- #1

neu

- 227

- 3

## Homework Statement

Have two Greenberger-Horne-Zeilinger (GHZ) states of qubits A, B, C and

D, E, F as follows:

[tex] \mid GHZ \rangle_{ABC} = \frac{1}{\sqrt{2}} \left( \mid 0 \rangle_{A}\mid 0 \rangle_{B}\mid 0 \rangle_{C} + \mid 1 \rangle_{A}\mid 1 \rangle_{B}\mid 1 \rangle_{C}\right)[/tex]

and

[tex] \mid GHZ \rangle_{DEF} = \frac{1}{\sqrt{2}} \left( \mid 0 \rangle_{D}\mid 0 \rangle_{E}\mid 0 \rangle_{F} + \mid 1 \rangle_{D}\mid 1 \rangle_{E}\mid 1 \rangle_{F}\right)[/tex]

If you perform a measurement in the Bell basis on the qubits A and D, and obtain

the outcome: [tex] \mid \Psi^{+} \rangle_{AD} = \frac{1}{\sqrt{2}} \left(\mid 0 \rangle_{A}\mid 1 \rangle_{D} + \mid 1 \rangle_{A}\mid 0 \rangle_{D}\right) [/tex]

Write down the state to which qubits B, C, D and F are projected?

**Attempted Solution**

Total state is : [tex] \mid GHZ \rangle_{ABCDEF}= \mid GHZ \rangle_{ABC} \mid GHZ \rangle_{DEF} [/tex]

Projector operator for measurement of A and D is [tex] \mid Bell \rangle \langle Bell \mid_{AD} [/tex]

so [tex] \mid Bell \rangle \langle Bell \mid_{AD} \mid GHZ \rangle_{ABCDEF} = \mid Bell \rangle_{AD} \mid \Psi^{+} \rangle_{AD} \mid \Psi^{+} \rangle_{BCEF} [/tex]; is this right?

Here I get confused. I think the method is simply:

[tex] \langle \Psi^{+} \mid_{AD}\mid GHZ \rangle_{ABCDEF} = \mid \Psi^{+} \rangle_{BCEF} [/tex]

But if so, why?

Last edited: