A picture of a laptop on a desk. There is a plant to the right of the laptop and a lamp positioned behind. In the back ground is a bookcase.Thinking about the hardware, the improvement in computing power over the past 50 years has been enormous. The number of transistors on a microprocessor chip has doubled approximately every two years, a fact known as Moore’s law. Unfortunately, this trend cannot continue indefinitely, because one of the main ways that these improvements are realised is by shrinking the size of the circuit elements.Improving the algorithms is the second possibility. The power of an algorithm can be expressed by stating how the ‘run-time’ which is the number of steps required to implement the algorithm, scales with the size of the task. Some algorithms have a polynomial run-time meaning that if a procedure is to be carried out on a number, n, of elements, the run-time scales as nx where x is typically a small integer. Other algorithms have an exponential run-time, scaling as en.The algorithms with polynomial run-time are considered to be much more useful than algorithms with exponential run-time, because exponential scaling means that only modest increases in the task size can exhaust available resources. Throughout this course there are a series of exercises for you to work through. Some of these exercises you can supply your answer to in the response boxes provided. Others will require you to work through your calculations on paper.Exercise 1Consider some algorithms that are carried out on two data sets – one with n = 10 elements and another with n = 100 elements. The run-time of one algorithm scales as n3 and the run-time of another algorithm scales as en. How do the run-times of the two algorithms compare for the smaller data set? How do the run-times compare for the larger data set?For the smaller data set, the run-times for the two algorithms are in the ratio \rm{e}^{10} / 10^3 \sim 22, whereas for the larger data set, the run-times for the two algorithms are in the ratio \rm{e}^{100} / 100^3 \sim 2.7 \times 10^{37}. The time to carry out the algorithm with the exponential run-time soon becomes unfeasibly long as the size of the data set increases.1.2 What can quantum computers do?Quantum computing offers a new computing approach that is based on the idea that the rules of quantum mechanics can allow shortcuts to solutions for certain tasks. There are quantum algorithms with polynomial run-times that can solve problems where the only known classical algorithms have exponential run-times. Quantum computers make use of the fact that quantum states can be made of linear combinations of individual states and when a measurement is taken, one of the individual states will be measured with a given probability. You will see that entanglement is another fundamental resource for quantum computing. The integer factorisation problem, which seeks to find the prime number factors of an n-digit integer, is of particular interest because it is the basis of a lot of information security. The run-time of the classical algorithm scales exponentially so as n increases a classical computer takes longer and longer to factorise the integers as n increases. The integer factorisation problem is an example of a a problem that quantum computers can solve much more efficiently. If quantum computing was able to solve the integer factorisation problem with a polynomial run-time then there would be a major problem for information security. Another example of a problem with an exponential run-time is the travelling salesperson problem, which seeks to find the shortest route between a number of cities subject to visiting each city only once and returning to the starting city at the end of the journey (see Figure 2). This is a typical optimisation problem which can be used to demonstrate the power of quantum computing as well as being a problem which delivery companies would like to be able to solve quickly.
Figure 2 An illustration of the travelling salesperson problem. The image on the left shows 7 cities (in blue) and a trial of all 360 possible routes between them (in red). The image on the right shows the optimum distance after trialling all possible routes.
The image on the left shows shows 7 blue dots in a grid with red lines connecting them. The red lines change position as different connections are trialled. The image on the right shows the same 7 blue dots with a green line connecting the shortest possible distance between them.Exercise 2Write a sentence or two to summarise in general terms the context in which quantum computers are considered to be an improvement on classical computers.Your answer will not be the same but a possible answer is:Quantum computers may be able solve some problems more quickly than classical computers if problem solving algorithms which have exponential run-times on a classical computer can be written to have polynomial run-times on a quantum computer.Your answer should include the same conclusions.2 Background mathematics and terminologyIn this section, some mathematics and terminology which will be needed to learn about quantum computing are introduced. This includes matrices, eigenvalue equations and complex numbers.2.1 Matrix multiplicationA matrix, \mathrm{A}, is a rectangular array of numbers arranged in rows and columns. (Matrices is the plural of the word matrix.) You will concentrate on two-dimensional situations, and so consider matrices of the form:\begin{bmatrix} {A_{11}} & {A_{12}} \\ {A_{21}} & {A_{22}} \end{bmatrix} \quad \quad 2 \times 2 \; \text{square matrix}\begin{bmatrix} {A_{11}} & {A_{12}} \end{bmatrix} \quad \quad 1 \times 2 \; \text{row matrix}\begin{bmatrix} {A_{11}} \\ {A_{21}} \end{bmatrix} \quad \quad 2 \times 1 \; \text{column matrix}In order to multiply two matrices, the number of columns of first matrix should be equal to the number of rows in the second matrix. Matrices of the right shape can be multiplied together as follows: \mathrm{C} = \mathrm{A} \mathrm{B}.To calculate \mathrm{C}, where \mathrm{C}_{ij} is the element in the i th row and j th column of \mathrm{C}, you go along the ith row of \mathrm{A} and down the jth column of \mathrm{B}, multiplying corresponding elements and adding the results. For two 2 \times 2 matrices, this pattern may be visualized as follows:\begin{bmatrix}{*}&{\phantom{xx}}\\{}&{} \end{bmatrix} =
\begin{bmatrix}{\rightarrow}& {\rightarrow}\\ {}& {} \end{bmatrix} \begin{bmatrix}{\downarrow}&{\phantom{xx}}\\{\downarrow}&{}\end{bmatrix}\begin{bmatrix}{\phantom{xx}}&{*}\\{}&{}\end{bmatrix} =
\begin{bmatrix} {\rightarrow}&{\rightarrow}\\{}&{}\end{bmatrix} \begin{bmatrix} {\phantom{xx}}&{\downarrow}\\{}&{\downarrow}\end{bmatrix} \begin{bmatrix}{}&{}\\{*}&{\phantom{xx}}\end{bmatrix} =
\begin{bmatrix}{}&{}\\{\rightarrow}&{\rightarrow}\end{bmatrix} \begin{bmatrix}{\downarrow}&{\phantom{xx}}\\{\downarrow}&{}\end{bmatrix}\begin{bmatrix}{}&{}\\{\phantom{xx}}&{*}\end{bmatrix} =
\begin{bmatrix}{}&{}\\{\rightarrow}&{\rightarrow} \end{bmatrix} \begin{bmatrix}{}&{\downarrow}\\{\phantom{xx}}&{\downarrow} \end{bmatrix}where the * indicates a matrix element in the new matrix, \mathrm{C}, and the arrows show how matrix elements in the old matrices are processed to obtain this. In order to multiply two matrices, the number of columns of first matrix should be equal to the number of rows in the second matrix. In other words, each term C_{ij} is given by C_{ij} = A_{i1}B_{1j} + A_{i2}B_{2j}.Exercise 3What shape is the result of multiplying a 1 \times 2 row matrix by a 2 \times 2 square matrix?What shape is the result of multiplying a 2 \times 2 square matrix by a 1 \times 2 row matrix?What shape is the result of multiplying a 2 \times 2 square matrix by a 2 \times 1 column matrix?What shape is the result of multiplying a 1 \times 2 row matrix by a 2 \times 1 column matrix?The result is a 1 \times 2 row matrix.This operation cannot be performed because the number of columns of first matrix is not equal to the number of rows in the second matrix.The result is a 2 \times 1 column matrix.The result is a 1 \times 1 matrix (i.e. a scalar number).Exercise 4Evaluate the following combination of square matrices:\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}The product of two matrices is given by:\begin{bmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{bmatrix} \begin{bmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \end{bmatrix} = \begin{bmatrix} A_{11}B_{11} + A_{12}B_{21} & A_{11}B_{12} + A_{12}B_{22} \\ A_{21}B_{11} + A_{22}B_{21} & A_{21}B_{12} + A_{22} B_{22} \end{bmatrix}So, to find the matrix element in row i and column j of the product \mathrm{AB}, we multiply the elements in row i of \mathrm{A} with the corresponding elements in column j of \mathrm{B}, and add the results together.
