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Exercises/Session 4 - Quantum Information (with solutions).ipynb
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| { | ||
| "cells": [ | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "e4f44eb7", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "$\\def\\tcoreleft{\\underset{\\tiny\\mid}{\\textcolor{MidnightBlue}{⦸}}}$\n", | ||
| "$\\def\\tcorecenter{\\underset{\\tiny\\mid}{\\textcolor{RedOrange}{⦿}}}$\n", | ||
| "$\\def\\tcoreright{\\underset{\\tiny\\mid}{\\textcolor{MidnightBlue}{\\oslash}}}$\n", | ||
| "<h1 style=\"text-align: center;\"><b>TMQS Workshop 2024</b> @ Zuse Institute Berlin</h1>\n", | ||
| "<h2 style=\"text-align: center;\">Summer School on Tensor Methods for Quantum Simulation</h2>\n", | ||
| "<h2 style=\"text-align: center;\">June 3 - 5, 2024</h2>\n", | ||
| "<h1 style=\"text-align: center; background-color:#D6EAF8 ;padding:50px\">$\\tcoreleft - \\tcoreleft - \\tcoreleft - \\cdots - \\tcorecenter - \\cdots - \\tcoreright - \\tcoreright$</h1>\n", | ||
| "</br>" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "6f6cc702", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "***" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "74ce99d5-268a-4ad8-980b-de19d60b6be6", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "## **Session 4 - Quantum Information**" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "e8441c54-ffd4-45b4-ab04-ff8cf85e6474", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "***" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "a781c1c8", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "## Exercise 4.1\n", | ||
| "\n", | ||
| "Which of the following are valid quantum states?\n", | ||
| "\n", | ||
| "$\\hspace{1cm}$$\\begin{pmatrix} 0 \\\\ 1\\end{pmatrix}$, $\\quad \\begin{pmatrix} 1 \\\\ 1\\end{pmatrix}$, $\\quad \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 0 \\\\ -i\\end{pmatrix}$, $\\quad \\frac{1}{\\sqrt{3}}\\begin{pmatrix} 1 \\\\ 2\\end{pmatrix}$, $\\quad \\begin{pmatrix} \\sqrt{2/3} \\\\ i/\\sqrt{3}\\end{pmatrix}$\n", | ||
| "\n", | ||
| "What is the probability to measure $0$ and $1$ for the valid quantum states?" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "9c481a4d-7f0e-45a0-8ea6-19ce1accee7c", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "***" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "ffc18314-42a9-4adf-a0ad-bcf763e8899e", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "$\\textcolor{red}{\\textbf{SOLUTION:}}$" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "61ad94fb-c6d8-4b79-b143-4850e790d826", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "Consider the euclidean norm of each vector, we get\n", | ||
| "\n", | ||
| "$\\hspace{0.5cm}$$\\left\\lVert\\begin{pmatrix} 0 \\\\ 1\\end{pmatrix}\\right\\rVert_2=1$, \n", | ||
| "$\\quad \\left\\lVert\\begin{pmatrix} 1 \\\\ 1\\end{pmatrix}\\right\\rVert_2=\\sqrt{2}$, \n", | ||
| "$\\quad \\left\\lVert\\frac{1}{\\sqrt{2}}\\begin{pmatrix} 0 \\\\ -i\\end{pmatrix}\\right\\rVert_2 = \\frac{1}{\\sqrt{2}}$, \n", | ||
| "$\\quad \\left\\lVert\\frac{1}{\\sqrt{3}}\\begin{pmatrix} 1 \\\\ 2\\end{pmatrix}\\right\\rVert_2 = \\sqrt{\\frac{5}{3}}$, \n", | ||
| "$\\quad \\left\\lVert\\begin{pmatrix} \\sqrt{2/3} \\\\ i/\\sqrt{3}\\end{pmatrix}\\right\\rVert_2 = 1$.\n", | ||
