Chapters/5_Single_Qubits_&_Bloch_Sphere.ipynb

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    "# 5 Single Qubits & the Bloch Sphere  \n",
    "\n",
    "Having covered what quantum states are, this chapter will cover qubits and how we can control them. \n",
    "\n",
    "## 5.1 The unit of information\n",
    "\n",
    "Having introduced some of the basic mathematics to describe quantum computing, let's get started. \n",
    "\n",
    "Almost all articles or news reports talking about quantum computing start off by explaining how qubits are different from classical bits. This chapter will use the maths from previous chapters to give a better description of these qubits. \n",
    "\n",
    "Earlier, the coin example was introduced with the heads & tails outcomes we can observe. A qubit also has 2 outcomes called $\\ket{0}$ & $\\ket{1}$, we can just replace $\\ket{H}$ with $\\ket{0}$ and $\\ket{T}$ with $\\ket{1}$. \n",
    "\n",
    "$$\n",
    "\\ket{0} = \\begin{bmatrix} 1 \\\\ 0 \\end{bmatrix} ; \\ket{1} = \\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix} \n",
    "$$\n",
    "\n",
    "These two states are called the computational basis\n",
    "\n",
    "Just as with the coins, our qubits can be in a superposition of $\\ket{0}$ & $\\ket{1}$. We can have any probability of measuring $\\ket{0}$ & $\\ket{1}$ (so long as all the probabilities are real positive numbers which add to 1). \n",
    "\n",
    "\n",
    "\n",
    "## 5.2 The Bloch Sphere \n",
    "\n",
    "Dirac notation is really useful for doing algebra with quantum states. We can model how the qubits change using vectors and calulate probabilities with the inner product. One limitation of Dirac notation is that it is difficult to visualise quantum states as it does not have a diagram to represent the state vector on. \n",
    "\n",
    "The Bloch sphere is a useful tool used to represent the state of a qubit. Below, the quantum state $\\ket{\\psi}$ is represented on the Bloch Sphere. \n",
    "\n",
    "![Bloch_sphere](Images/Bloch_Sphere.png) [1]\n",
    "\n",
    "At the top of the sphere  is the state $\\ket{0}$, at the botom is the state $\\ket{1}$. Our state $\\ket{\\psi}$ is somewhere in between so it is a superposition of $\\ket{0}$ & $\\ket{1}$. The closer $\\ket{\\psi}$  is to the top, the greater the probability of measuring $\\ket{0}$, the same is true for the bottom with $\\ket{1}$. The angle $\\theta$ between the $\\ket{0}$ & $\\ket{\\psi}$ gives us these probabilities. \n",
    "\n",
    "Notice how $\\ket{0}$ and $\\ket{1}$ are seperated by $180\\degree (\\pi)$ on the sphere, but they are perpendicular, so the angle between them is actually $90\\degree (\\pi/2)$. The angles on the Bloch sphere are twice as large as the angles for the qubits.\n",
    "\n",
    "> The angle $\\theta$ on the Bloch sphere gives the probabilities with angle $\\theta/2$ \n",
    "\n",
    "$$\n",
    "P(\\ket{0}) = |\\braket{0|\\psi}|^2 = (\\cos{\\theta/2})^2\n",
    "$$\n",
    "Similarly for the probability of $\\ket{1}$\n",
    "\n",
    "$$P(\\ket{1}) = |\\braket{0|\\psi}|^2 = (\\sin{\\theta/2})^2$$\n",
    "\n",
    "\n",
    "The z-axis is where the $\\ket{0}$ & $\\ket{1}$ are, for that reason the z-axis is known as th computational basis. In addition we have the x & y axis which correspond to different states. \n",
    "\n",
    "A useful tool for understanding the Bloch sphere [can be found here](https://javafxpert.github.io/grok-bloch/)\n",
    "\n",
    "\n",
    ">### Exercise 5.1: Where on the sphere\n",
    ">\n",
    ">We know the superposition state $\\ket{+}$ has equal probabilities of $\\ket{0}$ & $\\ket{1}$. Where on the Bloch sphere would this state be represented? What is the angle $\\theta$ that would describe this state?\n",
    "\n",
    "## 5.3 It's not just a phase!\n",
    "\n",
    "To show probabilities, a single angle is enough. So why do we need a sphere? \n",
    "\n",
    "There is another important parameter that describes a qubit: the phase. \n",