To find the product of three matrices, \mathrm{ABC}, we first evaluate \mathrm{BC} and then form the product \mathrm{A(BC)}, taking care to preserve the order of the matrices.So the solution is \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 \times 1 + 0 \times 2 & 1 \times 2 + 0 \times (-1) \\ 0 \times 1 + (-1) \times 2 & 0 \times 2 + (-1) \times (-1) \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + 2 \times (-2) & 1 \times 2 + 2 \times 1 \\ 2 \times 1 + (-1) \times (-2) & 2 \times 2 + (-1) \times 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} -3 & 4 \\ 4 & 3 \end{bmatrix}2.2 Finding the eigenvalues and eigenvectors of a <InlineEquation><TeX>2 \times 2</TeX></InlineEquation> matrixFor a given square matrix, \mathrm{A}, it is possible to solve the equation \mathrm{A}\mathbf{v}= \lambda \mathbf{v} where \mathbf{v} are column vectors known as eigenvectors and \lambda is a scalar called an eigenvalue.A procedure to find the eigenvectors and eigenvalues of a 2 \times 2 square matrix \mathrm{A} = \begin{bmatrix}{a}&{b}\\{c}&{d}\end{bmatrix} isSolve the quadratic equation \lambda^2-(a+d)\lambda+ (ad -bc)=0 to find the two values of \lambda which are the required eigenvalues.For each eigenvalue found, write down the eigenvector equations (a-\lambda) x + by=0cx + (d-\lambda)y=0This pair of equations usually reduces to a single equation that is readily solved for x and y. The eigenvector is given by \mathbf{v} =\begin{bmatrix}{x}\\{y}\end{bmatrix} with x and y replaced by their solved values.It is often useful to normalise \mathbf{v} by writing it as a unit vector. It this case, the unit vector is given by \mathbf{v}_u = \frac{1}{\sqrt{x^2+y^2}} \begin{bmatrix}{x}\\{y} \end{bmatrix}.Exercise 5Find the eigenvalues and eigenvectors of the following matrix:\Lambda = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}Following the prescription described above: a=4, b=1, c=2 and d=3. So we first need to solve the quadratic equation \lambda^2 - (4+3) \lambda + ((4 \times 3) - (1 \times 2)) = 0which is simply \lambda^2 - 7 \lambda + 10 = 0This can be written as (\lambda - 2) (\lambda - 5) = 0So it has solutions \lambda_1 = 2 and \lambda_2 = 5. These are the two eigenvalues.We now write the two eigenvector equations:\begin{split}
(4 - \lambda) x + 1y & = 0 \\
2x + (3 - \lambda)y & = 0
\end{split}For eigenvalue \lambda_1 = 2 these reduce to
\begin{split}
2x + y & = 0 \\
2x + y & = 0
\end{split}Both equations imply that y=-2x, so x=1 and y=-2 and the first eigenvector is \mathbf{v}_1 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}For eigenvalue \lambda_2 = 5 these reduce to
\begin{split}
-x + y & = 0 \\
2x - 2y & = 0
\end{split}Both equations imply that y=x, so x=1 and y=1 and the second eigenvector is \mathbf{v}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}2.3 Complex numbersA complex number may be written in the form:z = x +\textrm{i}y,where x and y are real numbers and i is a special quantity with the property that \textrm{i}^2=-1.Each complex number z = x +\textrm{i}y has a real part, Re\{z\} = x, and an imaginary part, Im\{z\}=y. Complex numbers can be added,(a+b\textrm{i}) + (c+d\textrm{i}) = (a+c) +(b+d)\textrm{i},and multiplied, (a+b\textrm{i})(c+d\textrm{i}) = (ac-bd) +(ad+bc)\textrm{i},using the usual rules of algebra along with \textrm{i}^2=-1.The complex conjugate of z = x +\textrm{i}y is z^* = x -\textrm{i}y (pronounced "z star").This results in zz^* = (x +\textrm{i}y)(x -\textrm{i}y) = x^2 + y^2 showing that zz^* is a positive real number, (unless x=y=0).The modulus of the complex number z = x +\textrm{i}y is defined as |z| =\sqrt{zz^*} = \sqrt{x^2 + y^2} which is a real, non-negative quantity.Complex numbers can also be written in polar form,z=r\text{e}^{\textrm{i}\theta},where r and \theta are real numbers. The relationship between x and y, r and \theta is shown in Figure 3. Here r is the modulus of z as defined in Equation (4); \theta is known as the phase, and \textrm{e}^{\textrm{i}\theta} is a phase factor.
Figure 3 A diagram showing the relationship between Cartesian coordinates x and y, and polar coordinates r and \theta
The fighure shows a graph with two axes, labelled x-axis and y-axis, which intersect near the middle of the diagram, marked 0. An arrow is drawn from the origin at 0 into the upper right quadrant of the graph. The length of the arrow is labelled r and the angle it makes with the horizontal x-axis is labelled theta. A horizontal dashed line from the end of this arrow intersects the y-axis at a point marked y. A vertical dashed line from the end of this arrow intersects the x-axis at a point marked x.Exercise 6Consider the complex number z=3+3\mathrm{i}.
Write down its complex conjugate z^*.
Calculate the modulus of z.Write z in polar form.The complex conjugate is z^* = 3 - 3\mathrm{i}.The modulus of z is |z| = \sqrt{z z^*} = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}In polar form z=r \textrm{e}^{\textrm{i}\theta} where r=|z| = 3\sqrt{2} and \tan \theta = 3/3 = 1, so \theta = \pi/4 radians. Therefore we can write z = 3\sqrt{2}\text{e}^{\text{i}\pi/4}.
2.4 Operators and superpositionIn quantum mechanics, an operator is a mathematical entity which converts one function into another function and is written with a ‘hat’on, for example \mathrm{\widehat{A}}, (pronounced ‘A hat’). Given an operator \mathrm{\widehat{A}}, the eigenvalue equation for that operator is \mathrm{\widehat{A}}f(x)= \lambda f(x)Here, \lambda is a constant known as an eigenvalue, which may be complex, and f(x) is function known as an eigenfunction. There may be more than one eigenvalue and corresponding eigenfunction associated with each eigenvalue equation. The eigenvalue matrix equation, Equation (2) as described in Section 2.2 is an example of this type of equation with the operator written as a matrix and the eigenfunction as a column vector. Consider an operator \mathrm{\widehat{A}} with two eigenvectors and two corresponding eigenvalues so that\mathrm{\widehat{A}}f_1(x)= \lambda_1 f_1(x) \quad \text{and} \quad \mathrm{\widehat{A}}f_2(x)= \lambda_2 f_2(x).Since f_1(x) and f_2(x) are both eigenfunctions or solutions of the eigenvalue equation any linear combination of f_1(x) and f_2(x) is also a solution. Such a linear combination, known as a superposition, isf(x) = a_1f_1(x) + a_2 f_2(x)where a_1 and a_2 are complex numbers.Exercise 7The wave function of a free particle in quantum mechanics may be written as \Psi_\text{free}(x,t) = A \text{e}^{\text{i}(kx - \omega t)}Confirm that \Psi_\text{free}(x,t) is an eigenfunction of \text{i} \hbar \partial / \partial t (where \partial / \partial t indicates a partial derivative) and show that the eigenvalue is the energy of the free particle, \hbar \omega. (\hbar is the reduced Planck’s constant.)Operating on \Psi_\text{free}(x,t) with \text{i} \hbar \partial / \partial t we find that\begin{split} \text{i} \hbar \frac{\partial}{\partial t} \Psi_\text{free} (x,t) & =
\text{i} \hbar \frac{\partial}{\partial t}\left( A \text{e}^{\text{i}(kx - \omega t)} \right) \\
& = - \hbar \omega A \text{i}^2 \text{e}^{\text{i}(kx - \omega t)} \\
& = \hbar \omega \Psi_\text{free}(x,t) \end{split}
Thus the free-particle wave function \Psi_\text{free}(x,t) = A \text{e}^{\text{i}(kx - \omega t)} is an eigenfunction of \text{i} \hbar \partial / \partial t and the corresponding eigenvalue is the energy, \hbar \omega, associated with this wave.
3 Setting the scene in quantum physicsIn this section, some quantum physics needed to study quantum computing is introduced. You will learn about spin-½ particles, which have two fundamental basis states but can also exist in a superposition of these states. This concept is central to quantum mechanics and directly relates to quantum computing, where quantum bits (qubits) similarly have two basis states and can exist in any superpositions of these two states. In the context of spin-½ particles the two states are called spin-up and spin-down; in quantum computing the two states are referred to as logical states. At the end of this section, the essential concept for quantum computing of entanglement is illustrated using spin-states.
3.1 Spin-½ particlesExperiments show that electrons have an intrinsic property which is called spin. (Mass and charge are other examples of intrinsic properties of particles.) Spin is a type of angular momentum with a quantum number of ½ which means that a measurement of spin along an axis can only have values of +\hbar/2 or -\hbar/2 as an outcome. (Here \hbar = h/2 \pi where h is Planck’s constant, 6.626 × 10-34 J s.) Thus, electrons are referred to as spin-½ particles. The most important spin operators are the component of spin angular momentum in the z-direction, \mathrm{\widehat{S}}_z and the total spin angular momentum, \mathrm{\widehat{S}^2}. The eigenvalue equations for these operators are \begin{align*}
\mathrm{\widehat{S}}^2 | S,M_\text{s} \rangle & = S(S+1) \, \hbar^2 \, | S,M_\text{s} \rangle, \\
\mathrm{\widehat{S}}_z | S,M_\text{s} \rangle & = M_\text{s} \, \hbar \, |S,M_\text{s} \rangle.
