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@@ -49,12 +49,26 @@ A state $|\psi\rangle$ is entangled if it cannot be expressed as a product state
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Example: the Bell state $|\psi^+\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ is entangled.
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Assume it can be written as $|\psi\rangle=|\psi_1\rangle\otimes|\psi_2\rangle$ where $|\psi_1\rangle=a|0\rangle+b|1\rangle$ and $|\psi_2\rangle=c|0\rangle+d|1\rangle$. Then:
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$$|\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle$$
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$$
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|\psi\rangle=a|00\rangle+b|01\rangle+c|10\rangle+d|11\rangle
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$$
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Setting this equal to $|\psi^+\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$ gives:
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$$ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$$
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$$
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ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)
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$$
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This requires:
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$$ac=bd=\frac{1}{2}$$
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$$ad=bc=0$$
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$$
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ac=bd=\frac{1}{2}
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$$
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$$
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ad=bc=0
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$$
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This is a contradiction, so $|\psi^+\rangle$ is entangled.
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@@ -21,4 +21,6 @@ export default {
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},
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Math401_P1: "Math 401, Paper 1: Concentration of measure effects in quantum information (Patrick Hayden)",
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Math401_P1_1: "Math 401, Paper 1, Side note 1: Quantum information theory and Measure concentration",
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Math401_P1_2: "Math 401, Paper 1, Side note 2: Page's lemma",
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Math401_P1_3: "Math 401, Paper 1, Side note 3: Levy's concentration theorem",
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}
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