    "\n",
    "At last, our coins will not work to describe phase, there is no analogue in classical computing, nor in probability.\n",
    "\n",
    "Two quantum states can have the same probabilities, but be completely different states.\n",
    "\n",
    "For example, earlier the flip coin operator $F$ had this form:\n",
    "\n",
    "$$\n",
    "F =  \\frac{1}{\\sqrt{2}} \\begin{bmatrix} 1 & 1 \\\\ 1 & -1 \\end{bmatrix}\n",
    "$$\n",
    "\n",
    "In quantum computing, this operator is known as a __Hadamard gate__ with symbol $H$. \n",
    "\n",
    "If we apply $H$ to $\\ket{0}$ we get the superposition state \n",
    "\n",
    "$$\n",
    "H\\ket{0} = \\frac{1}{\\sqrt{2}} ( \\ket{0} + \\ket{1}) = \\ket{+}\n",
    "$$\n",
    "\n",
    "Applying it to $\\ket{1}$ gives us a different superposition state: \n",
    "\n",
    "$$\n",
    "H\\ket{1} = \\frac{1}{\\sqrt{2}} ( \\ket{0} - \\ket{1}) = \\ket{-}\n",
    "$$\n",
    "\n",
    "We call this state $\\ket{-}$ because of the minus sign between the $\\ket{0}$ & $\\ket{1}$\n",
    "\n",
    "The probability of getting $\\ket{0}$ or $\\ket{1}$ is the same for both. If we were to measure either state, we would get 0 half the time and 1 the other half the time. Since the measurement gives the same outcome you might think they are the same state. \n",
    " \n",
    "They are different states, and suprisingly there is an easy way for us to tell which one we have. \n",
    "\n",
    "What happens if we apply $H$ on these states? \n",
    "\n",
    "$$ H\\ket{+} = H \\frac{1}{\\sqrt{2}} ( \\ket{0} + \\ket{1})$$\n",
    "\n",
    "$$= \\frac{1}{\\sqrt{2}} (H\\ket{0} + H\\ket{1})$$\n",
    "\n",
    "$$ = \\frac{1}{\\sqrt{2}}\\left( \\frac{1}{\\sqrt{2}}(\\ket{0} + \\ket{1})+\\frac{1}{\\sqrt{2}}(\\ket{0} - \\ket{1}))  \\right)$$\n",
    "\n",
    "The $\\ket{1}$'s cancel each other out and the $\\ket{0}$ add together to give us\n",
    "\n",
    "$$ = \\ket{0} $$\n",
    "\n",
    "So if we apply $H$ once to $\\ket{0}$ we get $\\ket{+}$ and then applying it again goes back to $\\ket{0}$. \n",
    "\n",
    "This implies applying $H$ twice does nothing. $H^2 = I$.\n",
    "\n",
    "So if we ever have $\\ket{+}$ we apply $H$ and get $\\ket{0}$ 100% of the time. If we get $\\ket{1}$ we must have had another state to begin with. If we had $\\ket{-}$ would measure 1 100% of the time. \n",
    "\n",
    "\n",
    ">### Exercise 5.2 Back to the start\n",
    ">\n",
    ">Apply $H$ twice to $\\ket{1}$ and verify the state we end up with is $\\ket{1}$. \n",
    "\n",
    ">### Exercise 5.3 The coin doesn't work\n",
    ">\n",
    "> When we apply the Hadamard twice to a state, we get the same state back. Without doing any maths, consider throwing a coin. Why does throwing the coin twice not guarantee you get back the side you started with?\n",
    "\n",
    "\n",
    "In this example, where is the phase? \n",
    "\n",
    "The phase is contained in the + or - sign between the $\\ket{0}$ & $\\ket{1}$. \n",
    "\n",
    "The phase can be more than just $\\pm 1$, it can be any complex number with magnitude 1. In general, any quantum state can have a phase $\\phi$ \n",
    "\n",
    "$$ \\alpha\\ket{0} + \\beta e^{i\\phi}\\ket{1} $$\n",
    "\n",
    "With $ 0 \\leq \\phi < 2\\pi $\n",
    "\n",
    "We can use what we know from the Bloch Sphere to write $\\alpha$ & $\\beta$ in terms of the angle $\\theta$\n",
    "\n",
    "\n",
    "$$ \\cos{(\\theta/2)}\\ket{0} + \\sin{(\\theta/2)} e^{i\\phi}\\ket{1} $$\n",
    "\n",
    "\n",
    "Remember Euler's formula (section 3.1.2) which allows us to convert from the exponential to the complex number\n",
    "\n",
    "$$ e^{i \\phi} = cos(\\phi) + i sin(\\phi) $$\n",
    "\n",
    "If we choose $\\phi = 0$ we get $e^0 = 1$ which gives us \n",
    "\n",
