Solution
= Solution
An ideal <cross-Kerr interaction> between two single modes is $H_{\rm int}=\hbar\chi n_1n_2$, with $n_j=a_j^\dagger a_j$. The <number operators> commute with the interaction, so this coupling changes phases without transferring <photons> between the modes. To implement the specified positive conditional phase, use
$$
\boxed{H_{\rm int}=-\hbar g\,n_1n_{2V},\qquad U=e^{i\Theta n_1n_{2V}},\quad \Theta=gt,}
$$
where $n_{2V}=|V\rangle\langle V|$ within the one-photon polarization subspace. The horizontal branch has zero interacting occupation. Since $e^{i\Theta n}|\alpha\rangle=|e^{i\Theta}\alpha\rangle$, the specified conditional rotation follows. This is an ideal lossless single-mode model; an actual travelling-wave medium's bandwidth, loss and noise are not established by writing this <Hamiltonian>.
Use $X_1=(a+a^\dagger)/2$ and $X_2=(a-a^\dagger)/(2i)$. A <coherent state> has centre $(\operatorname{Re}\alpha,\operatorname{Im}\alpha)$ and <variance> $1/4$ in either <optical quadrature>. Multiplication of $\alpha$ by $e^{i\Theta}$ rotates that centre anticlockwise through $\Theta$, keeping the <coherent state> noise disk unchanged. For real positive $\alpha$ the two centres are $(\alpha,0)$ and $(\alpha\cos\Theta,\alpha\sin\Theta)$.
Normalize the initial <polarization of an electromagnetic wave> qubit. The output of the medium is
$$
\boxed{|\Psi\rangle=\frac1{\sqrt2}\left(|\alpha\rangle|H\rangle+e^{i\phi}|\beta\rangle|V\rangle\right),\qquad
\beta=e^{i\Theta}\alpha.}
$$
The joint state is generally an <entangled state>. The overlap magnitude is $|\langle\alpha|\beta\rangle|=e^{-2|\alpha|^2\sin^2(\Theta/2)}$. The state factorizes when the amplitudes coincide, while nearly distinguishable amplitudes make the polarization nearly maximally entangled with the optical mode. Before any polarization conditioning, tracing out mode 2 gives the mixture $(|\alpha\rangle\langle\alpha|+|\beta\rangle\langle\beta|)/2$, with no interference between branches.
Include the normalization factors in the <half-wave plate> map: $|H\rangle\mapsto(|H\rangle+|V\rangle)/\sqrt2$ and $|V\rangle\mapsto(-|H\rangle+|V\rangle)/\sqrt2$. The transformed joint state is
$$
\frac12\left[(|\alpha\rangle-e^{i\phi}|\beta\rangle)|H\rangle+
(|\alpha\rangle+e^{i\phi}|\beta\rangle)|V\rangle\right].
$$
Conditioning on a vertical result and setting $\Theta=\pi$ therefore prepares
$$
\boxed{|\psi\rangle=N\left(|\alpha\rangle+e^{i\phi}|-\alpha\rangle\right).}
$$
This pure <optical cat state> is the conditional output; it is not the unconditioned reduced state immediately after the Kerr medium. The measurement erases the polarization label, allowing the two amplitudes to interfere. The calculated <optical Wigner functions> below display the resulting fringes; their negative regions are quasiprobability, not negative detection probabilities. These illustrative plots choose $α=1.15$: the rotation panel uses $Θ=2π/3$, while the cat panels use the conditional Kerr rotation $Θ=π$ and three different relative phases $φ$.
The <coherent-state overlap> follows by expanding in the <Fock basis>:
$$
\langle\beta|\alpha\rangle=e^{-(|\alpha|^2+|\beta|^2)/2}\sum_{n=0}^\infty\frac{(\beta^*\alpha)^n}{n!}
=e^{-(|\alpha|^2+|\beta|^2)/2+\beta^*\alpha}.
$$
Thus $s=\langle\alpha|-\alpha\rangle=e^{-2|\alpha|^2}$ and
$$
\boxed{N=[2(1+s\cos\phi)]^{-1/2},\qquad p_V=\frac{1+s\cos\phi}{2}.}
$$
The <postselection probability for Kerr-cat preparation> includes both the normalized input qubit and the <wave plate>. At $\alpha=0,\phi=\pi$ the vertical branch has zero probability and no normalized conditional state. For the remaining calculation take $\alpha$ real, as required.