| "\n", | ||
| "Only the first and the last vector represent valid quantum states!" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "6e1b9898-10a5-422d-8cdb-a8f42bdc01c4", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "***" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "0e8243a2-a54e-4aaf-b460-f1f7f9e369b0", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "## Exercise 4.2\n", | ||
| "\n", | ||
| "**a)**$\\quad$Write down the state vector of two quantum states \n", | ||
| "\n", | ||
| "$\\hspace{1cm}$$\\displaystyle|\\Psi_1\\rangle = \\alpha_1 |0\\rangle + \\beta_1 |1\\rangle \\quad $ and $\\quad |\\Psi_2\\rangle = \\alpha_2 |0\\rangle + \\beta_2 |1\\rangle$, \n", | ||
| "\n", | ||
| "$\\hspace{0.35cm}$$\\quad$i.e. the tensor product, in the computational basis. Write down the basis vectors of the composite system.\n", | ||
| "\n", | ||
| "**b)**$\\quad$Consider the $2$-qubit state \n", | ||
| "\n", | ||
| "$\\hspace{1cm}$$\\displaystyle|\\Psi\\rangle = \\frac{1}{\\sqrt{2}} |00\\rangle + \\frac{1}{2}|01\\rangle + \\frac{1}{2} |11\\rangle$. \n", | ||
| "\n", | ||
| "$\\hspace{0.35cm}$$\\quad$What is the state after a measurement of the first qubit where you obtain $|0\\rangle$?\n", | ||
| "\n", | ||
| "$\\hspace{0.35cm}$$\\quad$Is this an entangled state?\n", | ||
| "\n", | ||
| "$\\hspace{0.35cm}$$\\quad$*Hint:* Quantum states are normalized!" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "f1b9d1e8-8c3a-4ac4-a26c-e285881cf7bf", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "***" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "b9d96ea3-335a-4219-a46e-a52aada2c31e", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "$\\textcolor{red}{\\textbf{SOLUTION:}}$" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "ac34fc7e-46d8-4b58-9025-2f88b3c93405", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "**a)**$\\quad$$|\\Psi_1\\rangle \\otimes |\\Psi_2\\rangle = \\left( \\alpha_1 |0\\rangle + \\beta_1 |1\\rangle \\right) \\otimes \\left( \\alpha_2 |0\\rangle + \\beta_2 |1\\rangle \\right) = \\alpha_1 \\alpha_2 | 00 \\rangle + \\alpha_1 \\beta_2 | 01 \\rangle + \\beta_1 \\alpha_2 | 10\\rangle + \\beta_1 \\beta_2 | 11\\rangle$\n", | ||
| "\n", | ||
| "$\\hspace{2.95cm}$$\\widehat{=}~ \\begin{pmatrix} \\alpha_1 \\alpha_2 & \\alpha_1 \\beta_2 & \\beta_1 \\alpha_2 & \\beta_1 \\beta_2\\end{pmatrix}^\\top$\n", | ||
| "\n", | ||
| "$\\hspace{0.35cm}$$\\quad$basis vectors: $\\quad \\begin{pmatrix} 1 & 0 & 0 & 0 \\end{pmatrix}^\\top, \\dots, \\begin{pmatrix} 0 & 0 & 0 & 1 \\end{pmatrix}^\\top$\n", | ||
| "\n", | ||
| "**b)**$\\quad$The classical state $|11\\rangle$ drops off, and we obtain (after normalization):\n", | ||
| "\n", | ||
| "$\\hspace{1.5cm}$$\\displaystyle \\sqrt{\\frac{4}{3}}\\left( \\frac{1}{\\sqrt{2}} |00\\rangle + \\frac{1}{2}|01\\rangle \\right) = \\sqrt{\\frac{2}{3}} |00\\rangle + \\sqrt{\\frac{1}{3}} |01\\rangle$\n", | ||
| "\n", | ||
| "$\\hspace{0.35cm}$$\\quad$This state cannot be written as a tensor-product term. \n", | ||
| "\n", | ||