    "The phase is also shown on the Bloch sphere. In the above diagram the state $\\ket{\\psi}$ is described by two angles $\\theta$ describing the probabilities of measuring 0 or 1 and $\\phi$ describing the phase. The angle $\\phi$ corresponds to a rotation by $\\phi$ around the z-axis in the Bloch sphere.  \n",
    "\n",
    "Phases can be used to encode information, such as with the quantum Fourier transform, or to mark out quantum states (as with Grover's algorithm). \n",
    "\n",
    "> Phases only matter when they are relative. This means a phase difference between the different states. \n",
    "\n",
    "## 5.4 Qubit gates \n",
    "\n",
    "As mentioned in Chapter 4, operators are what allow us to do things to quantum states. Below are a collection of the most common single qubit operators. \n",
    "\n",
    "\n",
    "Gate | Description| Effect on $\\ket{0}$| Effect on $\\ket{1}$ | Matrix \n",
    "----- | ----- | ----- | ----- | -----|\n",
    "$I$ | Identity: do nothing | $\\ket{0} \\rightarrow  \\ket{0} $ | $\\ket{1} \\rightarrow \\ket{1}$ | $\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix}$\n",
    "$X$ | Not: Swaps $\\ket{0}$ with $\\ket{1}$   |$\\ket{0} \\rightarrow \\ket{1}$ |$\\ket{1} \\rightarrow  \\ket{0}$ | $\\begin{bmatrix} 0 & 1 \\\\ 1 & 0 \\end{bmatrix}$|\n",
    "$Y$ | Y: Rotation by angle $\\pi$ around the y-axis of the Bloch sphere | $\\ket{0} \\rightarrow  i\\ket{1} $ |$\\ket{1} \\rightarrow -i\\ket{0} $ | $\\begin{bmatrix} 0 & -i \\\\ i & 0 \\end{bmatrix}$|\n",
    "$Z$ | Z: Adds a phase of $e^{i\\pi}$ to $\\ket{1}$|$\\ket{0} \\rightarrow \\ket{0}$|$\\ket{1} \\rightarrow -\\ket{1}$ |$\\begin{bmatrix} 1 & 0 \\\\ 0 & -1 \\end{bmatrix}$ |\n",
    "$H$ | Hadamard: Creates the superposition state | $\\ket{0} \\rightarrow \\frac{1}{\\sqrt{2}} ( \\ket{0} + \\ket{1})$ | $\\ket{1} \\rightarrow \\frac{1}{\\sqrt{2}} ( \\ket{0} - \\ket{1})$ | $\\frac{1}{\\sqrt{2}}\\begin{bmatrix} 1 & 1 \\\\ 1 & -1 \\end{bmatrix} $\n",
    "$S$ | Adds a phase of $e^{\\pi/2}$ betweeen $\\ket{0}$ & $\\ket{1}$|$\\ket{0} \\rightarrow  \\ket{0} $ |$\\ket{1} \\rightarrow e^{i\\pi/2}\\ket{1}$  | $\\begin{bmatrix} 1 & 0 \\\\ 0 & e^{i\\pi/2} \\end{bmatrix}$\n",
    "$T$ | Adds a phase of $e^{\\pi/4}$ betweeen $\\ket{0}$ & $\\ket{1}$ | $\\ket{0} \\rightarrow  \\ket{0} $| $\\ket{1} \\rightarrow e^{i\\pi/4}\\ket{1}$ |$\\begin{bmatrix} 1 & 0 \\\\ 0 & e^{i\\pi/4} \\end{bmatrix}$\n",
    "\n",
    "Sometimes it is easier to perform gates by knowing what they do to the $\\ket{0}$, $\\ket{1}$ states, sometimes it is easier to do the matrix multiplication. \n",
    "\n",
    "A more comprehensive list of gates [can be found here](https://raw.githubusercontent.com/qiskit-community/qiskit-textbook/main/content/ch-states/supplements/single-gates-cheatsheet.pdf) \n",
    "\n",
    "> ### Exercise 5.4 \n",
    "> \n",
    "> By considering the effect on $\\ket{0}$ & $ \\ket{1}$, or by doing matrix multiplication, calculate the following states:\n",
    ">\n",
    "> a. $ XZ\\ket{1}$\n",
    ">\n",
    "> b. $ XH\\ket{+} $\n",
    ">\n",
    "> c. $ XYZ\\ket{0} $\n",
    "\n",
    "\n",
    "### Aside: gates vs operators\n",
    "\n",
    "In the last chapter, operators were introduced as matrices that act on a quantum state. In this chapter, it seems quantum gates do the same thing. This is true, all quantum gates are operators, but not all operators are quantum gates. \n",
    "\n",
    "For instance, the creation operator $\\hat{a}^\\dagger$ creates a photon from the vaccuum but can't be represented by a matrix. The momentum operator $\\hat{p} = -i\\hbar\\frac{\\partial}{\\partial x}$ is a partial derivative. \n",
    "\n",
    "\n",
    "## 5.5 Parameterised quantum gates \n",
    "\n",
    "In addition to these fixed quantum gates, there are quantum gates we can specify ourselves. These gates are described by a parameter we can control to change what the gate does. These gates are very similar to nodes in a neural network where we can change the weights (we will see a lot more about these in a chapter on quantum machine learning).\n",
    "\n",
