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Write $\kappa =\sqrt{2m{V}_0}/\hslash$ and $q=\kappa a$. At low energy, the $S$ wave normally dominates. Under the additional stated near-resonance assumption $ka\ll |{\delta }_0|$, neglect $ka$ inside the cotangent of the matching condition and set $k^{\prime }\simeq \kappa$:Substitution in ${\sigma }_0=4\pi {\sin }^2{\delta }_0/k^2$ givesThis is a near-resonance square-well approximation, not a uniform formula for every well radius. A phase shift is defined modulo $\pi$; its representative must be chosen consistently when applying the comparison with $ka$.

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]

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

A constant electrostatic energy offset is removed locally by the quantum phase substitution $\psi =e^{-i{V}_{0}t/\hslash }{\psi }_0$, since $(i\hslash {\partial }_t-V_0)\psi =e^{-i{V}_{0}t/\hslash }i\hslash {\partial }_t{\psi }_0$. Therefore the two frequency branches becomeThe normalization uses the kinetic frequency ${\omega }_k$, while the canonical energy eigenvalues are $E=V_0\pm \hslash {\omega }_k$. The opposite-frequency branch is not obtained by replacing ${\omega }_k$ by the same shifted positive frequency in both terms. A globally constant offset produces no electric field, whereas different offsets on opposite sides of a step cannot be removed by a single smooth global phase.

The negative-kinetic-energy branch is interpreted as an antiparticle channel. In the overlap $m{c}^2<E<V_0-m{c}^2$, particles can escape to the left and oppositely charged antiparticles to the right. This interval exists only when $V_0>2m{c}^2$. The external electrostatic source supplies the energy for pair creation; the reflected particle flux can exceed the incident one because the outgoing state need not contain only the incident particle.

The conserved quantity is signed charge current, not a positive one-particle probability in both regions. Negative $\mathcal{T}$ denotes rightward transport of opposite charge, so it is consistent with $\mathcal{R}>1$ and $\mathcal{R}+\mathcal{T}=1$. A quantized charged field with particle and antiparticle creation operators gives the appropriate interpretation of this Klein paradox. This bosonic result must not be replaced by a fermionic reflection law or interpreted as a failure of charge conservation.^[ref]

Set ${\vartheta }_{\alpha }(t)={\hslash }^{-1}{\int }^t{E}_{\alpha }(s)ds$. Differentiating the expansion produces ${\dot{a}}_{\beta }|{\phi }_{\beta }⟩+a_{\beta }|{\dot{\phi }}_{\beta }⟩-i{E}_{\beta }{a}_{\beta }|{\phi }_{\beta }⟩/\hslash$ for each term. The energy term cancels $H|\Psi ⟩$ in the Schrödinger equation. Projecting the remainder with $⟨{\phi }_{\alpha }|$ and using orthonormality yieldsThe diagonal term is purely imaginary and supplies the geometric phase. Off-diagonal instantaneous eigenstate couplings transfer population; differentiating the eigenvalue equation gives $⟨m|\dot{n}⟩=⟨m|\dot{H}|n⟩/(E_n-E_m)$ when the levels are distinct.

The quantum propagator is the position-space kernel of the time-evolution operator: $\Psi (x,t)=\int K(x,t|x^{\prime },t^{\prime })\Psi (x^{\prime },t^{\prime })d{x}^{\prime }$. For a time-independent Quantum Hamiltonian, $U=e^{-iH(t-t^{\prime })/\hslash }$. Inserting a complete orthonormal energy eigenbasis givesA continuous spectrum requires the corresponding integral. At equal times, completeness makes this kernel a Dirac delta function.

Let $I=m{R}^2$, $T=t-t^{\prime }>0$, and $\Delta =\theta -{\theta }^{\prime }$. Periodic angular wave functions have modes $e^{il\theta }/\sqrt{2\pi }$ with integer $l$ and energies ${\hslash }^2{l}^2/(2I)$. Normalizing the kernel with respect to $d{\theta }^{\prime }$ givesIf the integration measure is arc length $Rd{\theta }^{\prime }$, the corresponding kernel is $K_{\theta }/R$. This measure choice accounts for the otherwise ambiguous radius prefactor.

Define $A=m{g}^2/(2\pi {\hslash }^2)$, which has units of energy times length. Substituting the result of (b) into (c) givesAbsorb the real cutoff shift into the physical detuning,It follows thatUsing $k=\sqrt{2m{E}_k}/\hslash$, identifyHere $\gamma$ has units of square root of energy. This is the small-cutoff, renormalized resonant-level amplitude; terms vanishing with $k\delta$ have been omitted. Taking $\delta \to 0$ at fixed measured detuning requires the bare ${ϵ}_0$ to depend on the cutoff.

The sign is checked by $f_k^{-1}=-(E_k-{\widetilde{ϵ}}_0)/A-ik$, which has the elastic $s$-wave unitarity form. Thus $\mathrm{Im}{f}_k=k|f_k{|}^2$, consistent with the optical theorem and ${\sigma }_{\mathrm{tot}}=4\pi |f_k{|}^2$. If the renormalized level is positive, at $E_k={\widetilde{ϵ}}_0$ the amplitude is $i/k$ and the scattering cross-section reaches $4\pi /k^2$.

