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Invert the operator definitions to obtain $a=\sqrt{m\omega /(2\hslash )}\hat{x}+i\hat{p}/\sqrt{2\hslash m\omega }$ and the adjoint formula for $a^{†}$. The canonical commutation relation yields $[a,a^{†}]=1$, with $[a,a]=[a^{†},a^{†}]=0$. Substitution givesWrite $N=a^{†}a\ge 0$. Since $[N,a]=-a$ and $[N,a^{†}]=a^{†}$, these are lowering and raising operators. Applying $a$ to a ground state would lower its energy unless it vanishes, so $a|0⟩=0$. From the normalized unique ground state constructThe commutator gives the factorial normalization and $a|n⟩=\sqrt{n}|n-1⟩$. For any energy eigenstate, repeated lowering cannot pass below $N=0$; positivity of each lowered norm forces its $N$ eigenvalue to be a nonnegative integer and lowering eventually reaches the ground state. Its uniqueness then yields the entire eigenstate ladder. The usual oscillator on the real line has discrete complete spectrum because its potential is confining, so these states span its Hilbert space.

For the perturbation $\lambda W$ with $W=\hslash \omega (a^2+a^{†2})$, nondegenerate time-independent perturbation theory gives $E_n^{\prime }=E_n+\lambda W_{nn}+{\lambda }^2{\sum }_{m\ne n}|W_{mn}{|}^2/(E_n-E_m)+O({\lambda }^3)$. The first-order term vanishes. Only $m=n-2,n+2$ contribute, with squared matrix elements $(\hslash \omega {)}^{2}n(n-1)$ and $(\hslash \omega {)}^2(n+1)(n+2)$ respectively. ThusThe absent downward transition for $n=0,1$ is automatically represented by $n(n-1)=0$. Direct substitution also givesFor $|\lambda |<1/2$ this is another oscillator with mass $m^{\prime }=m/(1-2\lambda )$ and frequency ${\omega }^{\prime }=\omega \sqrt{1-4{\lambda }^2}$. ThereforeIts expansion is $E_n[1-2{\lambda }^2+O({\lambda }^4)]$, agreeing with the perturbative result and sharpening its remainder. At the excluded endpoints confinement is lost; beyond them the quadratic Hamiltonian is not bounded below.

The microcanonical ensemble assigns equal probability to accessible quantum states in a narrow energy shell of an isolated system with fixed energy, particle number and external parameters. To derive the canonical ensemble, put a small system in weak thermal contact with a much larger isolated bath. If the small-system state has energy $E_i$, its probability is proportional to the number ${\Omega }_b(E_{\mathrm{total}}-E_i)$ of bath states. Expanding the bath entropy gives $S_b(E_{\mathrm{total}}-E_i)=S_b(E_{\mathrm{total}})-E_i/T+\cdots$, since $\partial S_b/\partial E=1/T$. Negligible interaction energy and an effectively constant bath temperature therefore give $p_i=e^{-E_i/(k_{B}T)}/Z$. The microcanonical ensemble suits an isolated fixed-energy system; the canonical ensemble suits a system exchanging energy with a thermostat at fixed temperature, with particle number still fixed.

For one atom, summing the finite geometric series over $m=-J,\dots ,J$ gives the partition functionAt $x=0$ the ratio is understood by continuity as $2J+1$. Logarithmic differentiation gives the mean magnetic dipole momentThis is odd in $x$, has small-$x$ expansion ${\mu }_{B}J(J+1)x/3+O(x^3)$, and saturates at $\pm J{\mu }_B$. The requested common graph is plotted in units of ${\mu }_B$: its ordinate is $⟨{\mu }_z⟩/(J{\mu }_B)$, so multiplication by ${\mu }_B$ gives $⟨{\mu }_z⟩/J$. The normalized slopes at zero are $(J+1)/3$ and the limits are $\pm 1$.

The magnetic susceptibility isEquivalently it is $N{\mu }_B^2\mathrm{Var}(m)/(k_{B}T)\ge 0$. For $k_{B}T\gg {\mu }_B|B|$, the apparent singular terms cancel, yielding Curie lawFor fixed nonzero $B$, $T\to 0$ and integer $J\ge 1$, the first excited magnetic level is separated from the aligned ground state by ${\mu }_B|B|$. Its exponentially small occupation givesThe moments are saturated, so changing the magnetic field hardly changes their alignment. This is a fixed-nonzero-field low-temperature limit. At exactly $B=0$, differentiating first gives the Curie expression for every $T>0$, which diverges as $T\to 0$; the two limits do not commute in this noninteracting model. For $J=0$ both moment and susceptibility vanish identically.

