected and transmitted amplitudes even though both regions are classically allowed. With unit incident amplitude and no wave arriving from the right,The reflected component has signed wavevector $-k_1$ and the same wavenumber magnitude as the incident component. A finite potential jump with constant mass and no delta interaction requires continuity of $\psi$ and ${\psi }^{\prime }$:The probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ gives probability reflection and transmission $\mathcal{R}=|r{|}^2$ and $\mathcal{T}=(k_2/k_1)|t{|}^2$, with $\mathcal{R}+\mathcal{T}=1$. The amplitude $t$ may exceed one without violating conservation of total probability, because the transmitted group speed is smaller.

For $0<E<V_0$, let $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The bounded right-hand solution is $t{e}^{-\kappa x}$, an evanescent wave tail with amplitude penetration scale $1/\kappa$. It carries no outgoing probability current in an infinitely long forbidden half-space, so reflection is total. This differs from transmission through a finite potential barrier, where a second boundary can support quantum tunnelling.

For a downward step, replace $k_2$ by $\sqrt{2m(E+V_0)}/\hslash >k_1$. The same boundary conditions give a nonzero negative $r$, corresponding to a reflected amplitude with phase reversal. If reflection were forbidden, $r=0$ would require both $t=1$ and $k_1=k_2$, impossible for a nonzero step. Reflection therefore does not require a classical potential barrier; the discontinuity of wavenumber mismatch is enough.

For the specified electron energies, $k_2/k_1=\sqrt{40/100}=\sqrt{0.4}$, so the requested amplitude coefficients areThe corresponding probability coefficients are approximately $\mathcal{R}=0.05069$ and $\mathcal{T}=0.94931$. electron mass cancels from these ratios; there is no need to invent an additional material effective mass.

The ground-state energies are $1.50\mathrm{eV}$ and $0.376\mathrm{eV}$, respectively. For a particle in a box of width $L$, the wave function must vanish at both infinite walls. The lowest nonzero standing wave therefore has wavelength $2L$, giving the kinetic energy eigenvalueUsing the electron mass and converting joules to electronvolts givesDoubling the confinement length divides this kinetic energy by four. A zero potential energy inside the box does not imply zero total energy: a constant, zero-momentum wave function cannot satisfy both wall conditions unless it vanishes everywhere. This is a concrete manifestation of quantum confinement and the Heisenberg uncertainty principle.

The one-dimensional time-independent Schrödinger equation isFor the infinite square well on $[0,a]$, $V=0$ inside and the wave function is zero outside. The boundary conditions are $\psi (0)=\psi (a)=0$. For positive integer $n$, ${\psi }_n=A\sin (n\pi x/a)$ satisfies them, and ${\psi }_n^{\prime \prime }=-(n\pi /a{)}^2{\psi }_n$. Substitution gives the energy levelsNormalization requires $1={\int }_0^a|{\psi }_n{|}^{2}dx=A^{2}a/2$. The requested real positive constant is therefore

For the second isotope, consecutive gap differences are approximately $56.4$ MHz, but the smallest gap is about $1.5$ times this value. The higher-$F$ values for the five gaps are therefore $3/2,5/2,7/2,9/2,11/2$. The six levels run from $F=1/2$ to $11/2$. The electronic state is unchanged, so $J=5/2$ as in d; the endpoints giveFor example, fitting ${\nu }_F$ to $F|A|/h$ with the same equal-weight convention gives $56.408$ MHz. The nearly constant gap differences are compatible with the same leading model of the magnetic-dipole hyperfine interaction and the printed rounding.

Within an unchanged electronic state, and neglecting isotope-dependent changes of the electronic wave function and hyperfine anomaly, $A\propto g_I$. Thusor the reciprocal magnitude is about $1.284$. This is a ratio of nuclear g-factors, not directly a ratio of nuclear magnetic moments: ${\mu }_I=g_{I}I{\mu }_N$, so the moment-magnitude ratio would additionally contain $I_{56}/I_{55}=3/(5/2)$ and is about $0.935$.

The positive transition frequencies do not fix either coupling sign. If a signed ratio is required, equal signs must be supplied as an additional assumption or determined by an ordering-sensitive measurement. No absolute nuclear g-factor is inferred without a calibrated electronic field coefficient. The second ladder in b illustrates the inferred half-integer $F$ sequence without turning that sign ambiguity into a falsely measured ordering.

