Since ${\Pi }_1=I-{\Pi }_0$, the success probability is

The operator $D$ vanishes on the orthogonal complement of $S=\mathrm{span}\{|{\alpha }_0⟩,|{\alpha }_1⟩\}$, so it has rank at most two. The states are distinct rays, hence they are linearly independent and $S$ has dimension two. In an orthonormal basis of $S$, its determinant isThus its two eigenvalues on $S$ are nonzero and have opposite signs. All remaining eigenvalues are zero.

Under the gauge transformation,because the curl of a gradient vanishes, andThe transformed Lagrangian isA total time derivative changes the action only by endpoint terms, whose variation vanishes for fixed endpoints, so the Euler-Lagrange equations are unchanged.

The new momentum isUse the type-two generating function for a canonical transformationIt gives $\mathbf{p}=\partial F_2/\partial \mathbf{r}=\tilde{\mathbf{p}}-q\nabla f$, $\tilde{\mathbf{r}}=\partial F_2/\partial \tilde{\mathbf{p}}=\mathbf{r}$, andThis equals $|\tilde{\mathbf{p}}-q\tilde{\mathbf{A}}{|}^2/(2m)+q\tilde{\varphi }$, so the gauge change is a time-dependent canonical transformation.

The canonical momentum is $\mathbf{p}=m\dot{\mathbf{r}}+q\mathbf{A}$, and the Legendre transform givesIts Hamilton's equations areand, componentwise,Combining the second equation with $\mathbf{p}=m\dot{\mathbf{r}}+q\mathbf{A}$ recovers the Lorentz force law.

The canonical momentum isIn components, the Euler-Lagrange equation isThusThe last term is the $i$th component of $\mathbf{v}\times (\nabla \times \mathbf{A})$, sothe Lorentz force law.

Apply the Hankel representation with $z$ replaced by $1-z$:Enlarge $H$ to a modified Hankel contour that encloses the simple poles $t=2\pi in$, $n\in \mathbb{Z}\setminus \{0\}$. In a strip where the deformation and residue sum converge, the outer pieces vanish and the orientation gives the Hankel integral as minus $2\pi i$ times the sum of the enclosed residues. Part (b) givesSubstitution therefore yieldsBoth sides are meromorphic functions of $z$, so the identity theorem extends the equality from the initial strip to every point where either side is defined.

For $n\ne 0$, the denominator has derivativeHence the simple-pole residue is

Take the Hankel contour about the negative real axis: it starts at $-\infty$ just below the cut, circles the origin counterclockwise, and returns to $-\infty$ just above the cut. The notation $(0+)$ records this return side of the branch point. Choose the branch $-\pi <\arg t<\pi$ for $t^{z-1}$.

For $\mathrm{Re}z>1$, collapse the contour onto the two banks of the cut. Their phase difference is determined byand the resulting sine factor, together with the Gamma reflection formula $\Gamma (z)\Gamma (1-z)=\pi /\sin (\pi z)$, recovers the integral representation (‡). The Hankel integral converges exponentially at its unbounded end and, after the standard local subtraction of the singular part at zero, depends holomorphically on $z$. The apparent poles of $\Gamma (1-z)$ are cancelled by zeros of the contour integral except at $z=1$. It therefore gives the analytic continuation of the Riemann zeta function to $\mathbb{C}\setminus \{1\}$.

True. The Huffman tree is a full binary tree, so its leaves form a complete prefix code. The Kraft inequality is therefore an equality:

False. For probabilities $(0.30,0.25,0.25,0.20)$, every length-two code is optimal. A code in which the $0.30$ and $0.20$ symbols are sibling leaves is optimal, but no Huffman construction produces that labelled tree, because its first merge must pair $0.20$ with a $0.25$ symbol.

False. For probabilities $(0.4,0.2,0.2,0.1,0.1)$, first merge the two $0.1$ symbols. Subsequent ties allow Huffman trees with length multisets $(2,2,2,3,3)$ and $(1,3,3,3,3)$, both of expected length $2.2$.

The conclusions of (b) and (c) remain true. Indeed, the unit circle is a path-connected space, and the upper semicircle supplies a path from $-1$ to $1$; restricting the proposed closed cover or two-valued continuous function to that path reduces either claim to the corresponding result on $I$.

The fixed-point conclusion (d) fails. For example, any nontrivial spatial rotation $f(z)=e^{i\theta }z$ of the unit circle has no fixed point when $e^{i\theta }\ne 1$; the antipodal map $f(z)=-z$ is the simplest example.

Suppose $I=A\cup B$, where $A$ and $B$ are disjoint closed sets with $-1\in A$ and $1\in B$. Since each is the complement of the other, both are also open in $I$. Define $f=-1$ on $A$ and $f=1$ on $B$. This locally constant function is continuous and contradicts part (c). Therefore $A\cap B\ne \varnothing$.

Conversely, suppose a continuous $f:I\to \{-1,1\}$ satisfies $f(-1)=-1$ and $f(1)=1$. The function $g=-f:I\to I$ is continuous, so part (d) gives $x=g(x)=-f(x)$. If $f(x)=1$, this forces $x=-1$, contradicting $f(-1)=-1$; if $f(x)=-1$, it forces $x=1$, contradicting $f(1)=1$. Hence no such $f$ exists.

Suppose that the continuous function $f:I\to I$ has no fixed point. Then $g(x)=f(x)-x$ is continuous and never zero, sois a continuous function from $I$ to $\{-1,1\}$. Since $f(-1)\ge -1$ and equality would be a fixed point, $h(-1)=1$. Similarly $h(1)=-1$. Multiplying $h$ by $-1$ contradicts part (c). Thus $f$ has a fixed point.

If such a continuous function existed, the inverse imageswould be closed, would cover $I$, and would contain $-1$ and $1$ respectively. Part (b) gives a point in $A\cap B$, which is impossible.

Suppose for a contradiction that $A\cap B=\varnothing$. Each grid point then belongs to exactly one of $A$ and $B$; colour it red in $A$ and green in $B$. By part (a), there are adjacent points $a_n\in A$ and $b_n\in B$, in either order, withThe interval $I$ is a compact set, so a subsequence of $(a_n)$ converges to some $x\in I$. Along the same subsequence $b_n\to x$. Since $A$ and $B$ are closed sets, $x\in A\cap B$, a contradiction. Hence $A\cap B\ne \varnothing$.

Read the $2n+1$ colours from left to right. If every neighbouring pair had the same colour, mathematical induction would make every point the same colour as $-1$, contradicting the different colour of $1$. Therefore at least one adjacent pair has different colours.

Condition on a position retained after basis reconciliation, so Bob used Alice's basis. If Eve does not intercept, no error occurs. If she intercepts, her uniformly chosen basis agrees with Alice's with probability $1/2$, again causing no error. With probability $1/2$ she uses the conjugate basis; her measurement and resent state then make Bob's bit uniformly random, giving an error with probability $1/2$.

Therefore the bit error rate in the retained key is

Alice sends the $m$ conjugate-coded qubits through the noiseless quantum channel. Bob independently chooses a uniformly random basis bit $y_i^{\prime }$ for each qubit and performs a projective measurement in the computational basis when $y_i^{\prime }=0$ or the Hadamard basis when $y_i^{\prime }=1$.

After all measurements, Alice and Bob publicly reveal $Y$ and $Y^{\prime }$ but keep the data bits secret. They discard every position with $y_i\ne y_i^{\prime }$ and retain the measured bits at positions where the bases agree. In the ideal channel those retained bits equal Alice's $x_i$ with certainty. Each pair of bases agrees with probability $1/2$, so the shared secret key has expected length $m/2$.