\Image[/past-exam-of-the-nonlinear-optics-and-quantum-states-of-light-course-of-the-university-of-cambridge/2015/paper-1-cat-states.png]
{title=Original coherent-state rotation and calculated phase-space interference of three conditional cat states}
Using $a|\pm\alpha\rangle=\pm\alpha|\pm\alpha\rangle$, put $D=1+s\cos\phi$. Directly evaluating diagonal and cross terms gives
$$
\langle a\rangle=-\frac{i\alpha s\sin\phi}{D},\qquad
\langle a^2\rangle=\alpha^2,\qquad
\langle n\rangle=\alpha^2\frac{1-s\cos\phi}{D}.
$$
The last quantity is the mean <photon number> in the conditional cat, which differs from the input <coherent state> mean $\alpha^2$. Since $aa^\dagger=n+1$,
$$
\langle X_2^2\rangle=\frac14\langle2n+1-a^2-a^{\dagger2}\rangle
=\frac14-\frac{\alpha^2s\cos\phi}{D}.
$$
However, its first moment is
$$
\boxed{\langle X_2\rangle=-\frac{\alpha s\sin\phi}{D},}
$$
which is \b[generally nonzero]. The original PDF's assertion of a zero mean for arbitrary $\phi$ is inconsistent with its stated pure state. Consequently its displayed expression is the second moment, not the general <variance>. The <phase-dependent cat-state quadrature variance> is
$$
\boxed{\operatorname{Var}X_2=\langle X_2^2\rangle-\langle X_2\rangle^2
=\frac14-\frac{\alpha^2s(\cos\phi+s)}{(1+s\cos\phi)^2}.}
$$
The printed result does become the <variance> for $\sin\phi=0$, notably the even and <odd cat states>. It could also describe an ensemble mixing equal weights of $+\phi$ and $-\phi$: that different state's mean vanishes while its second moment is unchanged. Such phase averaging is not part of the specified pure-state preparation.
For nonzero finite $\alpha$, $0<s<1$ and $D>0$. The correct <quadrature squeezing> condition in this quadrature is therefore
$$
\boxed{\cos\phi>-e^{-2\alpha^2}.}
$$
For a representative phase in $-\pi<\phi\leq\pi$, this is $|\phi|<\arccos(-e^{-2\alpha^2})$. If instead one applies the printed second moment as a <variance>, the resulting condition is $\cos\phi>0$, or $|\phi|<\pi/2$ modulo $2\pi$. That is sufficient, but not necessary, for the actual pure state's squeezing. For example, at $\phi=\pi/2$, $\langle X_2^2\rangle=1/4$ while \b[$\operatorname{Var}X_2=1/4-\alpha^2e^{-4\alpha^2}<1/4$]. This explicitly tests the difference.
For a fixed phase other than $\pi$ modulo $2\pi$, expand
$$
|\alpha\rangle+e^{i\phi}|-\alpha\rangle=e^{-\alpha^2/2}\left[(1+e^{i\phi})|0\rangle+\alpha(1-e^{i\phi})|1\rangle+O(\alpha^2)\right].
$$
As the input mean number $\alpha^2\to0$, the normalized state tends to vacuum, $\langle X_2\rangle\sim-\alpha\tan(\phi/2)$ and
$$
\operatorname{Var}X_2=\frac14-\frac{\alpha^2}{1+\cos\phi}+O(\alpha^4).
$$
This expansion is nonuniform near the odd-cat phase. At exactly $\phi=\pi$ the vacuum terms cancel and the normalized <odd coherent-state superposition> tends to $|1\rangle$, giving \b[$\operatorname{Var}X_2\to3/4$], while its postselection probability tends to zero. The <even coherent-state superposition> at $\phi=0$ instead contains only even <photon numbers> and tends smoothly to vacuum.
At large input mean number, $s\to0$, the two <coherent state> branches become nearly orthogonal, and \b[$\operatorname{Var}X_2\to1/4$] for every fixed phase. The mean tends to zero as well. This does not make the state an incoherent mixture: phase-space interference still distinguishes the ideal <quantum superposition>, although its contribution to this low-order moment is exponentially small. The <loss of coherence in an optical cat state> arises when ordinary loss reveals the branch to the environment and suppresses that interference, with factor $e^{-2(1-\eta)\alpha^2}$ for transmissivity $\eta$.
For the especially simple even cat,
$$
\operatorname{Var}X_2=\frac14-\frac{\alpha^2}{e^{2\alpha^2}+1}.
$$
It starts at $1/4$, falls to about \b[$0.1108$ at $|\alpha|\simeq0.7995$], and returns towards $1/4$ from below. The odd-cat curve decreases from $3/4$ towards $1/4$ from above. Other fixed-phase curves have the small- and large-amplitude limits derived above; for negative $\cos\phi$, squeezing ends when $e^{-2\alpha^2}=-\cos\phi$. In <balanced homodyne detection>, centre each quadrature distribution on its measured mean before inferring its <variance>.
\Image[/past-exam-of-the-nonlinear-optics-and-quantum-states-of-light-course-of-the-university-of-cambridge/2015/paper-1-cat-variance.png]
{title=Calculated cat-state quadrature means, true variances, and the amplitude-dependent squeezing boundary}