Consider first the two-qubit unitary
$$
U = \sqrt{\text{SWAP}}
$$
acting on $\mathbb{C}^2 \otimes \mathbb{C}^2$. One way to write this is
$$
U = \exp\Bigl( i \, \tfrac{\pi}{4} \, \text{SWAP} \Bigr),
$$
where $\text{SWAP}$ is the swap operator on two qubits. Now let us add an ancillary qubit $|0\rangle_A$. For the state $\rho$ received from Alice (which is $\rho = | \psi \rangle \langle \psi |$, where $|\psi\rangle = H^\theta|x\rangle$), define an isometry
$$
V: \mathbb{C}^2 \rightarrow \mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2
$$
by
$$
V (|\psi\rangle) = U_{(1,2)} \bigl(|\psi\rangle_1 \otimes |0\rangle_2 \otimes |0\rangle_A \bigr).
$$
In more detail, let $\{ |0\rangle, |1\rangle \}$ be the standard basis for each qubit, and let $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$ label two-qubit basis states. The swap operator is defined as
$$
\text{SWAP} |ab\rangle = |ba\rangle,
$$
so the partial operator $\sqrt{\text{SWAP}}$ acts as
$$
U |ab\rangle = \cos\left(\tfrac{\pi}{4}\right) |ab\rangle + i \sin\left(\tfrac{\pi}{4}\right) |ba\rangle.
$$
Writing out matrix elements, one sees that upon tensoring with the ancillary qubit $|0\rangle_A$ and tracing out the second subsystem, this isometry creates a correlated state between three qubits.
Define the following Kraus operators for the channel $\mathcal{E}$ that prepares the bipartite state for Bob and Charlie: pick an orthonormal basis $\{|k\rangle_A\}$ in the ancillary space and set
$$
K_k = \langle k |_A U_{(1,2, A)} \bigl(I_1 \otimes |0\rangle_2 \langle 0|\bigr).
$$
Here the bracket $\langle k |_A$ denotes projecting the ancillary qubit onto $|k\rangle$. If we want a single Kraus operator representing a pure isometric extension, we can keep the ancilla unmeasured; but for a physical channel to Bob and Charlie, the ancilla can be measured or discarded. The final bipartite state is found by tracing out the ancillary qubit:
$$
\rho_{BC} = \mathrm{Tr}_A \bigl[ V (|\psi\rangle) \langle \psi| V^\dagger \bigr].
$$
Next, let us label Bob’s register as B and Charlie’s register as C. Concretely, if one wants a “cloning-like” effect, one can assign the output qubit 1 to Bob and qubit 2 to Charlie, or vice versa, depending on how we label the registers. A direct calculation shows that $\rho_{BC}$ will be a mixed state whose fidelity with $|\psi\rangle \langle \psi|\otimes|\psi\rangle \langle \psi|$ is limited by the no-cloning bound. In particular, since the channel cannot produce perfect clones, one finds from the universal cloning fidelity formula that Bob’s and Charlie’s reduced states each have fidelity
$$
F = \frac{1}{2} + \frac{1}{\sqrt{2}} \frac{1}{2},
$$
leading to the well-known $\tfrac{1}{2} + \tfrac{1}{2\sqrt{2}}$ limit on each party’s success probability for guessing the original bit x when $\theta$ is revealed.
To see why this limits the winning probability, note that after Alice reveals $\theta$, both Bob and Charlie measure their qubits in the corresponding basis $\{|0\rangle, |1\rangle\}$ or $\{|+\rangle, |-\rangle\}$, where $|\pm\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle \pm |1\rangle)$. The probability that Bob and Charlie simultaneously guess x is
$$
P(\text{win}) = \mathrm{Tr}\bigl[\bigl(|x\rangle\langle x|\bigr)_B \otimes \bigl(|x\rangle\langle x|\bigr)_C \,\rho_{BC}\bigr],
$$
and one can show it is capped by
$$
P(\text{win}) \le \Bigl(\tfrac{1}{2} + \tfrac{1}{2\sqrt{2}}\Bigr)^2.
$$
This bound follows from the optimal universal cloning limit, together with the fact that the channel $\mathcal{E}$ must be completely positive and trace preserving. Thus, while the isometry based on $\sqrt{\text{SWAP}}$ creates a state shared between Bob and Charlie that is correlated with the original $|\psi\rangle$, both parties cannot exceed the known cloning fidelity bound, confirming the practical impossibility of perfectly splitting the BB84 state.