Put $g=k\hslash /(2m\omega )$, which has units of energy. In the original oscillator basis,Write $A=\tilde{a}$, $B=\tilde{b}$, $C_2=\cosh 2\theta$ and $S_2=\sinh 2\theta$. The transformed number and pair terms obeyHence the coefficient of $AB+A^{†}{B}^{†}$ is $g{C}_2-\hslash \omega S_2$, which vanishes for

This cancels the requested pair terms but does not diagonalize the printed Hamiltonian. Its number-conserving coupling also transforms:After the pair cancellation, the last creation terms have coefficient $-g{S}_2/2$, nonzero for $k\ne 0$. They create two quanta from the $A,B$ vacuum, so that vacuum is not an eigenstate of the literal Hamiltonian. This is an inconsistency in the ground-state assertion used by the next part, present in the printed PDF.

For the actual coordinate-coupled system, the normal coordinates $q_{\pm }=(x\pm y)/\sqrt{2}$ and their conjugate momenta diagonalize $H$ into independent oscillators withA normalizable ground state requires $|k|<m{\omega }^2$. Eliminating just the two indicated cross-pair terms is not sufficient to establish that ground state.

At fixed particle number $N$, write $n_1=n$ and $n_2=N-n$. The contact interaction between two ring modes has energy eigenvaluesFor $g>0$, the nonnegative product $n(N-n)$ is minimized at the endpoints. ConsequentlyFor $N\ge 1$ this is a two-dimensional degenerate ground state subspace. $N=0$ instead has the unique Fock vacuum. In this projected model, putting all particles into one traveling mode avoids the extra exchange contribution paid when both distinct modes are occupied.

For the alternative coefficients, put $\alpha =\beta =A<0$ and $\gamma =0$. Then $E(n)=A[N(N-1)-2n(N-n)]$ is again minimized at $n=0,N$, with $E_{\min }=AN(N-1)$. Evenness of $N$ is unnecessary for this result; the balanced Fock state maximizes this attractive-interaction energy instead of minimizing it.

Add the one-particle energy term $H_0={ϵ}_1{n}_1+{ϵ}_2{n}_2$. It is diagonal in the same Fock state basis as the contact interaction between two ring modes. For $N=2$, the ordered basis $(|2,0⟩,|1,1⟩,|0,2⟩)$ givesFor $g>0$, the middle eigenvalue exceeds the mean of the two endpoint eigenvalues by $g/L$, so it cannot be the smallest. The ground state is therefore $|2,0⟩$ if ${ϵ}_1<{ϵ}_2$, and $|0,2⟩$ if ${ϵ}_2<{ϵ}_1$, withIf ${ϵ}_1={ϵ}_2$, every normalized linear combination of the two endpoint Fock states is a ground state. The same endpoint argument works for general fixed $N$, since the interaction contribution is a concave quadratic function of the mode occupation number.

For distinct positive integers $p,q$, use normalized box modes ${\phi }_1=\sqrt{2/L}\sin (p\pi x/L)$ and ${\phi }_2=\sqrt{2/L}\sin (q\pi x/L)$. Their contact interaction overlaps obeyUnlike traveling ring modes, these standing modes also allow pair conversion. For the first two box modes, or any distinct pair without a third-harmonic resonance, the contact interaction between two box modes isThe last two terms change an individual mode occupation number by two while conserving total particle number. Thus the Hamiltonian operator is not diagonal in the Fock state basis, and minimizing diagonal entries alone does not find its ground state. For example, at $N=2$ with no dispersion, $\left(|2,0⟩-|0,2⟩\right)/\sqrt{2}$ has the lowest energy $g/(2L)$, below either unmixed endpoint's $3g/(2L)$.

The wording permits modes beyond the first two. For completeness, the identity ${\sin }^{3}u=(3\sin u-\sin 3u)/4$ givesWhen $q=3p$, add $-g(a_1^{†2}{a}_1{a}_2+a_1^{†}{a}_2^{†}{a}_1^2)/(2L)$; when $p=3q$, add $-g(a_1^{†}{a}_2^{†}{a}_2^2+a_2^{†2}{a}_1{a}_2)/(2L)$. These extra terms change an individual occupation number by one and require diagonalizing the corresponding fixed-$N$ matrix as well.

Specify the magnetic coupling before fixing its sign: let the magnetic moment be $\mathbf{\mu }=\gamma \mathbf{S}$, where $\gamma$ is a signed gyromagnetic ratio. The Zeeman effect gives $H_Z=-\gamma \mathbf{B}\cdot \mathbf{S}$. For spin one-half, $\mathbf{S}=\hslash \mathbf{\sigma }/2$, so with $B=|\mathbf{B}|$ and $u=\beta \gamma \hslash B/2$,This is the quantum Gibbs state. Its eigenvalues are $(1\pm \tanh u)/2$, so its purity is $(1+{\tanh }^{2}u)/2$. For nonzero level splitting it approaches a pure state as $|u|\to \infty$: positive inverse temperature selects the lower energy, while negative inverse temperature selects the upper one. It is a maximally mixed state at $u=0$, including infinite temperature or zero magnetic splitting. At zero field the Gibbs state remains $I/2$; lowering temperature alone does not select a unique pure state from that degeneracy.

