narip
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Where did the article say M/N is the probability of error? M/N in the article is only for the use of normalization. For example(2 qubits), after $$H^{\otimes 2}$$ acted on initial $$\mid 0\rangle^{\otimes 2}$$, the state becomes $$\mid\psi\rangle\equiv1/2(\mid 00\rangle+\mid01\rangle+\mid10\rangle+\mid11\rangle)$$ . If the answer is $$| 01\rangle$$, then the $$\mid\alpha\rangle$$ in the book will be $$1/\sqrt3(\mid 00\rangle+\mid10\rangle+\mid11\rangle)$$, hence $$\mid\psi\rangle=\sqrt{3/4}\mid\alpha\rangle+\sqrt{1/4}\mid 01\rangle$$. Hence, $$M=1$$ and $$N=4$$ here.

Since $$\sqrt{N-M/N}^2+\sqrt{M/N} = 1$$$$\sqrt{N-M/N}^2+\sqrt{M/N}^2 = 1$$, we can choose them to be $$sin\theta/2$$ and $$cos\theta/2$$, i.e., $$sin\theta/2=\sqrt{M/N}$$. It only has number meaning.

edited body
Mark S
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Where did the article say NM/MN is the probability of error? M/N in the article is only for the use of normalization. For example(2 qubits), after $$H^{\otimes 2}$$ acted on initial $$\mid 0\rangle^{\otimes 2}$$, the state becomes $$\mid\psi\rangle\equiv1/2(\mid 00\rangle+\mid01\rangle+\mid10\rangle+\mid11\rangle)$$ . If the answer is $$| 01\rangle$$, then the $$\mid\alpha\rangle$$ in the book will be $$1/\sqrt3(\mid 00\rangle+\mid10\rangle+\mid11\rangle)$$, hence $$\mid\psi\rangle=\sqrt{3/4}\mid\alpha\rangle+\sqrt{1/4}\mid 01\rangle$$. Hence, $$M=1$$ and $$N=4$$ here.

Since $$\sqrt{N-M/N}^2+\sqrt{M/N} = 1$$, we can choose them to be $$sin\theta/2$$ and $$cos\theta/2$$, i.e., $$sin\theta/2=\sqrt{M/N}$$. It only has number meaning.

narip
• 2.2k
• 2
• 6
• 28

Where did the article say N/M is the probability of error? M/N in the article is only for the use of normalization. For example(2 qubits), after $$H^{\otimes 2}$$ acted on initial $$\mid 0\rangle^{\otimes 2}$$, the state becomes $$\mid\psi\rangle\equiv1/2(\mid 00\rangle+\mid01\rangle+\mid10\rangle+\mid11\rangle)$$ . If the answer is $$| 01\rangle$$, then the $$\mid\alpha\rangle$$ in the book will be $$1/\sqrt3(\mid 00\rangle+\mid10\rangle+\mid11\rangle)$$, hence $$\mid\psi\rangle=\sqrt{3/4}\mid\alpha\rangle+\sqrt{1/4}\mid 01\rangle$$. Hence, $$M=3$$$$M=1$$ and $$N=4$$ here.

Since $$\sqrt{N-M/N}^2+\sqrt{M/N} = 1$$, we can choose them to be $$sin\theta/2$$ and $$cos\theta/2$$, i.e., $$sin\theta/2=\sqrt{M/N}$$. It only has number meaning.

narip
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