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For two identical fermions of mass $m$ in the same one-dimensional quantum harmonic oscillator, use center of mass $X=(x_1+x_2)/2$ and separation $r=x_1-x_2$. The corresponding oscillator masses are $2m$ and reduced mass $m/2$, with the same frequency $\omega$. If $N,n$ are their occupation numbers, the unperturbed energy is $\hslash \omega (N+n+1)$. The fermionic exchange symmetry requires even $n$ with a spin-one-half singlet state and odd $n$ with a spin-one-half triplet state.

For a delta potential $\lambda \delta (r)$ the first-order shift is $\lambda |{\phi }_n^{\mathrm{rel}}(0){|}^2$. Every odd relative wavefunction vanishes at contact, so these states remain exact unaffected eigenstates of the spatial problem. For the even relative ground state, the oscillator length is $\sqrt{2}\ell$, where $\ell =\sqrt{\hslash /(m\omega )}$, and the shift is $\lambda /(\sqrt{2\pi }\ell )$, independent of $N$. This separates interaction-sensitive singlets from contact-insensitive triplets.

With oscillator length $\ell =\sqrt{\hslash /(m\omega )}$, the eigenfunctions are proportional to $H_n(x/\ell )e^{-x^2/(2{\ell }^2)}$. They have $n$ real nodes, parity $(-1{)}^n$, and energy $\hslash \omega (n+1/2)$.

The oscillator length fixes the spatial scale of a quantum harmonic oscillator of mass $m$ and angular frequency $\omega$. Its normalized ground wavefunction is ${\phi }_0(x)={\pi }^{-1/4}{\ell }^{-1/2}\exp [-x^2/(2{\ell }^2)]$, so $|{\phi }_0(0){|}^2=1/(\sqrt{\pi }\ell )$ and $⟨x^2⟩={\ell }^2/2$. The definition of $\ell$ is distinct from the root-mean-square displacement $\ell /\sqrt{2}$.

With $y=x-x_0$ and oscillator length $a_0=\sqrt{\hslash /(m\omega )}$, the first two harmonic oscillator eigenfunctions and energies areThe ground state is even and nodeless; the first excited state is odd with one node of a wavefunction. Their rms widths are $a_0/\sqrt{2}$ and $\sqrt{3/2}{a}_0$. Also, $m{\omega }^2{a}_0^2\sim \hslash \omega$ gives the characteristic extent directly.

For the shaken harmonic trap, expanding about the unshifted Hamiltonian givesThe scalar term gives only a common phase and no transitions. In time-dependent perturbation theory, initially $c_0=1$, $c_{n\ne 0}=0$ and the oscillatory factor for $0\to 1$ is $e^{i(E_1-E_0)t/\hslash }=e^{i\omega t}$. ThusThe odd operator $x$ makes transitions to even $n>0$ vanish by parity. In fact the first-order selection rule for a linearly driven oscillator is stronger: since $x=a_0(a+a^{†})/\sqrt{2}$,All excited $n\ne 1$ amplitudes vanish at first order, including odd $n>1$. This assertion does not apply to the surviving $c_0$ or to higher perturbative orders.

Put $\delta =\omega -\Omega$, $\Sigma =\omega +\Omega$. The exact first-order time integral isNear resonance, the second term oscillates rapidly at approximately $2\omega$ and is bounded by $1/\Sigma$, whereas the first grows as $T/2$ on resonance. Thus $\omega T\gg 1$ justifies the rotating-wave approximation, $\cos \Omega t\mapsto e^{-i\Omega t}/2$. At zeros of the resonant term, the omitted small contribution can dominate the residual; the approximation has no uniform relative-error bound there.

Using the requested order-one approximation $⟨1|x|0⟩\simeq a_0$ and $m{\omega }^2{a}_0/\hslash =\omega /a_0$ gives the square-pulse resonance profileThe exact harmonic oscillator matrix element instead gives half this probability. The finite central maximum is at $\Omega =\omega$; first zeros are at $|\Omega -\omega |=2\pi /T$, with smaller side lobes and a $1/{\delta }^2$ envelope. Increasing $T$ narrows the peak as $1/T$ and increases its height as $T^2$, as long as weak resonant driving keeps $p_1\ll 1$. Small displacement alone does not ensure this for arbitrarily long $T$; near resonance one also needs $\epsilon \omega T/a_0\ll 1$.

Write the trap angular frequency as $\omega$ and the single-particle oscillator length as $\ell =\sqrt{\hslash /(m\omega )}$. The center of mass coordinate $X=(x_1+x_2)/2$ and relative coordinate $r=x_1-x_2$ give independent quantum harmonic oscillators, both of frequency $\omega$, with masses $2m$ and reduced mass $m/2$. Label their excitation numbers by $N$ and $n$. The unperturbed energy eigenvalue is $\hslash \omega (N+n+1)$.

Particle exchange leaves $X$ unchanged and sends $r$ to $-r$. The exchange symmetry of two identical spin-one-half fermions therefore requires even $n$ to accompany the spin-one-half singlet state, and odd $n$ to accompany a spin-one-half triplet state. For the contact shifts of two spin-one-half fermions in a harmonic trap, the delta potential depends only on $r$. For odd $n$, the relative wavefunction vanishes at $r=0$, so its contact shift is zero. Within the first-excited manifold, both perturbations are diagonal in the labels $(N,n,S,M)$: this is a perturbation diagonal within a degenerate eigenspace. For $n=0$, the relative oscillator length is $\sqrt{2}\ell$, giving the first-order nondegenerate perturbation theory shiftThe Zeeman splitting is $2{\mu }_B{B}_{z}M$ for total spin projection $\hslash M$. Thus the five lowest states, counted with multiplicity, haveunder the assumed conditions $|U|,2{\mu }_B|B_z|\ll \hslash \omega$. The three triplet states have slopes $-2{\mu }_B,0,2{\mu }_B$, while both singlet states are field independent at this order. The three triplet state branches coincide at zero magnetic field. Their crossing with the excited singlet state is allowed because the total-spin sectors do not mix. The odd relative wavefunctions are in fact unaffected by this delta potential at any coupling; the displayed singlet state shifts retain only first order.

For repulsive contact interactions in a three-dimensional isotropic harmonic trap, neglecting the kinetic term in the Gross–Pitaevskii equation gives $n(r)=(\mu -m{\omega }^2{r}^2/2)/U_0$ inside $R=\sqrt{2\mu /(m{\omega }^2)}$ and zero outside. Normalization yields $N=8\pi \mu R^3/(15U_0)$, hence $\mu =(\hslash \omega /2)(15Na/a_0{)}^{2/5}$ and $R=a_0(15Na/a_0{)}^{1/5}$, with oscillator length $a_0=\sqrt{\hslash /(m\omega )}$. The density is parabolic but the wavefunction is its square root. Neglected kinetic energy rounds the edge and supplies a tail.

A sinusoidal trap displacement with amplitude $\epsilon$ and duration $T$ produces a resonant first-order excitation amplitude proportional to $\epsilon \omega T/a_0$, where $a_0$ is the oscillator length. Keeping its square much smaller than one is the long-time condition for weak excitation. The small-displacement condition $\epsilon /a_0\ll 1$ alone does not ensure uniform validity as $T$ grows. Exact linear driving remains solvable as a coherent state after this first-order approximation fails.