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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.

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 $\sqrt{N-M/N}^2+\sqrt{M/N} = 1$, 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.

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 $\sqrt{N-M/N}^2+\sqrt{M/N}^2 = 1$, 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.

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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.

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, $M=1$ 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.

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 $\sqrt{N-M/N}^2+\sqrt{M/N} = 1$, 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.

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narip
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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.

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, $M=3$ and $N=4$ here.

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, $M=1$ 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.

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narip
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