Quantum Computing 3.2 -- Measurement of a Bell pair

Computer Science Level 4

A qubit pair is initially in the Bell state ∣ ψ ⟩ = 1 2 ( ∣ 00 ⟩ + ∣ 11 ⟩ ) |\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) . Now the Hadamard gate H H is applied to the first qubit and then it is measured. In our case, the result of this measurement is ∣ 1 ⟩ |1\rangle .

In which state is the second qubit ∣ ψ 2 ⟩ |\psi_2\rangle after the measurement?

Details: The Hadamard gate is given in the basis { ∣ 0 ⟩ , ∣ 1 ⟩ } \{|0\rangle,|1\rangle\} by the matrix H = 1 2 ( 1 1 1 − 1 ) H = \frac{1}{\sqrt{2}} \left( \begin{array}{cc} 1 & 1 \\ 1 & - 1 \end{array} \right)

∣ 1 ⟩ \mid 1\rangle ∣ 0 ⟩ \mid 0\rangle ∣ ⁣ + ⟩ = ∣ 0 ⟩ + ∣ 1 ⟩ 2 \mid\!+\rangle = \frac{\mid 0\rangle+\mid 1\rangle}{\sqrt 2} ∣ ⁣ − ⟩ = ∣ 0 ⟩ − ∣ 1 ⟩ 2 \mid\!-\rangle = \frac{\mid 0\rangle-\mid 1\rangle}{\sqrt 2}

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

Markus Michelmann
Jul 11, 2018

The Hadamard operation acts only on the first qubit: H 1 ∣ ψ ⟩ = ( H ⊗ I ) 1 2 ( ∣ 00 ⟩ + ∣ 11 ⟩ ) = 1 2 ( H ∣ 0 ⟩ ⊗ I ∣ 0 ⟩ + H ∣ 1 ⟩ ⊗ I ∣ 1 ⟩ ) = 1 2 ( 1 2 ( ∣ 0 ⟩ + ∣ 1 ⟩ ) ⊗ ∣ 0 ⟩ + 1 2 ( ∣ 0 ⟩ − ∣ 1 ⟩ ) ⊗ ∣ 1 ⟩ ) = 1 2 ( ∣ 00 ⟩ + ∣ 01 ⟩ + ∣ 10 ⟩ − ∣ 11 ⟩ ) = 1 2 ( ∣ 0 ⟩ ⊗ ( ∣ 0 ⟩ + ∣ 1 ⟩ ) + ∣ 1 ⟩ ⊗ ( ∣ 0 ⟩ − ∣ 1 ⟩ ) ) = 1 2 ( ∣ 0 ⟩ ⊗ ∣ + ⟩ + ∣ 1 ⟩ ⊗ ∣ − ⟩ ) \begin{aligned} H_1|\psi\rangle &= (H \otimes I) \frac{1}{\sqrt{2}} \left( |00\rangle + |11\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( H |0\rangle \otimes I |0\rangle + H |1\rangle \otimes I |1\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left(\frac{1}{\sqrt{2}} \left(|0\rangle + |1\rangle \right) \otimes |0\rangle + \frac{1}{\sqrt{2}} \left(|0\rangle - |1\rangle \right) \otimes |1\rangle \right) \\ &= \frac{1}{2} \left( |00\rangle + |01\rangle + |10\rangle - |11\rangle \right) \\ &= \frac{1}{2} \left( |0\rangle \otimes (|0\rangle + |1\rangle) + |1\rangle \otimes (|0\rangle - |1\rangle) \right) \\ &= \frac{1}{\sqrt{2}} \left( |0\rangle \otimes |+\rangle + |1\rangle \otimes |-\rangle \right) \end{aligned} Note that this is still an entangled state. Only when measuring this entanglement is destroyed and we get of the measurement result ∣ 1 ⟩ | 1 \rangle the overall state ∣ ψ ′ ⟩ = ∣ 1 ⟩ ⊗ ∣ − ⟩ |\psi'\rangle = |1\rangle \otimes |-\rangle Therefore, the second qubit is in the state ∣ ψ 2 ⟩ = ∣ − ⟩ |\psi_2 \rangle= |-\rangle after the measurement.

Alternative solution: Alternatively, one can also represent the Bell state in the { ∣ + ⟩ , ∣ − ⟩ } \{|+\rangle, | -\rangle\} basis. A Hadamard operation corresponds to a basis transformation between { ∣ 0 ⟩ , ∣ 1 ⟩ } \{|0\rangle, |1\rangle\} and { ∣ + ⟩ , ∣ − ⟩ } \{|+\rangle, |-\rangle\} . Thus, in our experiment, the first qubit is actually measured in the base { ∣ + ⟩ , ∣ − ⟩ } \{|+\rangle, |-\rangle\} . The result ∣ 1 ⟩ |1\rangle therefore corresponds to the state H ∣ 1 ⟩ = ∣ − ⟩ H|1\rangle = |-\rangle . The measurement therefore corresponds to the application of a projection operator P = ∣ − ⟩ ⟨ − ∣ ⊗ I P = |-\rangle \langle -| \otimes I P ∣ ψ ⟩ = ( ∣ − ⟩ ⟨ − ∣ ⊗ I ) 1 2 ( ∣ 00 ⟩ + ∣ 11 ⟩ ) = 1 2 ( ⟨ − ∣ 0 ⟩ ∣ − ⟩ ⊗ ∣ 0 ⟩ + ⟨ − ∣ 1 ⟩ ∣ − ⟩ ⊗ ∣ 1 ⟩ ) = 1 2 ( 1 2 ( ⟨ 0 ∣ − ⟨ 1 ∣ ) ∣ 0 ⟩ ∣ − ⟩ ⊗ ∣ 0 ⟩ + 1 2 ( ⟨ 0 ∣ − ⟨ 1 ∣ ) ∣ 1 ⟩ ∣ − ⟩ ⊗ ∣ 1 ⟩ ) = 1 2 ⋅ 1 2 ( ∣ − ⟩ ⊗ ∣ 0 ⟩ − ∣ − ⟩ ⊗ ∣ 1 ⟩ ) = 1 2 ∣ − ⟩ ⊗ 1 2 ( ∣ 0 ⟩ − ∣ 1 ⟩ ) = 1 2 ∣ − ⟩ ⊗ ∣ − ⟩ \begin{aligned} P |\psi\rangle &= (|-\rangle \langle -| \otimes I) \frac{1}{\sqrt 2} (|00\rangle + |11\rangle) \\ &= \frac{1}{\sqrt 2} (\langle - |0\rangle |-\rangle \otimes |0\rangle + \langle - |1\rangle |-\rangle \otimes |1\rangle) \\ &= \frac{1}{\sqrt 2} \left( \frac{1}{\sqrt 2} (\langle 0 | - \langle 1|) |0\rangle |-\rangle \otimes |0\rangle + \frac{1}{\sqrt 2} (\langle 0 | - \langle 1|) |1\rangle |-\rangle \otimes |1\rangle\right) \\ &= \frac{1}{\sqrt 2} \cdot \frac{1}{\sqrt 2} (|-\rangle \otimes |0\rangle - |-\rangle \otimes |1\rangle) \\ &= \frac{1}{\sqrt 2} |-\rangle \otimes \frac{1}{\sqrt 2} \left(|0\rangle - |1\rangle\right) \\ &= \frac{1}{\sqrt 2} |-\rangle \otimes |-\rangle \end{aligned} Again, the second qubit is in the state ∣ − ⟩ |-\rangle after the measurement.

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