Take parity symmetry in quantum field theory to preserve free time evolution, so the conjugate initial data reflect along with the field. Substitute the mode expansion of a free field into the defining parity operator relation and change the integration variable $\mathbf{p}\mapsto -\mathbf{p}$. Since $E_{-\mathbf{p}}=E_{\mathbf{p}}$, coefficient comparison, or comparison of both the field and its conjugate initial data, gives the intrinsic-even conventionHere parity symmetry in quantum field theory reverses spatial momentum and preserves time. The creation operator formulas follow by taking Hermitian adjoints. The given field relation has fixed the intrinsic phase to $+1$.

For one real scalar field, let $R$ act on one-particle wave functions by $(Rf)(\mathbf{p})=f(-\mathbf{p})$. It is a Hermitian operator with $R^2=1$, so $P_{-}=(1-R)/2$ is the orthogonal projector onto odd wave functions. Its second quantization is the odd-mode number operatorThe full momentum space is integrated in both expressions; the factors prevent double counting the pairs $\{\mathbf{p},-\mathbf{p}\}$. A suitable parity operator isFor verification, $[N_{\mathrm{odd}},a_{\mathbf{p}}]=-(a_{\mathbf{p}}-a_{-\mathbf{p}})/2$. Thus the even combination $a_{\mathbf{p}}+a_{-\mathbf{p}}$ is unchanged, and the odd combination $a_{\mathbf{p}}-a_{-\mathbf{p}}$ changes sign under conjugation by $e^{i\pi N_{\mathrm{odd}}}$, exactly reproducing $a_{\mathbf{p}}\mapsto a_{-\mathbf{p}}$. The same holds for creation operators. The odd-mode number operator has integer eigenvalues, so ${\mathcal{P}}^2=1$, and $\mathcal{P}|0⟩=|0⟩$. Multiplying by an overall constant phase would not change its conjugation action. This is parity as odd-mode number parity.

In the scalar two-particle model, the fermionic exchange symmetry and bosonic exchange symmetry conditions areFor particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi (x_1,x_2)=\Psi (x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi =E\Psi$ implies $H(S\Psi )=E(S\Psi )$. The componentsare respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero quantum state, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.

The normal modes of two harmonically coupled identical particles provide the separation. The coordinate transformation to $y_{+},y_{-}$ is orthogonal, with absolute Jacobian one. Exchanging the particles leaves $y_{+}$ fixed and changes $y_{-}$ to $-y_{-}$. Hence bosonic quantum states are even and fermionic quantum states odd in the relative coordinate. The chain rule givesAlso $x_1^2+x_2^2=y_{+}^2+y_{-}^2$ and $(x_1-x_2{)}^2=2y_{-}^2$. Combining the two trapping potentials with the interaction therefore gives the separated Hamiltonian operatorThese are independent oscillators of frequencies $\omega$ and $\omega \cos \alpha$. Both frequencies are positive in the stated range. Their product eigenstates have energiesThe parity of an oscillator level is $(-1{)}^n$. Consequently bosons allow even $n_{-}$ and fermions odd $n_{-}$, while $n_{+}$ is unrestricted.

The bosonic ground state uses $n_{+}=n_{-}=0$, and the fermionic ground state uses $n_{+}=0,n_{-}=1$. The oscillator length in the relative mode isUsing the normalized Gaussian for each mode givesIndeed ${\varphi }_{\tau }(y)=\sqrt{2}y{\psi }_{\tau }(y)/\tau$ by differentiating the Gaussian, so it is the normalized odd first excited state. The ground energies are $E_{\mathrm{B}}=\hslash \omega (1+\cos \alpha )/2$ and $E_{\mathrm{F}}=\hslash \omega (1+3\cos \alpha )/2$.

Choose normalized factors for a normalized product quantum state; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,If the supplied factors are not individually normalized, the general expression has an additional factor $\Vert \psi {\Vert }^2$; choosing normalized factors is the convention in the displayed formula.