\end{align*} where the angled bracket, | ~~ \rangle is used and is called a ket. | S, M_\text{s} \rangle represents the eigenfunction, known as an eigenstate with a spin quantum number of S and a spin magnetic quantum number of M_\text{s}. For a single electron S=\frac{1}{2} and M_\text{s}=\pm \frac{1}{2}. The state with M_\text{s}=+\frac{1}{2} is the spin-up state and the state with M_\text{s}=-\frac{1}{2} is the spin-down state.3.2 Representing a general spin stateA ket can also be considered as a vector. The spin-up state is often represented by the symbol | \uparrow \rangle and the spin-down state by the symbol | \downarrow \rangle. The spin-up state obeys \mathrm{\widehat{S}}_z | \uparrow \rangle = +\frac{\hbar}{2} | \uparrow \rangleand the spin-down state obeys \mathrm{\widehat{S}}_z | \downarrow \rangle = -\frac{\hbar}{2} | \downarrow \rangle.The spin vectors are normalised and orthogonal to one another, as represented by the following relations:\begin{split}
\langle {\uparrow}|{\uparrow} \rangle & = \langle {\downarrow}|{\downarrow} \rangle =1 \quad \text{(normalized)}\\
\langle {\uparrow}|{\downarrow} \rangle &= \langle {\downarrow}|{\uparrow} \rangle =0 \quad \text{(orthogonal)} \end{split}where these equations use additional notation. \langle ~~ | is known as a bra and the combination of the bra and ket together is an inner product, \langle ~~ | ~~ \rangle.A general spin state | A \rangle can be written as a linear combination of | \uparrow \rangle and | \downarrow \rangle, thus| A \rangle = a_1 | \uparrow \rangle + a_2 | \downarrow \ranglewhere a_1 and a_2 are complex numbers referred to as probability amplitudes.In other words, | \uparrow \rangle and | \downarrow \rangle provide an orthonormal basis for spin space. The vectors | \uparrow \rangle and | \downarrow \rangle are called basis vectors.
Because any spin state | A \rangle can be written as a linear combination of just two basis vectors, the spin space of a spin-½ particle is two-dimensional.For an atom in any spin state, | A \rangle, as given in Equation 9 the probability of the outcome of a measurement indicating spin-up is | a_1 |^2 and for spin-down is |a_2|^2. Since these are the only possible outcomes, the corresponding probabilities must sum to one, therefore |a_1|^2+|a_2|^2=1.Matrices can be used as an alternative representation of spin states to simplify calculations. | \uparrow \rangle and | \downarrow \rangle are represented by the following column vectors:| \uparrow \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \text{and} \quad | \downarrow \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix} These matrices have two elements because spin space is two-dimensional. Spin states that do not have definite values of S_z are expressed as linear combinations of | \uparrow \rangle and | \downarrow \rangle.Any vector | A \rangle in spin space may be written as a linear combination of | \uparrow \rangle and | \downarrow \rangle. This means that | A \rangle, as defined in Equation 9 becomes:| A \rangle = a_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + a_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} In this way any spin state of a spin-½ particle can be represented as a two-element matrix, which is called a spinor.The inner product in spin space of two vectors in matrix form is written\langle {A}|{B} \rangle =a_1^* b_1 + a_2^* b_2, which is consistent with the matrix multiplication: a_1^* b_1 + a_2^* b_2 = \begin{bmatrix} a_1^* & a_2^* \end{bmatrix} \begin{bmatrix} {b_1}\\ {b_2} \end{bmatrix},and thus the inner product of two spin vectors can be written in the matrix form:\langle A | B \rangle = \begin{bmatrix} a_1^* & a_2^* \end{bmatrix} \begin{bmatrix} {b_1}\\ {b_2} \end{bmatrix}.We can therefore identify the separate bra and ket vectors as follows:\langle A | = \begin{bmatrix} a_1^* & a_2^* \end{bmatrix} \quad \text{and} \quad | B \rangle = \begin{bmatrix} {b_1}\\ {b_2} \end{bmatrix}.A ket spin vector is represented by a column spinor, and a bra spin vector is represented by a row spinor. To convert a column spinor into the corresponding row spinor, the rule is to turn the column into a row and take the complex conjugate of all elements. So\text{if} \quad |A \rangle = \begin{bmatrix} {a_1}\\ {a_2} \end{bmatrix}, \quad \text{then} \quad \langle A | = \begin{bmatrix} a_1^* & a_2^* \end{bmatrix}.When using a matrix representation, the spin operators are also represented as matrices, for example,\mathrm{\widehat{S}}_z = \frac{\hbar}{2}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}Rewriting Equation 7 using matrices gives, \mathrm{\widehat{S}}_z \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{\hbar}{2} \begin{bmatrix} 1 \\ 0 \end{bmatrix} where you can see that carrying out the matrix multiplication gives the expected result.The operators for a spin-½ particle are each represented by 2 \times 2 matrices. Along the three axes these are: \mathrm{\widehat{S}}_x = \frac{\hbar}{2}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad \mathrm{\widehat{S}}_y = \frac{\hbar}{2} \begin{bmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{bmatrix}, \quad \mathrm{\widehat{S}}_z = \frac{\hbar}{2} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}It is common to define the so called Pauli operators, \widehat{\sigma}, as \mathrm{\widehat{S}} = \frac{\hbar}{2} \widehat{\sigma} such that we have the following Pauli operator matrices:\widehat{\sigma}_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad \widehat{\sigma}_y = \begin{bmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{bmatrix}, \quad \widehat{\sigma}_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}These will be useful in the context of quantum computing where they can be used to represent the action of quantum gates.Exercise 8Show that the spin vectors| U \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \quad \text{and} \quad | V \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} -1 \\ 1 \end{bmatrix}are normalized and orthogonal, i.e. \langle U | U\rangle = \langle V | V \rangle =1 and \langle U | V \rangle = 0.Using matrix multiplication\langle U | U \rangle = \frac{1}{2} \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{2} ( 1 + 1) = 1 \langle V | V \rangle = \frac{1}{2} \begin{bmatrix} -1 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \frac{1}{2} ( 1 + 1) = 1 \langle U | V \rangle = \frac{1}{2} \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \frac{1}{2} ( -1 + 1) = 0 The first two equations show that the vectors |U \rangle and | V \rangle are normalised; the third that they are orthogonal to each other. Hence, they are orthonormal.3.3 Spin observablesIn quantum mechanics measurable quantities are called observables. Spin is an example of an observable because it can be measured in an experiment. (Position and orbital angular momentum are other examples of observables.) Each observable is associated with an operator and, in general, the only possible outcomes of a measurement of an observable are any of the eigenvalues. When a measurement is performed on a quantum system with spin, the wavefunction collapses into one of the eigenstates of the observable being measured. For example, if we measure the spin of an electron along the z-axis, the quantum state collapses into one of the two basis states: | \uparrow \rangle or | \downarrow \rangle with probabilities determined by the initial state before measurement (see Equation 9). This collapse means and any superposition that existed before the measurement is lost. After the measurement, the electron will be in a new well-defined spin state, either spin-up or spin-down depending on the result of the measurement. The general spin state has collapsed into one of the eigenstates due to being measured. As long as the initial general spin state is not an eigenstate, the spin state after the measurement will be different from the spin state before the measurement.Exercise 9Particles are prepared in the spin state| A \rangle = \frac{\sqrt{3}}{2}| \uparrow \rangle + \frac{1}{2}| \downarrow \rangleIf a single particle is prepared in the state | A \rangle, what prediction can be made about the result of measuring S_z for this particle?If a million particles are prepared identically, all in the state | A \rangle, what prediction can be made about the results of measuring S_z for this collection of particles?No definite prediction can be made for a single particle in the given state, but a measurement of S_z will give either+\hbar/2or-\hbar/2; see Equations 7 and 8. In given state | A \rangle and using Equation 9, a_1 = \sqrt{3}/2 and a_2 = 1/2 so the probability of getting +\hbar/2 is (\sqrt{3}/2)^2 = 3/4 and the probability of getting -\hbar/2 is (1/2)^2 = 1/4. As expected these two probabilities sum to unity because for any measurement either one or the other outcome will be obtained. This shows that the value +\hbar/2 is more likely, but the value -\hbar/2 would not be that surprising.For a million particles, the expected outcome is that close to three-quarters or 750,000 measurements will give S_z = +\hbar/2, and the remainder will give S_z = -\hbar/2.3.4 Two-particle spin statesIf we have two indistinguishableParticles are indistinguishable when they are identical (i.e. they have the same intrinsic properties like mass, charge, and spin) and they are so close together that their wavefunctions overlap so that we cannot tell them apart, even in principle. electrons, we can define a two-particle spin state. Due to symmetry and the rules of quantum mechanical addition of angular momentum, there are four possible spin states in total. These spin states are represented using the quantum numbers S and M_\text{s}, as introduced in Section 3.1, but now the quantum numbers are the sum of the values for the individual electrons. Therefore the spin quantum number is S= \frac{1}{2} + \frac{1}{2} = 1 or S= \frac{1}{2} - \frac{1}{2} = 0 and, for the case when S=1, the spin magnetic quantum is M_\text{s} = \pm \frac{1}{2} \pm \frac{1}{2} = -1 \, \text{or} \, 0 \, \text{or} \, +1 , while for the case when S=0, we only have M_\text{s} = + \frac{1}{2} - \frac{1}{2} = 0.Such a two-particle spin state therefore can only have an overall spin function which is either symmetric or antisymmetric with respect to exchange of the electrons. The symmetric spin state is referred to as a triplet because there are three possible symmetric combinations:\begin{split} | 1, 1 \rangle & = | \uparrow \uparrow \rangle \\
| 1, 0 \rangle & = \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle ) \\
| 1, -1 \rangle & = | \downarrow \downarrow \rangle \end{split} where the first arrow in each ket refers to particle 1 and the second to particle 2. The antisymmetric spin state is referred to as a singlet because there is only one possible combination:| 0, 0 \rangle = \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle )You can see that the states | \uparrow \uparrow \rangle and | \downarrow \downarrow \rangle can be factorised into | \uparrow \rangle_1 | \uparrow \rangle_2 and | \downarrow \rangle_1 | \downarrow \rangle_2 respectively, where the subscripts label each particle. In contrast, the states \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle ) and \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle ) cannot be factorised into the product of a particle 1 state multiplied by a particle 2 state. Two-particle states which cannot be factorised are known as entangled states and said to exhibit entanglement.
