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I was reading the book ''Introduction to Optical Quantum Information Processing" by Pieter Kok and Brendon Lovett and on page 199 they talk about cluster states and the circuit model of Quantum Computation and I didn't understand the following

Assume the state of multiple entangled qubits (like in a cluster state) is given by \par

$a|0>^{n+1}+b|1>^{n+1}$

After the measurement in the computational basis 1 and 0 the state of the remaining qubits is

$a|1>^{n}+b|0>^{n}$ for outcome 1

$a|0>^{n}+b|1>^{n}$ for outcome 0

I don't see how measuring a single photon in this cluster states leaves the rest in those particular states.Thank you in advance.

glS
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physics22
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1 Answers1

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I don't have access to the book you mentioned but I tried to understand and decipher what you are asking. Let's start with the state

$$ a|0\rangle|0\rangle^{\otimes n} + b|1\rangle|1\rangle^{\otimes n}, $$

which can be rewritten as (up to a normalization factor)

$$ |+\rangle(a|0\rangle^{\otimes n} + b|1\rangle^{\otimes n})+|-\rangle(a|0\rangle^{\otimes n} - b|1\rangle^{\otimes n}). $$

If you measure the first qubit in the X basis, you will end in

$$ a|0\rangle^{\otimes n} + b|1\rangle^{\otimes n} $$ for +1 outcome and $$ a|0\rangle^{\otimes n} - b|1\rangle^{\otimes n} $$ for -1 outcome.

Honestly, I haven't seen a way to flip a multi-entangled state with just local operations. So I guess there might be some typos.

Yunzhe
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