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Update entanglement.ipynb
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jyotiraj-code authored and amir-naveh committed Jun 9, 2024
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"source": [
"# Quantum Entanglement with Classiq\n",
"\n",
"Entanglement is an important aspect to study the quantum algorithms. In this tutorial, we show how to create a bell pair state $|\\Phi^{+}\\rangle$ of 2 qubits, using the Hadamard and Controlled-NOT transformation."
"Entanglement is an important aspect to study the quantum algorithms. In this tutorial, we show how to create a bell pair state $|\\Phi^{+}\\rangle$ of 2 qubits, using the Hadamard and the Controlled-NOT transformation."
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"$$\\begin{pmatrix} 1 & 0 & 0 & 0 \\\\ 0 & 1 & 0 & 0 \\\\ 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 \\end{pmatrix} \\begin{pmatrix} \\frac{1}{\\sqrt{2}} \\\\ 0 \\\\ \\frac{1}{\\sqrt{2}}\\\\ 0 \\end{pmatrix} = \\begin{pmatrix} \\frac{1}{\\sqrt{2}} \\\\ 0 \\\\ 0 \\\\ \\frac{1}{\\sqrt{2}} \\end{pmatrix} = \\frac{1}{\\sqrt{2}} \\left| 00 \\right\\rangle + \\frac{1}{\\sqrt{2}} \\left| 11 \\right\\rangle$$\n",
"\n",
"In this situation the qubit's of Alice and Bob are correlated to each other.<br>\n",
"If we measure the both the qubits we will either get the state $|00\\rangle$ or the state $|11\\rangle$ with equal probability.\n",
"If we measure both the qubits we will either get the state $|00\\rangle$ or the state $|11\\rangle$ with equal probability.\n",
"\n",
"It is described as:\n",
"- When the Alice oberseve her state as $|0\\rangle$ then the state of Bob's qubit collapses to the state $|0\\rangle$\n",
"- When the Alice oberseve her state as $|1\\rangle$ then the state of Bob's qubit collapses to the state $|1\\rangle$ \n",
"- When Alice oberseve her state as $|0\\rangle$ then the state of Bob's qubit collapses to the state $|0\\rangle$\n",
"- When Alice oberseve her state as $|1\\rangle$ then the state of Bob's qubit collapses to the state $|1\\rangle$ \n",
"\n",
"The resultant entangled state is designated as a bell pair state $|\\Phi^{+}\\rangle$\n",
"$$|\\Phi^{+}\\rangle = \\frac{1}{\\sqrt{2}} \\left( |00\\rangle + |11\\rangle \\right)$$\n",
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