Exercise 10Verify that the three spin kets \begin{split} | 1, 1 \rangle & = | \uparrow \uparrow \rangle , \\
| 1, 0 \rangle & = \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle ), \\
| 1, -1 \rangle & = | \downarrow \downarrow \rangle \end{split}are symmetric with respect to swapping the labels of the particles.Starting with\frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle ) = \frac{1}{\sqrt{2}} ( | \uparrow \rangle_1 | \downarrow \rangle_2 + | \downarrow \rangle_1 | \uparrow \rangle_2 ), exchanging the particle labels and then rearranging gives\begin{split} \frac{1}{\sqrt{2}} ( | \uparrow \rangle_2 | \downarrow \rangle_1 + | \downarrow \rangle_2 | \uparrow \rangle_1 )
& = \frac{1}{\sqrt{2}} ( | \downarrow \rangle_1 | \uparrow \rangle_2 + | \uparrow \rangle_1 | \downarrow \rangle_2 ) \\
& = \frac{1}{\sqrt{2}} ( | \uparrow \rangle_1 | \downarrow \rangle_2 + | \downarrow \rangle_1 | \uparrow \rangle_2 ) \\
& = \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle ). \end{split}Since this final expression is identical to the initial expression, this shows \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle ) is symmetric to swapping particle labels.For | \uparrow \uparrow \rangle = | \uparrow \rangle_1 | \uparrow \rangle_2 and | \downarrow \downarrow \rangle = | \downarrow \rangle_1 | \downarrow \rangle_2 the particle labels are interchanged and re-ordered (perfectly acceptable!) to get the same expressions as required.3.5 EntanglementEntanglement is a consequence of quantum theory and leads to a correlation between the outcomes of measurements which cannot be explained by classical physics. Two-particle states which show such non-classical correlations are known as entangled states. Consider the following thought experiment. Suppose you have a two-particle system in the spin state |0,0\rangle as given in Equation 10. Before the experiment you know that particle 1 can be either spin-up or spin-down with equal probability. However, if you measure particle 1 to be spin-up then you know that particle 2 is spin-down as the measurement means that the two-particle state has collapsed into the | \uparrow \downarrow \rangle arrangement of spins. In contrast, if you measure particle 1 to be spin-down then you know that particle 2 is spin-up as the two-particle state has collapsed into the | \downarrow \uparrow \rangle arrangement of spins. This type of prediction is quite puzzling because the two entangled particles can be as far apart as possible and when a measurement is made on particle 1 then it is known simultaneously what the outcome of a measurement on particle 2 will be.Entanglement is essential for quantum computing. Entangled states are generated as part of the workings of a quantum computer, as you will see later.Exercise 11 Confirm that the following states are normalised and determine whether the states are entangled.| B \rangle = \frac{1}{\sqrt{2}} | \uparrow \downarrow \rangle - \frac{1}{\sqrt{2}} | \uparrow \uparrow \rangle| C \rangle = \frac{1}{2} | \uparrow \uparrow \rangle - \frac{1}{\sqrt{2}} | \uparrow \downarrow \rangle - \frac{\text{i}}{2} | \downarrow \downarrow \rangleThe normalisation condition for a general two-particle state is that the sum of the squares of the probability amplitudes is equal to 1.The states are entangled if the two-particle state cannot be factorised into the product of a particle 1 state multiplied by a particle 2 state.Checking the normalisation of state | B \rangle: \left|\frac{1}{\sqrt{2}}\right|^2 + \left|-\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2} + \frac{1}{2} = 1,showing state | B \rangle is normalised. | B \rangle can also be factorised:\begin{split} | B \rangle & = \frac{1}{\sqrt{2}} | \uparrow \downarrow \rangle - \frac{1}{\sqrt{2}} | \uparrow \uparrow \rangle \\
& = \frac{1}{\sqrt{2}} ( | \uparrow \rangle_1 | \downarrow \rangle_2 - | \uparrow \rangle_1 | \uparrow \rangle_2 )\\
& = \frac{1}{\sqrt{2}} | \uparrow \rangle_1 \, ( | \downarrow \rangle_2 - | \uparrow \rangle_2) \end{split}which is a particle 1 state multiplied by a particle 2 state so state | B \rangle is not entangled.Checking the normalisation of state | C \rangle:\left|\frac{1}{2}\right|^2 + \left|-\frac{1}{\sqrt{2}}\right|^2 + \left| \frac{\text{i}}{2} \right|^2 = \frac{1}{4} + \frac{1}{2} + \frac{1}{4} = 1,showing state | C \rangle is normalised. State | C \rangle cannot be factorised and so is an entangled state. 4 Classical computingIn this section, you will be introduced to a few aspects of classical computing to give you a reference frame for discussing quantum computing. All computers work by taking input information and processing it using gates to give output information. By the end of the section, you will be familiar with the classical NOT and CNOT gates including their truth tables, so that you can compare them with the non-classical output of the quantum versions of these gates described in Section 5.
4.1 Classical bits and logic gatesIn classical computing, the smallest piece of information is called a bit. A bit may take only one of two logical values: either 0 or 1. Strings of bits are used to represent information as numbers, which can be stored, copied and processed by the computer. The processing of information is accomplished by logic gates, which take strings of bits as their input and produce an output value for each bit that they act on. A diagram showing how a logic gate works is given in Figure 4.
Figure 4 A diagram to illustrate how a logic gate works
The figure shows a schematic of a logic gate with a string of binary numbers as input and a different string of binary numbers as output.4.2 Classical Boolean gatesA single-bit gate acts on one bit at a time. In classical computing, there are two universal single-bit gates, the NOT gate and the Reset gate which, either acting alone or in a sequence, can generate all possible transformations of a single bit. The NOT gate (Figure 5) simply flips the value of the bit to the alternative value, so a 0 becomes a 1, and a 1 becomes a 0.
Figure 5 The symbol for a NOT gate
The figure shows a horizontal line labelled as input leading to the base of a right-pointing equilateral triangle. At the right hand tip of the triangle is a small circle, followed by another horizontal line labelled as output.A truth table is a convenient way of summarising the action of a gate. The truth table for a NOT gate is given in Table 1.
Table 1 Classical NOT gate truth table
Input
Output
0
1
1
0
The Reset gate sets a bit to value 0, regardless of the input state.Single-bit gates are not sufficient to perform computing: it is also necessary have conditional gates in which an operation on a target bit depends on the state of one or more control bits. It will be helpful to write the two input bits as an ordered pair of values, CT (for example, 01 has C = 0 and T = 1), where C represents the control bit and T the target bit. An important two-bit gate is the CNOT gate (controlled NOT gate), which performs a NOT operation on the target bit conditional on the state of the control bit being 1; if the control bit is 0, then no operation is applied. The state of the control bit is unchanged by the CNOT operation. Table 2 is the truth table for the CNOT gate.
Table 2 Classical CNOT gate truth table
Input
Output
0 0
0 0
0 1
0 1
1 0
1 1
1 1
1 0
To perform classical computing, the bits are set to some initial values and the gates are applied to the bits in an ordered sequence to make an algorithm. Quantum computing has different sets of universal gates, which include quantum versions of the NOT and CNOT gates.5 Qubits and quantum gatesQuantum computing is based on units of information called qubits (quantum bits, and pronounced kew-bits), which obey the laws of quantum mechanics.