The intrinsic parity $\eta$ of a particle is the parity eigenvalue of its one-particle state at rest:For the particles in this question, $\eta =\pm 1$. A two-particle state has multiparticle parity ${\eta }_1{\eta }_2(-1{)}^{\ell }$, where $\ell$ is its relative orbital angular momentum.

The pionic-deuterium ground state has atomic $\ell =0$. Since the pion has spin zero, its total angular momentum is therefore the deuteron spin, $J=1$, and its parity is ${\eta }_d{\eta }_{\pi }={\eta }_{\pi }$. The two neutrons are identical fermions. Their antisymmetric spin-one-half singlet state would require even orbital $\ell$, but $S=0$ and $J=1$ would instead require $\ell =1$. Hence they must occupy the symmetric spin-one-half triplet state, so the orbital state is antisymmetric and $\ell$ is odd. The addition of angular momentum with $S=J=1$ permits $\ell =0,1,2$, leaving $\ell =1$. Thus parity conservation givesso the pion has intrinsic parity $\boxed{{\eta }_{\pi }=-1}$.

In ${\Delta }^{-}\to {\pi }^{-}+n$, the final total spin is $S=1/2$. Coupling it to $\ell$ to obtain $J=3/2$ givesso the allowed orbital angular momenta are $\boxed{\ell =1,2}$. The observed angular probability is ${\cos }^2\theta =|P_1(\cos \theta ){|}^2$, where $P_1$ is a Legendre polynomial; it therefore selects the $\ell =1$ partial wave rather than $\ell =2$. The parity selection rule for a two-body decay now givesTherefore the delta particle has intrinsic parity $\boxed{{\eta }_{{\Delta }^{-}}=+1}$.

Write $m_i={\mu }_i^2$ and $r^2={\varphi }_1^2+{\varphi }_2^2$. The uniform Landau free energy isso stationarity requiresFor $m_1,m_2>0$, the unique ground state is $(0,0)$ and the ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$ discrete symmetry is unbroken. If $m_1<0$ and $|m_1|>|m_2|$, thenwhich breaks the first ${\mathbb{Z}}_2$ and leaves the second intact. This includes the case in which both masses are negative, because $m_1$ is then the more negative one. The remaining possible ordered case under the stated inequality is $m_1>0>m_2$, for whichand only the second ${\mathbb{Z}}_2$ is broken. The Hessian matrix in each ordered state is positive because the uncondensed direction has squared mass $m_k-m_j>0$, where $m_j$ is the condensed, more negative mass.

Near either continuous transition the nonzero order parameter is proportional to $(-m_i{)}^{1/2}$. Hence the mean-field critical exponents are

The lower critical dimension is the dimension at or below which fluctuations destroy the proposed finite-temperature ordered phase. Here the broken symmetry is discrete, so $d_{\mathrm{l}}=1$. In one dimension a domain wall interpolating between the two signs has finite energy, whereas its possible position gives an entropy growing as $\log L$. Domain walls therefore occur with nonzero density at every positive temperature and split the system into domains of finite typical length. Thus there is no finite-temperature ordered phase in one dimension.

The canonical commutation relations are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hslash {\delta }_{ij}$. Substitution into the definitions of the creation and annihilation operators gives

The normalized ground state is defined byRepeated use of $[A_i,A_i^{†}]=1$ gives the Cartesian number state of the three-dimensional isotropic harmonic oscillatorwith energy eigenvalue

Inverting the ladder-operator definitions yieldsThe antisymmetry of the Levi-Civita symbol then cancels the two-creation and two-annihilation terms in $\mathbf{L}=\mathbf{X}\times \mathbf{P}$, leaving the orbital angular momentum in oscillator ladder operatorsWrite $|j⟩=A_j^{†}|0⟩$. ThenSince the orbital angular momentum eigenvalue is $\ell (\ell +1){\hslash }^2$, every first-excited state hasFinally,Therefore the normalized magnetic quantum number $m=+1$ state can be chosen aswhich obeys $L_z|\psi ⟩=\hslash |\psi ⟩$ and $\Vert |\psi ⟩\Vert =1$.

For part (a), the adjoint operator isUsing the canonical commutation relation $[X,P]=i\hslash I$,Thus $A$ and $A^{†}$ are the annihilation and creation operators of the quantum harmonic oscillator.

For part (b), writeSince $[A,K]=A^{†}$ and $[A^{†},K]=A$, differentiation of the unitary conjugation givesThe initial conditions are $A(0)=A$ and $A^{†}(0)=A^{†}$. Solving this pair of linear ordinary differential equations gives the Bogoliubov transformationThis is the transformation implemented by the single-mode squeeze operator.