For the ground states, put $\rho (y)=|{\psi }_{\tau }(y){|}^2=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$. Then $|{\varphi }_{\tau }(y){|}^2=2y^2\rho (y)/{\tau }^2$ and ${\rho }^{\prime }=-2y\rho /{\tau }^2$. Thus, for differentiable $f$ with suitable Gaussian integrability and vanishing boundary terms,This is integration by parts, and every polynomial satisfies the required endpoint conditions. The relative-coordinate moments in even and odd oscillator ground states follow by substituting $f(y)=y^n$:For odd $n$, both expectations vanish by parity, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular ${\mathbb{E}}_{\mathrm{B}}[y_{-}^2]={\tau }^2/2$ and ${\mathbb{E}}_{\mathrm{F}}[y_{-}^2]=3{\tau }^2/2$, so the fermionic root-mean-square particle separation is $\sqrt{3}$ times the bosonic value. The fermionic probability density vanishes at coincidence $y_{-}=0$, expressing Pauli exclusion principle in this spatial symmetry sector. This is an exchange constraint on the wave function, not an additional classical force. As $\alpha$ approaches $\pi /2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator ground state.

For identical fermions the full wave function is antisymmetric:In particular it vanishes when the two complete one-particle arguments coincide. Applying this to the position-only model assumes that no independent spin symmetry changes the required spatial parity.

For identical bosons the full wave function is symmetric under the particle exchange operator:In the given position-only model this is the spatial parity under the particle exchange operator. If additional spin variables are present, the symmetry applies to the complete state including them.

The orthogonal coordinate change preserves the kinetic energy and takes the particle exchange operator to $(x_{+},x_{-})\mapsto (x_{+},-x_{-})$. Thus the spatial wave function is even in $x_{-}$ for bosons and odd for fermions.

The total potential energy and Hamiltonian separate asAssume $1+2\kappa >0$ so the relative oscillator is confining. Its spectrum is $E_{n_{+},n_{-}}=\hslash \omega (n_{+}+1/2)+\hslash \Omega (n_{-}+1/2)$, and its parity in the two-particle relative coordinate is $(-1{)}^{n_{-}}$.

The lowest boson state uses $n_{+}=n_{-}=0$, whereas the lowest fermion state uses $n_{+}=0$, $n_{-}=1$. ThereforeFor an oscillator of frequency $\Omega$, $⟨x_{-}^2{⟩}_n=(n+1/2)\hslash /(m\Omega )$. Since $(x_1-x_2{)}^2=2x_{-}^2$,The factor three is the extra relative-coordinate excitation required by antisymmetry. The supplied oscillator wave functions correspond to kinetic energy $P^2/(2m)$; the extra factor $1/2$ printed before that term in the hint is inconsistent with those functions and the stated energies. The calculation uses the conventional kinetic energy dictated by particle mass $m$.

For a normalized state the position probability density is $\boxed{f=|\psi {|}^2}$. With a real potential, the Time-dependent Schrödinger equation and its conjugate giveThis is the continuity equation $\boxed{f_t+j_x=0}$. Integrating over the line yields $d\Vert \psi {\Vert }^2/dt=-j(\infty )+j(-\infty )=0$ under the stated decay. A real-valued wave function gives $\overline{\psi }{\psi }_x-\psi \overline{{\psi }_x}=0$, so its instantaneous current vanishes; the same is true up to a spatially constant phase.

Put $\alpha =m\omega /\hslash$. Complete the square in the overlap:The Gaussian integral is $\sqrt{\pi /\alpha }$, also for complex displacement by analytic continuation. HenceThe product of square roots fixes the branch convention implicit in the printed square root of $\overline{u}s$. At $u=s\ne 0$ the norm squared is $|N{|}^2|s|\sqrt{\pi /\alpha }{e}^{2\alpha |s{|}^2}$, which becomes one for the stated $N$. In particularThe centre is $-2\mathrm{Re}s$ and its position variance is $1/(2\alpha )$.

When differentiating in $s$, keep $N$ fixed on the chosen constant-radius orbit. ThenSubstitution gives $HG=\hslash \omega s{G}_s$. Since $\dot{s}=-i\omega s$, the chain rule gives $\boxed{i\hslash G_t=HG}$. Continue $\sqrt{s(t)}$ continuously along the time orbit; its sign after a full orbit supplies the usual oscillator zero-point phase. The case of zero displacement is obtained as a limit, since the separate printed $N$ and $\sqrt{s}$ expressions are singular at $s=0$.

