15.
If |0 and |1 represent an orthonormal basis for the states of a single qubit, which of the
following represents an entangled state of two qubits A and B? (The subscript labels the
state of the respective subsystem A or B.)
a. |0 ! + |1 !
b. (|0 ! + |1 !)/ 2
c. |0 !⨂|1 !
d. (|0 !⨂|1 ! + |1 !⨂ 0 ! / 2
The correct option is (d), because
(|0⟩A ⊗ |1⟩B + |1⟩A ⊗ |0⟩B)/√2
is a non‑factorizable superposition of product states and therefore represents an
entangled state of two qubits A and B.
Introduction: Entangled State of Two Qubits
In quantum information, an entangled state of two qubits is a joint state that cannot be written as a simple product of a state of qubit A and a state of qubit B.
Such non‑separable states are central to quantum computing and quantum communication because they exhibit correlations that have no classical counterpart.
The question gives four expressions built from the single‑qubit basis |0⟩ and |1⟩ and asks which one is entangled for subsystems A and B.
Product vs Entangled States
For two subsystems A and B with Hilbert spaces HA and HB, any product (separable) state has the form
|ψ⟩AB = |α⟩A ⊗ |β⟩B, where each factor is a single‑qubit state.
A state is entangled if it cannot be written in this product form for any choice of |α⟩A and |β⟩B.
Option (a): |0⟩A + |1⟩B
Option (a) is written as |0⟩A + |1⟩B, which is not a well‑formed two‑qubit state in standard notation, because the two kets belong to different subsystems and are not combined via a tensor product.
A valid two‑qubit state must live in the tensor‑product space HA ⊗ HB, with basis states like |0⟩A|0⟩B, |0⟩A|1⟩B, |1⟩A|0⟩B, or |1⟩A|1⟩B, so option (a) is rejected.
Option (b): (|0⟩A + |1⟩B)/√2
Option (b) adds a normalization factor 1/√2 but keeps the same structurally incorrect sum |0⟩A + |1⟩B. [web:39]
Normalization cannot fix the fact that the terms correspond to different subsystems and are not tensor products, so this expression still does not represent a legitimate two‑qubit state and is not entangled.
Option (c): |0⟩A ⊗ |1⟩B
Option (c) is |0⟩A ⊗ |1⟩B, or |0⟩A|1⟩B, which has the exact product form |ψ⟩AB = |α⟩A ⊗ |β⟩B with |α⟩A = |0⟩A and |β⟩B = |1⟩B.
Because it is separable and contains no intrinsically quantum correlations beyond classical probabilities, option (c) is not an entangled state.
Option (d): (|0⟩A|1⟩B + |1⟩A|0⟩B)/√2
Option (d) is
|ψ⟩AB = (|0⟩A|1⟩B + |1⟩A|0⟩B)/√2,
a superposition of two different product states with equal amplitudes.
Assuming it could be factorized as (a|0⟩A + b|1⟩A) ⊗ (c|0⟩B + d|1⟩B) leads to contradictions when matching coefficients, because the |00⟩ and |11⟩ terms cannot be simultaneously forced to zero while keeping |01⟩ and |10⟩ nonzero.
Since no product decomposition exists, this state is entangled and is, in fact, one of the standard Bell states (|Ψ+⟩ up to notation), which are maximally entangled states of two qubits.
Hence, option (d) is the unique correct answer among the four choices.
Why This Bell-Type State Is Genuinely Entangled
Bell states such as |Ψ+⟩ = (|01⟩ + |10⟩)/√2 exhibit perfect correlations: if qubit A is measured as 0 in the computational basis, qubit B is guaranteed to be 1, and vice versa.
Yet each subsystem alone is described by a maximally mixed reduced density matrix, meaning maximal knowledge of the whole system does not determine the state of the parts, a hallmark of entanglement.
Final Answer
Among the given options, only option (d),
(|0⟩A ⊗ |1⟩B + |1⟩A ⊗ |0⟩B)/√2,
represents a valid entangled state of two qubits A and B; the others are either invalid two‑qubit expressions or simple product states.