A qubit is the quantum analogue of a classical bit. The classical bit values 0 and 1 are replaced by the orthonormal basis states of the quantum-mechanical qubit |0 \rangle and | 1 \rangle. The basis states are given the name logical states, since they correspond to the classical bits upon which the logic gates operate. The key difference between qubits and classical bits is that qubits can exist in a superposition of the |0 \rangle and |1 \ranglestates, which means qubits can be prepared in the superposition state:| \psi \rangle = a_0 | 0 \rangle + a_1 | 1 \rangleRemarkably, you can take logic gates similar to the Boolean logic gates of classical computing and apply them to the qubits. In doing so, the input state of the qubits is transformed into the output state. To obtain the result of the computation, the value (0 or 1) of each qubit is measured. As you will see, the resulting outputs can include entangled states of two or more qubits.Quantum entanglement is a fundamental resource for quantum computing as it involves the distribution of information in a fundamentally non-classical way.In this section, you will learn the definition of a qubit and be introduced to some single-qubit and two-qubit logic gates. The quantum CNOT gate is an important gate as it can entangle and disentangle a pair of qubits. By the end of this section you will have been introduced to quantum circuits and there is a final activity to test your understanding.5.1 Defining a qubitA qubit is defined by the equation| \psi \rangle = a_0 | 0 \rangle + a_1 | 1 \ranglewhere | \psi \rangle is a two-state quantum system and |0 \rangle and |1 \rangle are logical states. Note that |0 \rangle has a logical value of 0 and |1 \rangle has a logical value of 1. a_0 and a_1 are the probability amplitudes which may be complex numbers and satisfy the normalisation condition |a_0|^2 + |a_1|^2 = 1.Examples of qubits are the general spin states of spin-½ particles as described in Section 3.2. You can see that the spin-up state has been replaced by logical state |0 \rangle and the spin-down state by logical state |1 \rangle and you can see that Equation 9 and Equation 11 have a similiar form.To fully specify the two complex amplitudes or four real numbers are required: the real and imaginary parts of each. However, the number of values can be reduced by two, one because of the normalisation condition, and one because the phase of one basis state can be set to zero without changing anything.This leads to the equation, | \psi \rangle = \cos(\theta/2)\, | 0 \rangle + \sin(\theta/2)\, \mathrm{e}^{\text{i} \phi} | 1 \rangle.where \theta and \phi are real numbers with 0\leq\theta\leq \pi and 0\leq\phi\leq 2\pi.A qubit state can therefore also be represented as a column vector | \psi \rangle = \begin{bmatrix} {\cos(\theta/2)} \\ {\sin(\theta/2)\, \mathrm{e}^{\text{i} \phi}} \end{bmatrix}.This representation is like the spinor representation introduced in Section 3.2. The column vector representation is useful because the operators corresponding to single-qubit gates and observables may be written as 2 × 2 matrices.
The qubit basis states are defined as|0 \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \text{and} \quad |1 \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix} The column vectors of Equations 12 are eigenstates of a 2 × 2 matrix operator known as the Pauli-Z operator \widehat{\sigma}_z or \widehat{\mathrm{Z}}, which as mentioned earlier is defined as\widehat{\sigma}_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}Exercise 12 Show that the qubit basis states |0 \rangle and |1 \rangle are eigenvectors of the Pauli-Z operator and find the corresponding eigenvalues. Determine the relationship between the eigenvalues and the logical values of the basis states. Use the symbol, m to represent the logical value.Noting | 0 \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix} and has logical value, m = 0; and | 1 \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix} and has logical value, m = 1 and that \widehat{\sigma}_z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}First, for basis state, |0 \rangle the eigenvalue equation (Equation 2) becomes\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \lambda_0 \begin{bmatrix} 1 \\ 0 \end{bmatrix}Doing the matrix multiplication gives\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \lambda_0 \begin{bmatrix} 1 \\ 0 \end{bmatrix}showing that \lambda_0 = 1Similarly, for basis state |1 \rangle\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ -1 \end{bmatrix} = -\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \lambda_1 \begin{bmatrix} 0 \\ 1 \end{bmatrix}showing that \lambda_1 = -1Comparing the eigenvalues with the logical values: when \lambda_0 = 1, m = 0, and when \lambda_1 = -1, m = 1. The relationship between the eigenvalue and the logical value is thereforem = \frac{1 - \lambda_m}{2}5.2 Single qubit gatesIn quantum computing, a gate is a reversible transformation of a qubit state | \Psi \rangle to another qubit state | \Phi \rangle, represented by an operator \widehat{\mathrm{U}}. It is convenient to write the single-qubit gate operators as 2 × 2 matrices. Therefore, the action of the gate \widehat{\mathrm{U}} can be written as follows:\widehat{\mathrm{U}} | \Psi \rangle = | \Phi \rangle.The word reversible is important because it is a reminder that a gate operation is of a different nature from a measurement. The operation of the gate can be reversed so that it is possible to get back to the state \Psi, whereas, in general, a measurement makes an irreversible change to the qubit state. \widehat{\mathrm{U}}^{\dagger} is defined as the operator needed to reverse the gate action and transform | \Phi \rangle back to | \Psi \rangle:\widehat{\mathrm{U}}^{\dagger} | \Phi \rangle = | \Psi \rangle.therefore\widehat{\mathrm{U}}^{\dagger} \widehat{\mathrm{U}} | \Psi \rangle = | \Psi \rangle,which means that\widehat{\mathrm{U}}^{\dagger} \widehat{\mathrm{U}} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \widehat{\mathrm{I}}where \widehat{\mathrm{I}} is the identity operator, represented by the 2 × 2 matrix appearing in Equation 13. An operator, \widehat{\mathrm{U}} that obeys Equation 13 is called a unitary operator, hence gates are represented by unitary operators. The identity operator is itself a gate, denoted \widehat{\mathrm{I}}, and its symbol is shown in Figure ‘6.
Figure 6 The symbol used in a quantum circuit for an identity gate, \widehat{\mathrm{I}}
The figure shows a horizontal line on the left leading to a square box with the capital letter I in it, then a horizontal line leading from the box to the right. You will now look at some gates, starting with the quantum NOT gate. (From now on the prefix quantum will be omitted as long as it is obvious the gates are quantum gates and not classical gates from the context.)
5.2.1 The NOT gateThe NOT gate is denoted by the operator, \widehat{\mathrm{X}} and, in the basis of the logical qubits |0 \rangle and |1 \rangle, is represented by the matrix\widehat{\mathrm{X}} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.which is also the Pauli-X operator, \widehat{\sigma}_x, mentioned earlier. In a circuit diagram representing a quantum algorithm, the symbol for the NOT gate is shown in Figure 7.
Figure 7 The symbol used in a quantum circuit for a NOT gate, \widehat{\mathrm{X}}
The figure shows a horizontal line on the left leading to a circle with a large plus sign in it, then a horizontal line leading from the box to the right. In quantum computing, the truth table specifies outcomes for the logical states |0 \rangle and |1 \rangle. In Table 3 the general superposition state is also included for reference.
Table 3 Quantum NOT gate truth table
Input
Output
|0 \rangle
|1 \rangle
|1 \rangle
|0 \rangle
a_0 |0 \rangle + a_1 | 1 \rangle
a_0 |1 \rangle + a_1 | 0 \rangle
5.2.2 The Hadamard gate The Hadamard gate is a single-qubit gate defined by a matrix \widehat{\mathrm{H}}\widehat{\mathrm{H}} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}.In a circuit diagram representing a quantum algorithm, the symbol for the Hadamard gate is shown in Figure 8.
Figure 8 The symbol used in a quantum circuit for a Hadamard gate, \widehat{\mathrm{H}}
The figure shows a horizontal line on the left leading to a square box with a capital letter H in it, then a horizontal line leading from the box to the right. Consider the action of a Hadamard gate on logical state | 0 \rangle. First, the gate and logical state are written as matrices,\widehat{\mathrm{H}} | 0 \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix}Next, the matrices are multiplied to obtain the final state matrix\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}Finally, the final state matrix is rewritten in terms of the logical states | 0 \rangle and | 1 \rangle:\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} ( | 0 \rangle + | 1 \rangle )You can see that the final output state, \frac{1}{\sqrt{2}} ( | 0 \rangle + | 1 \rangle ) is a superposition state. This calculation shows that a Hadamard gate allows the transformation of the logical qubit state into a superposition state.Exercise 13Use matrices to work out the action of a Hadamard gate on logical state | 1 \rangle.Noting, logical state | 1 \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix} and the Hadamard gate is \widehat{\mathrm{H}} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}. Using matrices gives\begin{split} \widehat{\mathrm{H}} | 1 \rangle & = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \\
& = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 0 \end{bmatrix} - \frac{1}{\sqrt{2}} \begin{bmatrix} 0 \\ 1 \end{bmatrix} \\
& = \frac{1}{\sqrt{2}} ( | 0 \rangle - | 1 \rangle ) \end{split}You can see that the Hadamard gate has transformed logical state | 1 \rangle into a superposition state.The effect on a general state can be determined by combining the results of the action of a Hadamard gate on logical states | 0 \rangle and | 1 \rangle. \begin{split} \widehat{\mathrm{H}} ( a_0 | 0 \rangle + a_1 | 1 \rangle ) & = a_0 \widehat{\mathrm{H}} | 0 \rangle + a_1 \widehat{\mathrm{H}} | 1 \rangle \\
& = \frac{a_0}{\sqrt{2}} ( | 0 \rangle + | 1 \rangle ) + \frac{a_1}{\sqrt{2}} ( | 0 \rangle - | 1 \rangle ) \\
& = \frac{1}{\sqrt{2}} ( a_0 + a_1) | 0 \rangle + \frac{1}{\sqrt{2}} ( a_0 - a_1) | 1 \rangle \end{split}All these results are summarised in the truth table for the Hadamard gate in Table 4.