Finally $j=(\hslash /m)\mathrm{Im}(\overline{G}{G}_x)=-2\omega \mathrm{Im}s|G{|}^2$. For real nonzero $a$ this isThe printed final current omits the Gaussian normalization factor $\sqrt{m\omega /(\pi \hslash )}$. Retaining that factor is required both by the normalized state and by the continuity equation. This current is the Gaussian probability density times the velocity of its moving centre, $2a\omega \sin \omega t$.

For bound hydrogen atom states, the radial wave function is regular and finite at the origin, and decays at infinity so that $4\pi {\int }_0^{\infty }{r}^2|\Psi (r){|}^{2}dr<\infty$. Equivalently the reduced radial function $w(r)=r\Psi (r)$ has $w(0)=0$ and is square integrable on the half-line. A singular $1/r$ branch is not an admissible regular Coulomb eigenfunction; its derivative energy diverges and it introduces an origin singularity absent from the Hamiltonian domain.

Write $\kappa =e^2/(4\pi {ϵ}_0)$, so $a={\hslash }^2/(m\kappa )$. Multiplication by $r$ turns the radial equation intoSince $w=a{e}^{-\lambda R}f(R)$ and $r=aR$, substitution givesChoose $\boxed{\lambda =\mu a>0}$ to remove the constant coefficient and factor out the decaying exponential. For $f={\sum }_{k\ge 0}{a}_k{R}^{k+c}$, the lowest power has coefficient $c(c-1)a_0$, so $\boxed{c(c-1)=0}$. The $c=0$ behavior would give $\Psi \sim a_0/R$, violating regularity; choose $c=1$.

Matching successive powers then yieldsA nonterminating tail eventually has one sign, and its coefficient ratio is asymptotic to $2\lambda /k$. More precisely, for any $0<\delta <\lambda$, the ratio is eventually at least $(2\lambda -\delta )/(k+1)$. Its same-sign tail consequently dominates a positive constant times the exponential series of rate $2\lambda -\delta$, up to finitely many terms. The factor $e^{-\lambda R}$ cannot then make $\Psi$ decay. Thus square integrability forces termination.

If the last nonzero term is $a_{n-1}{R}^n$, termination requires $\lambda n=1$. ThereforeFor $n=1$, $\lambda =1$, $f$ is proportional to $R$, andThe radial volume element is $4\pi r^{2}dr$, so the mean radius is the ratio of the radius-weighted norm integral to the norm integral. With $r=aR$, the common angular and normalization factors cancel:In the ground state $\boxed{⟨r⟩=3a/2}$.

A permutation $\pi$ of identical particles multiplies a fermionic state by $\mathrm{sgn}(\pi )$; a bosonic state is invariant. Particle exchange keeps $y_{+}$ fixed and sends $y_{-}$ to $-y_{-}$. Consequently the spatial wave function is even in $y_{-}$ for the stated bosonic model and odd for the stated fermionic model. This assumes no additional spin variables in the supplied model; exchange of full particle states is the underlying rule.

The coordinate change is orthogonal, with unit Jacobian determinant, so the kinetic energy becomeson the square $|y_{+}|,|y_{-}|<a$ with zero boundary values. Product states ${\psi }_r(y_{+}){\psi }_s(y_{-})$ have energyThe one-dimensional eigenfunction is even for odd $s$ and odd for even $s$. Bosons therefore allow odd $s$, whereas fermions require even $s$. The lowest bosonic pair is $(r,s)=(1,1)$; its first excited pair is $(2,1)$. The lowest fermionic pair is $(1,2)$. ThusThe printed bosonic ground-energy claim omits the factor ${\pi }^2/4$; the value above follows directly from the supplied one-dimensional spectrum and kinetic energy.

The separation is $|x_1-x_2|=\sqrt{2}|y_{-}|$. The normalized $y_{+}$ factor integrates out. In the fermionic ground state, the relative-coordinate density is $a^{-1}{\sin }^2(\pi y_{-}/a)$, givingFor the bosonic ground state it is $a^{-1}{\cos }^2(\pi y_{-}/(2a))$, givingBoth integrals follow by replacing squared trigonometric functions by their constant and cosine parts and integrating once by parts. Henceand the difference is $\boxed{2\sqrt{2}a/{\pi }^2}$. The odd relative state vanishes at coincidence, which increases its expected separation.