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For $f(x)=V_{0}x/L$, direct integration givesandThe reflected expression increases up to threshold and the transmitted expression decreases above it. The supremum is therefore approached from below at $E=V_0$ and equals

Ignoring the small radiative back-reaction, conservation of energy gives $\dot{x}=\sqrt{2(E-f(x))/m}$ and Newton's equation gives $\ddot{x}=-f^{\prime }(x)/m$. Integrating the Larmor formula with $dt=dx/\dot{x}$ yields, for $E>V_0$,If $E<V_0$, the particle turns at $x_{\ast }$ defined by $f(x_{\ast })=E$ and traverses the accelerated region twice, so the same formula holds with twice the integral from $0$ to $x_{\ast }$.

For a point charge, $p=qx$, so the radiation fields are transverse and satisfy $|E_{\mathrm{Rad}}|=c|B_{\mathrm{Rad}}|$. The radial Poynting vector isChoosing the acceleration as polar axis and integrating ${\sin }^2\theta$ over the sphere gives $\int {\sin }^2\theta d\Omega =8\pi /3$. Hence the Larmor formula is

In the continuum limit the number outside the ground state isAt the largest allowed chemical potential, $\mu =0$, the integrand behaves near zero as $E^{d/\alpha -2}$. The integral is finite exactly when $d/\alpha >1$. Excess particles must then occupy the ground state, so a Bose-Einstein condensate exists precisely for $\alpha <d$.

For a homogeneous dispersion $E(p)=A|p{|}^{\alpha }$, the kinetic expression for pressure givesThus $pV=DE$ with $D=\alpha /d$.

The Bose-Einstein distribution isPositivity requires $\mu$ not to exceed the ground-state energy, here zero. For a finite system above the condensation limit one has $\mu <0$; it approaches zero from below at Bose-Einstein condensation.

One quantum state occupies volume $(2\pi {)}^d/V$ in wavevector space. Counting the shell of radius $k$ and including spin degeneracy givesUsing $k=(E/A{)}^{1/\alpha }/\hslash$ gives the stated power with

TakeTheir integer span consists exactly of the cubic points together with the body-centre points, so the BCC set is a Bravais lattice. Using the inverse matrix supplied in the question, reciprocal primitive vectors areThese generate a face-centred cubic lattice; thus the reciprocal lattice of BCC is FCC, with the displayed scale.

A Bravais lattice is the integer span of three linearly independent primitive vectors. Its reciprocal lattice consists of vectors $G$ satisfying $G\cdot r\in 2\pi \mathbb{Z}$ for every lattice vector $r$.

Translations by lattice vectors commute with the Hamiltonian. Their simultaneous eigenvalues have the form $e^{ik\cdot r}$, so an energy eigenfunction may be chosen withThus ${\psi }_k=e^{ik\cdot x}{u}_k$ with lattice-periodic $u_k$, which is Bloch theorem. Wavevectors differing by a reciprocal-lattice vector describe the same translation character, and the Brillouin zone is a fundamental Wigner-Seitz cell of the reciprocal lattice.

Parity conservation and the negative parity of $X$ require odd orbital angular momentum $L$. Since the two $Y$ particles are identical bosons, their spin state must then be antisymmetric, so their total spin is $S=1$. Total angular momentum zero forces $L=S=1$.

Conditioning on spin projection $m_S=1$ leaves orbital projection $m_L=-1$. The angular density is therefore proportional to $|Y_{1,-1}(\theta ,\phi ){|}^2\propto {\sin }^2\theta$. Including the solid-angle measure, the requested probability is

For two equal spins, exchange acts on the coupled state $|j,m⟩$ by the Clebsch-Gordan coefficient symmetry factor $(-1{)}^{2j_1-j}$. It is therefore $-1$ when $j=2j_1-1$, making every state in that multiplet antisymmetric, and $+1$ when $j=2j_1-2$, making the next multiplet symmetric.

The addition of angular momentum rule givesFor each such $j$, measurement along any axis has magnetic quantum number $m=-j,-j+1,\dots ,j$.

Substituting the prolonged formulas showsso the nonlinear Schrödinger equation is invariant. Applying the symmetry to the given solitary wave produceswhich travels with speed $s$.

Direct substitution gives $g^s{g}^r=g^{s+r}$. Differentiating at $s=0$ gives $t{\partial }_x+ix(\psi {\partial }_{\psi }-{\psi }^{\ast }{\partial }_{{\psi }^{\ast }})$.

Writing the transformed field in transformed coordinates givesConsequentlywhere $\Phi =s\tilde{x}-s^{2}t/2$. These formulas, together with their complex conjugates and the analogous mixed derivatives, give the second prolongation. Differentiating at zero yields

The flow equations of the displayed vector field are ${\dot{\psi }}_1=-{\psi }_2$ and ${\dot{\psi }}_2={\psi }_1$. Hence $d\psi /ds=i\psi$, whose solution is $\psi (s)=e^{is}\psi (0)$. The vector field therefore generates the stated phase rotation.

The transformations obey $g_1^s{g}_1^r=g_1^{s+r}$ and $g_1^0$ is the identity. If $\tilde{x}(\tilde{t})=x(t)+st$, then $d^2\tilde{x}/d{\tilde{t}}^2=d^{2}x/d{t}^2$. Thus solutions of the free-particle equation are mapped to solutions, so this is a one-parameter group of Lie symmetries.

For $\nu =\mu +1>0$ small, $(1,0)$ has two negative eigenvalues and is a stable node. The two nearby equilibria satisfy $u=\pm \sqrt{\nu }+O(\nu )$ and are saddles. Their stable and unstable separatrices form the local phase portrait; trajectories in the wedge between the stable manifolds approach $(1,0)$, while trajectories crossing an unstable direction leave the neighbourhood.

Put $x=1+u$, $y=v$, and $\mu =-1+\nu$, adjoining $\dot{\nu }=0$. ThenThe zero eigenvalue has centre direction $v=-u$. Solving the center manifold invariance equation givesSubstitution yields the reduced equationIts equilibria are $u=0$ and $u=\pm \sqrt{\nu }+O(\nu )$ for $\nu >0$, exactly matching the three branches meeting at the pitchfork in part (b).

At a non-axis fixed point the determinant and trace areThese formulas determine the solid stable and dashed unstable branches in the bifurcation diagram. The pair $x_{\pm }$ is born in a saddle-node bifurcation at $\mu =-(\beta +1{)}^2/4$. For $\beta \ne 1$, it exchanges branches with $x=1$ at $\mu =-\beta$ and with $x=0$ at $\mu =0$, both transcritical bifurcations. If $0<\beta <1$, the saddle-node lies to the left of $x=1$; if $\beta >1$, it lies to the right. For $\beta =1$, the saddle-node and the collision at $x=1$ combine into a pitchfork bifurcation at $\mu =-1$.

The second equation gives $y=x(1-x)$. The first then says either $y=0$, giving the persistent equilibrium points $(0,0)$ and $(1,0)$, orgiving $x_{\pm }=((\beta +1)\pm \sqrt{(\beta +1{)}^2+4\mu })/2$ and $y_{\pm }=x_{\pm }(1-x_{\pm })$. These two points exist for $\mu \ge -(\beta +1{)}^2/4$.

At the origin the Jacobian matrix has trace $-1$ and determinant $-\mu$, so it is stable for $\mu <0$ and a saddle point for $\mu >0$. At $(1,0)$ the trace is $-1$ and determinant $\mu +\beta$, so it is stable for $\mu >-\beta$ and a saddle for $\mu <-\beta$. Stationary bifurcations occur atwith the first two coinciding when $\beta =1$.