5.2.3 Sequences of gatesThe next step towards quantum computing is to combine gates sequentially to give a quantum circuit. In a circuit, successive operations are applied to a single qubit.If you first apply the NOT gate, and then apply the Hadamard gate, the mathematical expression representing this is written as a sequence of operators, operating from right to left on the qubit | \psi \rangle:| \phi \rangle = \widehat{\mathrm{H}}\widehat{\mathrm{X}} | \psi \rangleThis ordering of operators should not come as a surprise; reading the expression on the right hand side of Equation 14, first gate \widehat{\mathrm{X}} is applied to qubit | \psi \rangle to give an intermediate qubit, say | \alpha \rangle, and then gate \widehat{\mathrm{H}} is applied to | \alpha \rangle to give the final resultant | \phi \rangle. Alternatively, if calculating the outcome of
\widehat{\mathrm{H}}\widehat{\mathrm{X}} on qubit | \psi \rangle, the matrices representing \widehat{\mathrm{H}} and \widehat{\mathrm{X}} can be multiplied together to give a resultant matrix, which can be considered a new operator, \widehat{\mathrm{W}} = \widehat{\mathrm{H}} \widehat{\mathrm{X}}. Then operator, \widehat{\mathrm{W}} can be thought to act on | \psi \rangle to give the resultant | \phi \rangle.The diagram representing this sequence is shown in Figure 9.
Figure 9 A circuit for applying a NOT gate and a Hadamard gate in sequence
The figure shows a horizontal line on the left leading to a circle with a large plus sign in it, then a further horizontal line from the right of the circle leading to a square box with a capital letter H in it, then a horizontal line leading from the box to the right. A circuit diagram represents the logical flow of the circuit from the left (the initial state) to the right (the final state) of the diagram. Therefore, in the circuit shown in Figure 8, the elements are ordered from left to right: the NOT, which acts first, is on the left. Note that the ordering of gates in the circuit diagram is the opposite to the ordering of gates when the circuit is written as a sequence of operators acting on a ket.By matrix multiplication, any sequence of single-qubit gates can be represented by a single 2 × 2 matrix found by forming an ordered product of the matrices representing the gates. You have to be careful because, in general, the single-qubit operators do not commute. In the next exercise you will see that the method using matrix multiplication to combine gates is equivalent to applying the gates sequentially.Exercise 14 Given the initial qubit state | 0 \rangle, show that the sequence of gates in Equation 14 produces the final state| \psi \rangle = \frac{1}{\sqrt{2}} ( | 0 \rangle - | 1 \rangle ).Show that you get the same result by applying either the gates in sequence or by using matrix multiplication to combine gates. Applying the gates in sequence from right to left:\begin{split} | \psi \rangle & = \widehat{\mathrm{H}} \widehat{\mathrm{X}} | 0 \rangle \\
& = \widehat{\mathrm{H}} | 1 \rangle \quad \text{by definition of a NOT gate} \\
& = \frac{1}{\sqrt{2}} (| 0 \rangle - | 1 \rangle ) \quad \text{using row 2 of Table 4} \end{split}Using matrix multiplication to combine gates: First, calculate the product of the matrices \widehat{\mathrm{H}} \widehat{\mathrm{X}}:\widehat{\mathrm{H}} \widehat{\mathrm{X}} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} then apply this matrix to the column vector for | 0 \rangle:\begin{split} | \psi \rangle & = \widehat{\mathrm{H}} \widehat{\mathrm{X}} | 0 \rangle \\
& = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} \\
& = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \\
& = \frac{1}{\sqrt{2}} ( | 0 \rangle - | 1 \rangle ) \end{split}which is the same final state as applying the gates in sequence, as expected.5.3 Two-qubit gatesThe gates discussed in Section 5.2 are examples of single-qubit gates, but these are insufficient for quantum computing. Just as the classical computing must include conditional gates that act on two bits, so a working quantum computer needs both single-qubit and two-qubit gates. A quantum CNOT gate is introduced and you will see that it is able to produce entangled states of two qubits.5.3.1 Two-qubit statesA straightforward way to define the two-qubit states is to build the two-qubit basis states from product states:| \Psi \rangle = | q_1 \rangle | q_2 \rangle = | q_1 q_2 \rangle.There are four possible product states of the usual single-qubit basis states:|00 \rangle, |01 \rangle, |10 \rangle, | 11 \rangleA general two-qubit state, therefore can be expressed in terms the product states as | \Psi \rangle = a_{00} |00 \rangle + a_{01} |01 \rangle + a_{10} |10 \rangle + a_{11} |11 \rangle, where | \Psi \rangle is normalised in the usual way:| a_{00}|^2 + | a_{01}|^2 + | a_{10}|^2 + | a_{11}|^2 =1To specify the state of the two-qubit system, six real numbers must be given. Counting them in the same way as for a single qubit, these are the eight numbers specifying the real and imaginary parts of each complex number a_{mn} (where the indices are mn = 00, 01, 10, 11). Eight is reduced by one because of the normalisation condition, given in Equation 15 and by one more, to six because the phase of one of the basis states can be set to zero without changing anything physical. In general, an n-qubit system requires (2^{n+1} - 2) real numbers to specify the state, which is an exponential scaling in the number of qubits.If the two-qubit gate \widehat{\mathrm{G}} is an operation, an equation can be written which represents the transformation from a two-qubit state | \Psi \rangle into a new two-qubit state | \Phi \rangle: \widehat{\mathrm{G}} | \Psi \rangle = | \Phi \rangle = b_{00} |00 \rangle + b_{01} |01 \rangle + b_{10} |10 \rangle + b_{11} |11 \rangle.Multiple qubits are represented using the tensor product, which combines their states into a larger system. The resulting matrices represent the full system and can be used to calculate outputs by multiplying them with quantum state vectors. For instance the example above can be written as\begin{bmatrix} g_{00} & g_{01} & g_{02} & g_{03} \\ g_{10} & g_{11} & g_{12} & g_{13} \\ g_{20} & g_{21} & g_{22} & g_{23} \\ g_{30} & g_{31} & g_{32} & g_{33} \end{bmatrix} \begin{bmatrix} a_{00} \\ a_{01} \\ a_{10} \\ a_{11} \end{bmatrix} = \begin{bmatrix} b_{00} \\ b_{01} \\ b_{10} \\ b_{11} \end{bmatrix} This representation shows that a two-qubit gate can be expressed as a 4 × 4 matrix with elements g_{ij}, which act on a 4 × 1 column vector representing the quantum state. This allows you to compute the output state using matrix multiplication. However, we won’t be using this formalism in our discussions, as we will focus on other intuitive approaches to understanding multi-qubit systems and gates.5.3.2 How the CNOT gate worksIn the same way as a classical CNOT gate, described in Section 4.2, acts on two bits, a control bit and a target bit; the quantum CNOT gate also acts on two qubits, a control qubit and a target qubit. The CNOT gate, represented by an operator \widehat{\mathrm{CX}}_{C,T}, acts on a target qubit | \phi_T \rangle depending
on the state of a control qubit | \psi_C \rangle. Note that the single ‘hat’ over the \widehat{\mathrm{CX}} operator tells you that this is one operator in contrast to the sequential operators, e.g. \widehat{\mathrm{H}}\widehat{\mathrm{X}} as in Equation 14, which are two operators. In the following, the two-qubit state is assumed to be the product | \psi \phi \rangle = | \psi \rangle_C | \phi \rangle_T . The quantum CNOT gate follows the same rules as the classical CNOT gate: if the state of the control qubit is | 0 \rangle, then it leaves the target qubit unchanged. If the state of the control qubit is | 1 \rangle, then it applies the NOT gate to the target qubit. Thus the CNOT gate would act on the state | 00 \rangle as follows:\widehat{\mathrm{CX}}_{C,T} | 00 \rangle = | 00 \rangle \quad \text{(state of the target is unchanged)}and on the state | 1 0 \rangle as follows:\widehat{\mathrm{CX}}_{C,T} | 10 \rangle = | 11 \rangle \quad \text{(state of the target is flipped)}The transformations on the kets | 01 \rangle and | 11 \rangle can be worked out in the same way. The results of these transformations are collected in the truth table in Table 6 and are identical to the classical rules given in Table 2. The quantum CNOT gate, however, can also act on superposition states, which is completely beyond the capabilities of the classical CNOT gate. So now consider how the CNOT gate transforms superposition states, starting from the situation where the control state is prepared in the superposition state| \psi \rangle_C = \frac{1}{\sqrt{2}} ( | 0 \rangle_C + | 1 \rangle_C )and the target qubit is in the state | 0 \rangle_T. First, here is the initial two-qubit state:\begin{split} | \Psi \rangle & = \frac{1}{\sqrt{2}} ( | 0 \rangle_C + | 1 \rangle_C )| 0 \rangle_T \\
& = \frac{1}{\sqrt{2}} ( | 0 \rangle_C | 0 \rangle_T + | 1 \rangle_C | 0 \rangle_T ) \\
& = \frac{1}{\sqrt{2}} ( | 0 0 \rangle + | 1 0 \rangle ) \end{split}.Then applying the CNOT operator gives:\begin{split} \widehat{\mathrm{CX}}_{C,T} | \Psi \rangle & = \widehat{\mathrm{CX}}_{C,T} \frac{1}{\sqrt{2}} ( | 00 \rangle + | 10 \rangle ) \\
& = \frac{1}{\sqrt{2}} \left( \widehat{\mathrm{CX}}_{C,T} | 00 \rangle + \widehat{\mathrm{CX}}_{C,T} | 10 \rangle \right) \\
& = \frac{1}{\sqrt{2}} ( | 00 \rangle + | 11 \rangle ) \end{split}. You can see that the final state is an entangled state because it cannot be factorised. If the control is in the superposition state orthogonal to the state described in Equation 16, i.e. | \psi \rangle_C = ( |0 \rangle - | 1 \rangle ) / \sqrt{2}, then the negative sign simply propagates so that:\widehat{\mathrm{CX}}_{C,T} \frac{1}{\sqrt{2}} ( | 00 \rangle - | 10 \rangle ) = \frac{1}{\sqrt{2}} ( | 00 \rangle - | 11 \rangle ).The quantum CNOT gate is depicted graphically in Figure 10 and the full CNOT truth table including the quantum-mechanical results is given in Table 6.