For part (c), invert the definition of $A$:The transformation above impliesThe ground state has zero position and momentum means and variances $\hslash /(2m\omega )$ and $m\hslash \omega /2$. Hence the squeezed vacuum state $|\gamma ⟩=S(\gamma )|0⟩$ hasThe reciprocal squeezing factors cancel, soThe state therefore continues to saturate the Heisenberg uncertainty relation.

Before the change, the normalized ground state of the infinite square well isFor an allowed post-quench energy $E\in (0,U_0)$ satisfying part (i), define the unnormalized eigenfunctionIts normalization factor isThe Born rule therefore givesFor a value of $E$ that is not an eigenvalue, this probability is zero. The sudden change leaves the wavefunction fixed, while the energy eigenbasis changes.

The unitary time evolution isFirst suppose that time reversal $T$ is linear and unitary. Differentiatingat $t=0$ givesand linearity permits the scalar $i$ to pass through $T$. HenceIf $H|E⟩=E|E⟩$, thenThus every energy has its negative as an energy. The Coulomb Hamiltonian has arbitrarily large positive energies in its continuum, so this symmetry would force energies arbitrarily far below zero. It would have no lowest energy eigenvalue and therefore no stable ground state. This is the unitary time reversal reverses the energy spectrum obstruction.

Now let $T$ be antiunitary. It is conjugate linear, soDifferentiating the same intertwining equation givesThereforeTime reversal now maps an energy eigenstate to another state with the same energy:The spectrum is preserved rather than reflected through zero, so the lower-bounded Coulomb spectrum and its stable ground state are compatible with time-reversal symmetry. This is why antiunitary time reversal preserves the energy spectrum.

Since ${\hat{x}}_i=P_{1,i}$ and $I-{\hat{x}}_i=P_{0,i}$, the diagonal Hamiltonian iswith the identity on every unlisted qubit. Each computational-basis state is an eigenstate, and its eigenvalue is minus the cost of the corresponding cut. Part i supplies cost four, while the assumed bound $|{\lambda }_{\min }|\le 4$ rules out a lower energy. Thus the ground-state energy is $-4$. For the assignment $(x_1,x_2,x_3,x_4,x_5)=(0,1,0,0,1)$, one ground state isIts bitwise complement $|10110⟩$ is another ground state, as are the basis states corresponding to the other maximum cuts.

Assume $eB>0$; reversing the sign only exchanges the spin labels. Since $\mathbf{B}\cdot \mathbf{\sigma }=B{\sigma }_3$, the Pauli term shifts the spinless energies by $\pm \hslash {\omega }_c/2$. ThusAfter collecting equal energies, the spin splitting of Landau levels isThe level $\ell =0$ contains only $(n,sigma)=(0,-1)$ and has degeneracy $D$. Every level $\ell \ge 1$ contains $(\ell ,-1)$ and $(\ell -1,+1)$ and has degeneracy $2D$.

For $N$ noninteracting electrons, the Pauli exclusion principle fills these states from the bottom. If $0\le N\le D$, every electron fits in the zero-energy level andFor $N>D$, writeThen the first $k$ positive levels are full and $r$ states in level $k+1$ are occupied. The ground-state energy isEquivalently, onThe graph is continuous and piecewise linear: it is flat up to $N=D$, then has slopes $\hslash {\omega }_c,2\hslash {\omega }_c,3\hslash {\omega }_c,\dots$, with kinks at

Separation with Dirichlet walls givesFor $a<b<c$, the ground state $(1,1,1)$ has a nondegenerate energy. The cheapest excitation changes the quantum number in the longest direction, so the first excited state $(1,1,2)$ also has a nondegenerate energy.

If $a<b=c$, the ground-state energy remains nondegenerate, but $(1,2,1)$ and $(1,1,2)$ have equal first-excited energy, giving degeneracy two.

A global symmetry is spontaneously broken when it preserves the action but does not preserve a chosen ground state. If $G$ is broken to the stabilizer $H$, the degenerate vacua form a vacuum manifold $G/H$. The Goldstone theorem states that a relativistic theory has one massless scalar mode for each broken continuous internal generator, so the standard counting gives $\dim G-\dim H$ Goldstone bosons.

A real nonnegative normalized filter $w$ defines $g_Z=\int w(t)e^{itH}{h}_Z{e}^{-itH}dt$. In an energy eigenbasis, it multiplies each matrix element by $\hat{w}(E_i-E_j)$. A Fourier cutoff at the spectral gap removes couplings between a unique ground state and excitations. The filtered terms still sum to $H$, because the total Hamiltonian is unchanged by its own unitary time evolution. Each term commutes with the ground projector, but need not be minimized there.

Spontaneous symmetry breaking occurs when the action and equations have a symmetry but a chosen ground state is invariant only under a proper subgroup. Acting with the broken symmetry generates a degenerate vacuum manifold.