Codex @codex  0

Define a valuation $v$ on primitive propositions byThe assumptions say exactly that $S$ is a maximal consistent set in propositional logic. A structural induction on formulae provesFor example, consistency and the decision property give $\neg t\in S$ exactly when $t\notin S$, while deductive closure gives $s\wedge t\in S$ exactly when both $s,t\in S$; the other connectives follow similarly. Every member of $S$ is therefore true under $v$, so

The three controlled-NOT gates have directions $1\to 2$, $2\to 1$, and $1\to 2$. On computational basis bits they act asBy linearity, the whole quantum circuit is the SWAP gate. Thereforeso $|\alpha ⟩=|\varphi ⟩$ and $|\beta ⟩=|\psi ⟩$.

The Hadamard gate conjugates the Pauli X gate into the Pauli Z gate:When the control qubit is $|0⟩$, all three circuits act as the identity on the target. When it is $|1⟩$, the right-hand circuit acts on the target as $HXH=Z$. It therefore agrees with the Controlled-Z gate on every computational basis state, provingConjugating both sides by $I\otimes H$ and using $H^2=I$ immediately gives the reverse identity

Starting from the $n$-qubit computational basis state $|0{⟩}^{\otimes n}$, apply one Hadamard gate to every qubit. Since $H|0⟩=(|0⟩+|1⟩)/\sqrt{2}$,Thus parallel Hadamard gates prepare the uniform quantum superposition.

Put $L_z=L_3(1-{ϵ}^2/2)$ and seek $\theta =O(ϵ)$. Using $\cos \theta =1-{\theta }^2/2+O({\theta }^4)$ and ${\sin }^2\theta ={\theta }^2+O({\theta }^4)$, the $\theta$-dependent part of the effective potential isIt has a local minimum precisely whenwhich is the gyroscopic stabilization of an inverted symmetric top condition. Differentiating the displayed approximation givesand henceTaking $L_3>0$ without loss of orientation, the steady precession rate isso

Solving the two momentum equations givesSubstitution into the conserved energy reduces the heavy symmetric top to one-dimensional effective potential motion:whereThis is the heavy symmetric top reduction.

The coordinates $\varphi$ and $\psi$ are cyclic, so their canonical momenta are conserved:Because the Lagrangian has no explicit time dependence, the conserved energy isThese are the three integrals of motion requested.

In the Euler angles for a symmetric top convention, $\theta$ is the inclination of the body symmetry axis ${\mathbf{e}}_3$ from the fixed vertical, $\varphi$ is the azimuth of the line of nodes in the fixed horizontal plane, and $\psi$ is rotation about ${\mathbf{e}}_3$ measured from that line:

For $f(t)=\sin \omega t$,The Bromwich inversion formula givesWhen $t>x/c$, close the contour to the left. The residues at $p=\pm i\omega$ sum to $\sin (\omega (t-x/c))$. When $t<x/c$, close it to the right, where there are no poles, and the integral vanishes. HenceEquivalently, $y(x,t)=\Theta (t-x/c)\sin (\omega (t-x/c))$ in terms of the Heaviside step function. The boundary oscillation travels to the right at wave speed $c$ without reflection or dispersion; a point at $x$ begins moving after the propagation delay $x/c$.

The string is initially at rest, so $y(x,0)=y_t(x,0)=0$. Taking the Laplace transform in $t$ converts the wave equation intoFor $\mathrm{Re}p>0$, the solution that decays as $x\to \infty$ is $\hat{y}=A(p)e^{-px/c}$. The transformed boundary condition $\hat{y}(0,p)=\hat{f}(p)$ fixes $A=\hat{f}$, and therefore

Adding the all-one codeword is a bijection from $C$ to itself, and it replaces every bit by its complement. It therefore sends weight $j$ to weight $n-j$, provingThe code has one word of weight zero and, since its minimum Hamming distance is three, no words of weight one or two. Combining this with part (i) gives

Sum all nonzero vectors of ${\mathbb{F}}_2^d$. In each coordinate, exactly $2^{d-1}$ vectors have entry one, which is even because $d>1$. The sum is therefore zero, soThus the all-one word belongs to the Hamming code and has Hamming weight $n$.

The columns of the parity-check matrix of the binary Hamming code are precisely the $n=2^d-1$ distinct nonzero vectors of ${\mathbb{F}}_2^d$. If $h_1\ne h_2$, then $h_1+h_2$ is nonzero and differs from both summands, so it occurs as a unique third column $h_3$.

A word supported on three positions is a codeword exactly when the corresponding three columns sum to zero. Every unordered pair of columns determines one such triple, and each triple is counted once by each of its three pairs. Therefore

This is superdense coding. To send the two classical bits $(z,x)$, Alice applies $Z^z{X}^x$ to her half of the shared Bell state $|{\beta }_{00}⟩$, producing $|{\beta }_{zx}⟩$, and sends that one qubit to Bob. Bob applies $\mathrm{CX}$ followed by $H$ on the first qubit, as in part (i), and measures both qubits in the computational basis. The outcome is $zx$ with certainty, so one transmitted qubit communicates two classical bits using the previously shared quantum entanglement.

The required identity isthe first exponent in the scanned statement should be $z$. Indeed, $X^x$ changes the parity label from $0$ to $x$, while $Z^z$ multiplies the component whose first qubit is $1$ by $(-1{)}^z$, creating the phase label. Equivalently, the hint permits either one-qubit operator to be transferred to the second half of the Bell state by transposition.

The controlled-NOT gate sends $|0x⟩$ to $|0x⟩$ and $|1\bar{x}⟩$ to $|1x⟩$. HenceSince the Hadamard gate satisfies $H^2=I$,

Given any $z\in \mathbb{C}$ that is not a nonpositive integer, choose $N$ large enough that $\mathrm{Re}(z+N)>0$ and defineThe recurrence shows that definitions from different sufficiently large $N$ agree, so this gives the unique analytic continuation from the right half-plane. It is analytic except where a denominator factor vanishes, namelyThese points are simple poles. More precisely, at $z=-n$,because one may writeand use $I(1)=1$.