Figure 10 The symbol for the quantum CNOT gate, with the control (C) and target (T) qubits labelled
The figure shows two parallel horiontal lines. The upper one is labelled C and the lower one is labelled T. In the middle of the upper line is a large dot. A vertical line leads from this dot to the lower line where it joins a large circle with a large plus sign in it.
There are other useful states that can be generated using a CNOT gate; another is introduced in the next exercise.Exercise 15Find the two-qubit output state produced by the CNOT operation if the control qubit is prepared in the state | \psi \rangle_C = (| 0 \rangle_C + | 1 \rangle_C)/\sqrt{2}, and the target qubit is prepared in the state | \phi \rangle_T = | 1 \rangle. State whether the output state is entangled or not. First, writing the input two-qubit state\begin{split} | \Psi \rangle & = \frac{1}{\sqrt{2}} ( | 0 \rangle_C + | 1 \rangle_C ) | 1 \rangle_T \\
& = \frac{1}{\sqrt{2}} ( | 0 \rangle_C | 1 \rangle_T + | 1 \rangle_C |1 \rangle_T ) \\
& = \frac{1}{\sqrt{2}} ( | 01 \rangle + | 11 \rangle ) \end{split}. Next, applying the operator:\begin{split} \widehat{\mathrm{CX}}_{C,T} | \Psi \rangle & = \widehat{\mathrm{CX}}_{C,T} |\frac{1}{\sqrt{2}} ( | 01 \rangle + | 11 \rangle ) \\
& = \frac{1}{\sqrt{2}} \left( \widehat{\mathrm{CX}}_{C,T} | 01 \rangle + \widehat{\mathrm{CX}}_{C,T} | 11 \rangle \right) \\
& = \frac{1}{\sqrt{2}} ( | 01 \rangle + | 10 \rangle ) \end{split}The output state cannot be factorised so it is an entangled state.To complete this section, consider the case when one of the entangled outputs from Table 6, | \Phi \rangle_+ = \frac{1}{\sqrt{2}} ( | 00 \rangle + | 11 \rangle ) is used as an input state.
\begin{split} \widehat{\mathrm{CX}}_{C,T} | \Phi \rangle & = \widehat{\mathrm{CX}}_{C,T} |\frac{1}{\sqrt{2}} ( | 00 \rangle + | 11 \rangle ) \\
& = \frac{1}{\sqrt{2}} \left( \widehat{\mathrm{CX}}_{C,T} | 00 \rangle + \widehat{\mathrm{CX}}_{C,T} | 11 \rangle \right) \\
& = \frac{1}{\sqrt{2}} ( | 00 \rangle + | 10 \rangle ) = \frac{1}{\sqrt{2}} ( | 0 \rangle_C + | 1 \rangle_C ) | 0 \rangle_T \end{split}The final state is now a product state (i.e. it is disentangled), where the control qubit is in the superposition state (|0 \rangle_C + | 1 \rangle_C)/\sqrt{2}. Thus the CNOT gate can disentangle a pair of qubits as well as entangle them.5.4 Quantum circuitsIn this section, you will make use of what you have already learnt about single-qubit and two-qubit gates to construct and interpret quantum circuits. A quantum circuit performs an algorithm in the sense that it executes a predefined sequence of quantum operations that embody an algorithm’s logic. However, unlike classical circuits, quantum circuits rely on principles like superposition and entanglement, and their output is often probabilistic rather than deterministic.By the end of this section you will be able to predict the outcome of the application of a quantum circuit to a given input state. As well as combining a series of gates, measurements can also be made in quantum circuits when one of the eigenstates of the measurement operator will be obtained with a certain probability. Therefore, the prediction of an output from a quantum circuit may involve more than one possible outcome and the associated probabilities of these outcomes. At the end of this section there is an activity to test your understanding of quantum circuits.5.4.1 Circuits with multiple gatesSingle-qubit gates and two-qubit gates can be combined in a structured sequence to give a quantum circuit containing multiple quantum gates. Consider as an example the circuit shown in Figure 11, which depicts a sequence of gates acting on two qubits, labelled | q_1 \rangle and | q_2 \rangle. Remember that each qubit enters its circuit from the left; first, a Hadamard gate is applied to qubit | q_1 \rangle. Then a CNOT is applied to | q_1 \rangle and | q_2 \rangle, where the control C is | q_1 \rangle and the target T is | q_2 \rangle. Finally, a NOT gate is applied to | q_2 \rangle.
Figure 11 An example of a quantum circuit with multiple quantum gates
The figure shows two parallel horiontal lines. The upper one is labelled q1 and the lower one is labelled q2. In the left part of the upper line is a square box with a capital letter H in it. Further along the upper line is a large dot. A vertical line leads from this dot to the lower line where it joins a large circle with a large plus sign in it. to the right of this on the lower line is another large circle with a large plus sign in it.The circuit in Figure 11 is a sequence of operations applied to qubits and can be analysed using the methods already introduced, with extra subscripts labelling the single-qubit gates and operations so that the qubit each operation is acting on is clear. Thus \widehat{\mathrm{H}}_1 acts only on the qubit q_1, leaving q_2 unchanged. Therefore, using | 0 0 \rangle as a sample input the calculation becomes:| q_1 q_2 \rangle_\text{final} = \widehat{\mathrm{X}}_2 \widehat{\mathrm{CX}}_{1,2} \widehat{\mathrm{H}}_1 | 0 0 \rangle.Now \widehat{\mathrm{H}}_1 acts only on q_1, so\begin{split} \widehat{\mathrm{H}}_1 | 0 0 \rangle & = \widehat{\mathrm{H}}_1 | 0 \rangle_1 | 0 \rangle_2 \\
& = \frac{1}{\sqrt{2}} ( | 1 \rangle_1 + | 0 \rangle_1 ) | 0 \rangle_2 \\
& = \frac{1}{\sqrt{2}} ( | 10 \rangle + | 00 \rangle ) \end{split}This means that| q_1 q_2 \rangle_\text{final} = \widehat{\mathrm{X}}_2 \widehat{\mathrm{CX}}_{1,2} \frac{1}{\sqrt{2}} (|10 \rangle + | 00 \rangle ).Next comes the effect of the CNOT gate, courtesy of \widehat{\mathrm{CX}}_{1,2} and gives| q_1 q_2 \rangle_\text{final} = \widehat{\mathrm{X}}_2 \frac{1}{\sqrt{2}} (|11 \rangle + | 00 \rangle ).Finally, observe that \widehat{\mathrm{X}}_2 acts on q_2 only, so flip q_2 of each ket in the superposition: | q_1 q_2 \rangle_\text{final} = \frac{1}{\sqrt{2}} (|10 \rangle + | 01 \rangle ).5.4.2 MeasurementsMeasurements are different from gate operations in a very important way, since rather than transforming a qubit from one definite state | \Psi \rangle to another definite state | \Phi \rangle, the final state after measurement is one of the two eigenstates of the measurement operator, which are obtained with some probability. Therefore, measurements are not reversible. The results of measurements are real numbers, so they can be stored as bits (rather than qubits) in a modern memory cell. The circuit symbol for the measurement operation is shown in Figure 12.