    "For instance, we might want to get more than a 50% chance of measuring an output. We can see on the Bloch sphere that if we rotate our quantum state by some angle towards $\\ket{0}$ or $\\ket{1}$ we can boost the probability of measuring those. The matrix that does this rotation is known as $R_x$\n",
    "\n",
    "$$\n",
    "R_y(\\theta) = \\begin{bmatrix} \\cos(\\theta /2) & -\\sin(\\theta /2) \\\\ \\sin(\\theta /2) & \\cos(\\theta /2) \\end{bmatrix}\n",
    "$$\n",
    "\n",
    "> The angle we give as input is $\\theta$ the angle we perform the rotation is $\\theta/2$.\n",
    "\n",
    "We also have rotations for the x & y axis too. In fact, the X,Y gates are special cases of $R_x(\\theta)$ & $R_y(\\theta)$ with $\\theta = \\pi$. \n",
    "\n",
    "$R_z(\\phi)$ (also known as P) is also useful because it allows us to give any phase to a qubit. \n",
    "\n",
    "$$\n",
    "R_z(\\phi) = P(\\phi) = \\begin{bmatrix} 1 & 0 \\\\ 0 & e^{i\\phi} \\end{bmatrix}\n",
    "$$\n",
    "\n",
    "This gives a phase shift of $\\phi$ between $\\ket{0}$ & $\\ket{1}$\n",
    "\n",
    "\n",
    "> ### Exercise 5.5 \n",
    "> By using $\\theta = \\pi/2$ apply $R_y(\\pi/2)$ to the state $\\ket{1}$. What gate does $R_y(\\pi/2)$ do? \n",
    " \n",
    "> ### Exercise 5.6 \n",
    "> The X gate turns $\\ket{0}$ to $\\ket{1}$ and $\\ket{1}$ to $\\ket{0}$. By considering angles on the Bloch sphere, what is the angle $\\theta$ such that $R_y(\\theta) = X$\n",
    "\n",
    "\n",
    "## 5.6 The Universal Quantum Gate \n",
    "\n",
    "Just as we can represent any single qubit state with two parameters, we can also represent any single qubit gate using parameters. \n",
    "\n",
    "$$\n",
    "U(\\theta, \\phi, \\lambda) = \\begin{bmatrix} \\cos(\\theta/2) & -e^{i\\lambda}\\sin(\\theta/2) \\\\ e^{i\\phi}\\sin(\\theta/2) & e^{i(\\phi + \\lambda)}\\cos(\\theta) \\end{bmatrix} \n",
    "$$\n",
    "\n",
    "Just as we can reach any point on the Bloch Sphere with $\\theta$ & $\\phi$ we can get any quantum gate using $U$. \n",
    "\n",
    "The set of all gates that can be made from $U$ are known as $SU(2)$. \n",
    "\n",
    ">### Exercise 5.7 \n",
    "> Which gate in the list above is represented by  $U(\\theta, \\phi, \\lambda) = U(0,0,0)$ \n",
    "\n",
    "\n",
    "## 5.7 Bonus: Beyond Bits \n",
    "\n",
    "For this course, we work with qubits- 2 level quantum systems. There is considerable research into using more than 2 levels for quantum computation. Such systems are called qudits- d-level quantum systems that we can use to process quantum information. \n",
    "\n",
    "Qudits are much less commonly used than qubits for a few reasons\n",
    "\n",
    "- Having multiple levels is much more complicated. Think about the X gate, for a qudit, this gate becomes much more complicated\n",
    "\n",
    "- Qudits present even more engineering challenges than qubits. For instance, a qubit in a trapped ion consists of the ground level and the first excited state. Higher energy levels have smaller gaps in energy, so they are more difficult to control precisely. \n",
    "\n",
    "- Quantum error correction for qudits is much more complicated. With qubits $\\ket{0}$ can only go to $\\ket{1}$ but for a 4 level system there are 3 possible states it could go to. And that's just one qudit\n",
    "\n",
    "\n",
    "# Chapter 5 Summary \n",
    "\n",
    "- The state of a qubit can be represented on the Bloch sphere\n",
    "- The BS shows the probability and phase of a qubit \n",
    "- The phase of a qubit is the $e^{i\\theta}$ difference between the $\\ket{0}$ & $\\ket{1}$ states\n",
    "- There are many qubit gates we can use to perform operations on qubits\n",
    "- Some qubit gates are rotations specified by a parameter which describes what the gate does\n",
    "### References \n",
    "\n",
    "[1] Chetan Wake, 2019, Introduction to quantum computing part -1 Representation of qubit using Bloch sphere"
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