= Solution
{c}
At $N=M$, the two <spin> occupations per site are $x$ and $1-x$. With $\tau=|t|>0$, the trial <energy> per site is
$$
\boxed{e(x)=\frac{\langle H\rangle}{M}=-\frac{4\tau}{\pi}\sin(\pi x)+V\left[x^2+(1-x)^2-\frac{2\sin^2(\pi x)}{\pi^2}\right],\quad 0\leq x\leq1.}
$$
Its <derivative> is $e'(x)=-4\tau\cos(\pi x)+V[4x-2-(2/\pi)\sin(2\pi x)]$. Since $x=0$ is a boundary point, its <local minimum> test is one-sided: $e'(0)=-4\tau-2V$. At $V=-2\tau$, where this <derivative> vanishes, its <Taylor series> begins $e(x)-e(0)=(2\pi^2\tau/3)x^3+O(x^4)>0$ for small positive $x$. Hence
$$
\boxed{x=0\text{ is a strict one-sided local minimum exactly when }V/\tau\leq-2.}
$$
The same applies to $x=1$ by <spin> interchange.
The <symmetry> $e(x)=e(1-x)$ makes $x=1/2$ a stationary point for every $V$. Here $e''(1/2)=4\pi\tau+8V$, so the <second-derivative test> settles all cases except equality. Writing $y=x-1/2$, at $V=-\pi\tau/2$ the <Taylor series> is $e(1/2+y)-e(1/2)=(\pi^3\tau/6)y^4+O(y^6)$. Thus
$$
\boxed{x=\tfrac12\text{ is a strict local minimum exactly when }V/\tau\geq-\pi/2;\quad V/\tau<-\pi/2\text{ makes it a local maximum}.}
$$
\b[The minimizing branches interpolate continuously.] To find all of them, put $u=\pi(1/2-x)$ on $0<x<1/2$. An interior stationary point obeys
$$
\frac V\tau=-\frac{\pi\sin u}{u+\sin u\cos u},\qquad 0<u<\frac\pi2.
$$
The right side decreases strictly from $-\pi/2$ to $-2$, since
$$
\frac{d}{du}\frac{\sin u}{u+\sin u\cos u}
=\frac{\cos u\,(u-\sin u\cos u)}{(u+\sin u\cos u)^2}>0.
$$
The positivity follows from $u-\sin u\cos u=\int_0^u2\sin^2v\,dv$. There is therefore exactly one such stationary point on each side of $1/2$ when $-2<V/\tau<-\pi/2$. The signs of $e'$ show that both are <global minima>, whereas $1/2$ is then a <local maximum>. The complete answer is
$$
\boxed{x_{\min}=\begin{cases}
0\text{ or }1,&V/\tau\leq-2,\\
\tfrac12\mp u/\pi,&-2<V/\tau<-\pi/2,\quad V/\tau=-\pi\sin u/(u+\sin u\cos u),\\
\tfrac12,&V/\tau\geq-\pi/2.
\end{cases}}
$$
The pair merges through a <supercritical pitchfork bifurcation> at $V/\tau=-\pi/2$. Comparing only $e(0)$ and $e(1/2)$ would miss the intervening <ground states> with partial <spin polarization>.
\Image[/past-exam-of-the-quantum-matter-course-of-the-university-of-oxford/original2025-polarization.png]
{title=Minimizing spin fractions in the restricted Fermi-sea approximation}
{description=Both symmetry-related global minima are plotted. The intermediate branches join the fully polarized endpoints continuously at $V/|t|=-2$ and merge at $V/|t|=-\pi/2$.}
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