Negative temperature is meaningful for an internally equilibrated spin sector with a bounded energy spectrum and population inversion. It must be sufficiently isolated from unbounded motional degrees of freedom: the oscillator's unbounded spectrum, for instance, makes its partition sum diverge at negative inverse temperature. The two-level spin partition sum itself stays finite for either sign.

At the origin, the normalized bound state in a delta potential contributes $|{\phi }_0(0){|}^2=\kappa$, and the odd scattering eigenfunctions contribute zero. The even continuum functions from the preceding part haveTheir energy is $E_k={\hslash }^2{k}^2/(2m)$. Inserting these weights into the spectral resolution of a quantum propagator, and using evenness to extend the integral to the whole real axis, yields for $T>0$This is the quantum propagator of an attractive delta potential. The printed continuum exponent has an extra factor of $\hslash$: its time phase must be $-i{E}_{k}T/\hslash =-i\hslash k^{2}T/(2m)$. This correction also makes the exponent dimensionless. Oscillatory integrals are understood with an infinitesimal positive convergence factor before taking its limit.

In the ordered product basis $(|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩)$, the two-spin XY Hamiltonian isThe transverse term exchanges opposite spins and leaves parallel spins unchanged. Diagonalizing its two-dimensional block gives the complete orthonormal eigenbasisThe antisymmetric state is the singlet state; the other three are triplet states. Accidental equality of energies does not invalidate this eigenbasis.

Using the consistently normalized Jordan–Wigner transformation from part (c), write ${\sigma }_1^x{\sigma }_2^x+{\sigma }_1^y{\sigma }_2^y=2(s_1^{+}{s}_2^{-}+s_1^{-}{s}_2^{+})$. Since $f_1^{†}{P}_1=f_1^{†}$ and $f_1{P}_1=-f_1$, the parity string givesThereforeThe orthogonal transformation $b=(f_1+f_2)/\sqrt{2}$, $c=(f_1-f_2)/\sqrt{2}$ preserves the canonical anticommutation relations and diagonalizes this bilinear Hamiltonian:The Fock vacuum is $|\downarrow \downarrow ⟩$, with energy $2h$. Occupying $b$ alone gives the symmetric opposite-spin state with energy $-2J$; occupying $c$ alone gives the antisymmetric state with energy $2J$; occupying both gives $|\uparrow \uparrow ⟩$, up to a global phase, with energy $-2h$. This reproduces the entire spectrum in part (b).

Take the two magnetic moments along $z$ to be $\pm \mu$, so $H=-\mu B_z{\sigma }_z$ and the two energies are $\mp \mu B_z$. The quantum Gibbs state at inverse temperature $\beta =1/(k_{B}T)$ isIts Bloch vector is $(0,0,\tanh \beta \mu B_z)$. At $B_z=0$, or in the high-temperature limit, it approaches $I/2$; at $T\to 0^{+}$ with nonzero $\mu B_z$, it approaches the lower-energy pure state. Here $\mu$ is the magnitude or signed coefficient multiplying ${\sigma }_z$ in the magnetic Hamiltonian, rather than a moment with an additional unstated factor of two.

For a uniform classical spin $S\mathbf{n}$ with ${\mathbf{n}}^2=1$, the energy per bond isThe strict inequalities among the couplings make the $x$ direction the unique easy axis. ThusThese uniform configurations are also global classical minima: the largest singular value of the diagonal exchange matrix is $J^x$, so each bond has energy at least $-J^x{S}^2$, and both configurations saturate that bound.

A global rotation by $\pi$ about the $z$ axis sends $(S^x,S^y,S^z)$ to $(-S^x,-S^y,S^z)$, preserves every exchange product, and exchanges the two ground states. A $\pi$ rotation about $y$ does the same. The broken discrete symmetry can be represented by either operation.

Allow a freely chosen center $x_0$, set $u=k(x-x_0)$ and $z=e^u$. Direct differentiation of the proposed Sine-Gordon kink givesThe angle identity givesSince $k^2/b=a$, these formulas show $\boxed{{\varphi }^{\prime \prime }=a\sin (b\varphi )}$, precisely the static equation of motion.

For $x\to -\infty$ and $x\to +\infty$, the field approaches $0$ and $2\pi /b$ respectively. These are neighboring minima of the scalar potential when $ab>0$, and ${\varphi }^{\prime }$ vanishes exponentially at either end. The energy is therefore finite and the topological charge $Q=b[\varphi (+\infty )-\varphi (-\infty )]/(2\pi )$ is one. Reversing $x-x_0$ gives the corresponding antikink. If $ab<0$, the printed real exponential profile is unavailable; its formal imaginary scale is not a real finite-energy solution of the stated form.