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Use deterministic two-qubit entanglement dilution, retaining the specified complex probability amplitudes. Alice implements a quantum instrument on $A$ with Kraus operatorsThe Kraus representation completeness condition holds even for complex $\alpha ,\beta$:To realize this using the permitted quantum gates and a basic measurement, Alice initializes an ancilla qubit $C$ to $|0⟩$. Conditional on her computational-basis state being $j$, she applies to $C$ the unitary matrixEach has orthonormal columns. Therefore $W=|0⟩⟨0{|}_A\otimes V_0+|1⟩⟨1{|}_A\otimes V_1$ is a unitary gate, withA computational-basis measurement of $C$ implements the desired quantum instrument. Its two unnormalized branches on $AB$ areEach has probability $1/2$. Alice sends her outcome $m$ as one classical bit. For $m=0$, both parties keep their qubits unchanged. For $m=1$, both apply a Pauli X gate, givingThus both outcomes produce the exact target using one classical bit and local operations. The construction includes $\alpha =0$ or $\beta =0$ and arbitrary relative quantum phases; it requires no quantum channel between the parties.

Yes, the conversion can be deterministic using local operations and classical communication. Put $s=\sin \theta$ and $c=\cos \theta$. The initial state's squared Schmidt coefficients are $(1/2,1/2)$, while the target's decreasingly ordered squared Schmidt coefficients are $(c^2,s^2)$. Since $0<\theta <\pi /4$, one has $1/2<c^2$ and both pairs sum to one. ThusThe direction of this majorization is the allowed one in Nielsen's pure-state conversion theorem: maximal entanglement can be reduced by deterministic LOCC.

Here is an explicit protocol, so no conversion theorem need be assumed. The sender first applies a Pauli Z gate to remove the initial Bell-state minus sign. Starting from $|{\Phi }^{+}⟩$, she performs a two-outcome local measurement with Kraus operatorsThey define a valid measurement because $M_0^{†}{M}_0+M_1^{†}{M}_1=I$. The two unnormalized states after measurement areEach outcome has probability $1/2$. For outcome zero, the normalized state is already $s|00⟩+c|11⟩$. For outcome one, both parties apply Pauli X gates, obtaining the same state. The sender communicates only which outcome occurred. Finally the receiver applies a Pauli X gate, giving $s|01⟩+c|10⟩$ in either branch. This is deterministic two-qubit entanglement dilution, with total success probability one; it does not rely on postselection.