| "$\\hspace{0.35cm}$$\\quad$You can also see this from the fact that we get different probabilities for the outcomes of the second qubit depending on the different outcomes of the first qubit." | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "818bf57f-be0c-408a-9776-414578b7763a", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "***" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "bed85201-1bfb-45a2-a57a-44332f370276", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "## Exercise 4.3\n", | ||
| "\n", | ||
| "Suppose you have $n + 1$ qubits. We will write $|\\vec{x}\\rangle$ to mean the $n$-qubit classical state given by the number $x$ in binary. For instance, if $n = 2$ then:\n", | ||
| "\n", | ||
| "$\\hspace{0.5cm}$$|\\vec{0}\\rangle = |00\\rangle , \\quad |\\vec{1}\\rangle = |01\\rangle, \\quad |\\vec{2}\\rangle = |10\\rangle, \\quad |\\vec{3}\\rangle = |11\\rangle$.\n", | ||
| "\n", | ||
| "Assume the qubits are in the state\n", | ||
| "\n", | ||
| "$\\hspace{0.5cm}$$\\displaystyle |\\Psi\\rangle = \\frac{1}{\\sqrt{2^n}} \\sum_{x=0}^{2^n -1} | \\vec{x} \\rangle \\otimes | x~\\text{mod}~2 \\rangle$.\n", | ||
| "\n", | ||
| "**a)**$\\quad$What is the resulting state if we measure the last qubit and obtain $|0\\rangle$?\n", | ||
| "\n", | ||
| "**b)**$\\quad$What is the resulting state if we measure the last qubit and obtain $|1\\rangle$?" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "1166e9c0-7a64-4cf6-a269-0c83743afa43", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "***" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "464bcad9-fe4c-420e-8b41-dad09248a94f", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "$\\textcolor{red}{\\textbf{SOLUTION:}}$" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "markdown", | ||
| "id": "bb28814f-8792-4365-ac53-15be05d0549f", | ||
| "metadata": {}, | ||
| "source": [ | ||
| "**a)**$\\quad$If we measure $|0\\rangle$, all states with last qubit $|1\\rangle$ disappear from the summation and we have\n", | ||
| "\n", | ||
| "$\\hspace{1cm}$$\\displaystyle\\frac{1}{\\sqrt{2^{n-1}}} \\sum_{x=0, x~\\text{even}}^{2^n -1} | \\vec{x} \\rangle \\otimes | 0 \\rangle = \\frac{1}{\\sqrt{2^{n-1}}} \\sum_{x=0}^{2^{n-1} -1} | \\overrightarrow{2x} \\rangle \\otimes | 0 \\rangle $\n", | ||
| "\n", | ||
| "**b)**$\\quad$If we measure $|1\\rangle$, all states with last qubit $|0\\rangle$ disappear from the summation and we have\n", | ||
| "\n", | ||
| "$\\hspace{1cm}$$\\displaystyle\\frac{1}{\\sqrt{2^{n-1}}} \\sum_{x=0, x~\\text{odd}}^{2^n -1} | \\vec{x} \\rangle \\otimes | 0 \\rangle = \\frac{1}{\\sqrt{2^{n-1}}} \\sum_{x=0}^{2^{n-1} -1} | \\overrightarrow{2x+1} \\rangle \\otimes | 0 \\rangle $" | ||
| ] | ||
| }, | ||
| { | ||
| "cell_type": "code", | ||
| "execution_count": null, | ||
| "id": "c5f602d6-775a-4c77-a737-c2e6168794a5", | ||
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| "metadata": { | ||
| "kernelspec": { | ||
| "display_name": "Python 3 (ipykernel)", | ||
| "language": "python", | ||
| "name": "python3" | ||
| }, | ||
| "language_info": { | ||
| "codemirror_mode": { | ||
| "name": "ipython", | ||
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| "file_extension": ".py", | ||
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| "pygments_lexer": "ipython3", | ||
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Exercises/Session 5 - Quantum Computing (with solutions).ipynb
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