Solution

Let Alice's two bits be

(x, z)

and Bob's bit be

b

. Alice applies

X_{A}^{x} Z_{A}^{z},

while Bob applies

X_{B}^{b} .

They then send their qubits to Charlie. Acting on

∣ χ_{1}^{+} ⟩ = \frac{∣000 ⟩ + ∣111 ⟩}{2},

these operations produce, up to an irrelevant global phase,

\frac{∣ x , b , 0 ⟩ + ( - 1 ) ^{z} ∣1 \oplus x , 1 \oplus b , 1 ⟩}{2} .

The complete encoding table is

(x, z) (0, 0) (0, 1) (0, 0) (0, 1) (1, 0) (1, 1) (1, 0) (1, 1) b 00110011 state measured by Charlie ∣ χ_{1}^{+} ⟩ ∣ χ_{1}^{-} ⟩ ∣ χ_{2}^{+} ⟩ ∣ χ_{2}^{-} ⟩ ∣ χ_{4}^{+} ⟩ ∣ χ_{4}^{-} ⟩ ∣ χ_{3}^{+} ⟩ ∣ χ_{3}^{-} ⟩

where a possible overall minus sign in the two

x = z = 1

cases has no observable effect.

Charlie performs a projective measurement in the eight-state GHZ basis. The sign

+

-

reveals

z

. The basis index reveals

(x, b)

according to

1 \leftrightarrow (0, 0), 2 \leftrightarrow (0, 1), 3 \leftrightarrow (1, 1), 4 \leftrightarrow (1, 0) .

He therefore reconstructs Alice's two bits and Bob's one bit with certainty, which is the three-party dense coding with a GHZ state protocol.

Solved by gpt-5.6-sol high.

 Ancestors (11)