Figure 12 The symbol for a measurement operation
The figure shows a horizontal line. In the middle of the line is a square box. In the box is a curved line and an arrow that resemble a measuring meter.The probabilistic nature of measurement is a feature of quantum computing that must be accounted for when evaluating the performance of a quantum algorithm. ExampleA circuit is set up as shown in Figure 13 and the input qubits are both | 1 \rangle. Calculate the output qubits and hence the possible results of the measurements and their probabilities.
Figure 13 A circuit incorporating both a single-qubit and a two-qubit gate
The figure shows two parallel horiontal lines. The upper one is labelled q1 and the lower one is labelled q2. In the middle of the upper line is a large circle with a large plus sign in it. Further along the upper line is a square box with a curved line and an arrow in it. A vertical line leads from the large circle on the upper line to the lower line where it joins a large dot. To the left of this on the lower line a square box with a capital H in it; to the right of the large dot on the lower line is a square box with a curved line and an arrow in it.Writing the sequence of operations applied to the input qubits and using subscripts to label the qubits and the operations to show which qubit the gates are operating on, gives| q_1 q_2 \rangle_\text{final} = \widehat{\mathrm{CX}}_{2,1} \widehat{\mathrm{H}}_2| 1 1 \rangle.\widehat{\mathrm{H}}_2 acts on q_2 so\begin{split} \widehat{\mathrm{H}}_2 | 1 1 \rangle & = \widehat{\mathrm{H}}_2 | 1\rangle_1 | 1 \rangle_2 = | 1 \rangle_1 \widehat{\mathrm{H}}_2 | 1 \rangle_2 \\
& = | 1 \rangle_1 \frac{1}{\sqrt{2}} ( | 0 \rangle_2 - | 1 \rangle_2 ) \\
& = \frac{1}{\sqrt{2}} | 1 \rangle (| 0 \rangle - | 1 \rangle ) \end{split}so now| q_1 q_2 \rangle_\text{final} = \widehat{\mathrm{CX}}_{2,1} \frac{1}{\sqrt{2}} | 1 \rangle (| 0 \rangle - | 1 \rangle ) Note that q_2 is the control qubit and q_1 is the target qubit. Consequently, when \widehat{\mathrm{CX}}_{2,1} operates on | q_1 q_2 \rangle, look at | q_2 \rangle to decide whether | q_1 \rangle is flipped. Again, adding subscripts to identify the qubits,\begin{split}
| q_1 q_2 \rangle_\text{final} & = \frac{1}{\sqrt{2}} \widehat{\mathrm{CX}}_{1,2} ( | 1 \rangle_1 | 0 \rangle_2 - | 1 \rangle_1 | 1 \rangle_2 ) \\
& = \frac{1}{\sqrt{2}} \left( \widehat{\mathrm{CX}}_{1,2} (| 1 \rangle_1 | 0 \rangle_2 - \widehat{\mathrm{CX}}_{1,2} | 1 \rangle_1 | 1 \rangle_2 \right) \\
& = \frac{1}{\sqrt{2}}( | 1 \rangle_1 | 0 \rangle_2 - | 0 \rangle_1 | 1 \rangle_2 ) \\
& = \frac{1}{\sqrt{2}} | 1 \rangle | 0 \rangle - \frac{1}{\sqrt{2}} | 0 \rangle| 1 \rangle \end{split}
This is the final state which is measured. It is an entangled state. There are two possible outcomes; either q_1 is measured as | 1 \rangle and q_2 is measured as | 0 \rangle or q_1 is measured as | 0 \rangle and q_2 is measured as | 1 \rangle. From the 1/\sqrt{2} coefficients, the conclusion is that each outcome has a probability of 1/2. 5.4.3 ActivityThis activity is in two parts. In the first, you are given a quantum circuit and input qubits. Your task is to work out the output qubits and to determine whether the final output states, before measurements are taken, are entangled. In the second task you will design your own circuit for given input and output qubits. Part 1Consider the quantum circuit shown in Figure 14. The input qubits are both | 0 \rangle. Determine if the output two-qubit state, before measurements are taken, is entangled. Calculate the possible measurements and the probability of each possibility.
Figure 14 A circuit to be analysed for the Activity
The figure shows two parallel horiontal lines. The upper one is labelled q1 and the lower one is labelled q2. Near the left end of both lines are square boxes each containing a capital letter H. Further along the upper line is a large dot. A vertical line leads from this dot to the lower line where it joins a large circle with a large plus sign in it. Still further along the upper line is a large circle with a large plus sign in it. Even further along the upper line is another large circle with a large plus sign in it. A vertical line leads from this circle to the lower line where it joins a large dot. Near the right hand ends of both the upper and lower lines are square boxes each with a curved line and an arrow in it.Writing the sequence of operations applied to the input qubits and using subscripts to label the qubits and the operations to show which qubit the gates are operating on, gives| q_1 q_2 \rangle_\text{final} = \widehat{\mathrm{CX}}_{2,1} \widehat{\mathrm{X}}_1 \widehat{\mathrm{CX}}_{1,2} \widehat{\mathrm{H}}_2 \widehat{\mathrm{H}}_1| 0 0 \rangle.\widehat{\mathrm{H}}_1 acts on q_1 and \widehat{\mathrm{H}}_2 acts on q_2. So looking first at q_1,\widehat{\mathrm{H}}_1 | 0 \rangle_1 = \frac{1}{\sqrt{2}} ( | 0 \rangle_1 + | 1 \rangle_1 ) So q_1 is now in a superposition state. The effect on q_2 is similar, \widehat{\mathrm{H}}_2 | 0 \rangle_2 = \frac{1}{\sqrt{2}} ( | 0 \rangle_2 + | 1 \rangle_2 ) and q_2 is also in a superposition state. So now we have\begin{split} | q_1 q_2 \rangle_\text{final} & = \widehat{\mathrm{CX}}_{2,1} \widehat{\mathrm{X}}_1 \widehat{\mathrm{CX}}_{1,2} \frac{1}{\sqrt{2}} ( | 0 \rangle_1 + | 1 \rangle_1 ) \frac{1}{\sqrt{2}} ( | 0 \rangle_2 + | 1 \rangle_2 ) \\
& = \widehat{\mathrm{CX}}_{2,1} \widehat{\mathrm{X}}_1 \widehat{\mathrm{CX}}_{1,2} \frac{1}{2} ( |0\rangle_1 |0 \rangle_2 + |0\rangle_1 |1 \rangle_2 + |1 \rangle_1 |0 \rangle_2 + |1 \rangle_1 |1 \rangle_2) \end{split} Note that for the first CNOT gate, q_1 is the control qubit and q_2 is the target qubit. Consequently, when \widehat{\mathrm{CX}}_{1,2} operates on | q_1 q_2 \rangle, look at | q_1 \rangle to decide whether | q_2 \rangle is flipped. Again, adding subscripts to identify the qubits,\widehat{\mathrm{CX}}_{1,2} ( |0\rangle_1 |0 \rangle_2 + |0\rangle_1 |1 \rangle_2 + |1 \rangle_1 |0 \rangle_2 + |1 \rangle_1 |1 \rangle_2) = ( |0\rangle_1 |0 \rangle_2 + |0\rangle_1 |1 \rangle_2 + |1 \rangle_1 |1 \rangle_2 + |1 \rangle_1 |0 \rangle_2) The result is an entangled state. Next the NOT gate acts on qubit q_1 to give\widehat{\mathrm{X}}_1( |0\rangle_1 |0 \rangle_2 + |0\rangle_1 |1 \rangle_2 + |1 \rangle_1 |1 \rangle_2 + |1 \rangle_1 |0 \rangle_2) = ( |1\rangle_1 |0 \rangle_2 + |1\rangle_1 |1 \rangle_2 + |0 \rangle_1 |1 \rangle_2 + |0 \rangle_1 |0 \rangle_2)So we now have | q_1 q_2 \rangle_\text{final} = \frac{1}{2} \widehat{\mathrm{CX}}_{2,1} ( |1\rangle_1 |0 \rangle_2 + |1\rangle_1 |1 \rangle_2 + |0 \rangle_1 |1 \rangle_2 + |0 \rangle_1 |0 \rangle_2)The second CNOT gate acts on q_2 as the control qubit and q_1 as the target qubit, so this time look at | q_2 \rangle to decide whether | q_1 \rangle is flipped. \widehat{\mathrm{CX}}_{2,1} ( |1\rangle_1 |0 \rangle_2 + |1\rangle_1 |1 \rangle_2 + |0 \rangle_1 |1 \rangle_2 + |0 \rangle_1 |0 \rangle_2) = ( |1\rangle_1 |0 \rangle_2 + |0\rangle_1 |1 \rangle_2 + |1 \rangle_1 |1 \rangle_2 + |0 \rangle_1 |0 \rangle_2)So we finally have the following, | q_1 q_2 \rangle_\text{final} = \frac{1}{2} ( |1 0 \rangle + |0 1 \rangle + |1 1 \rangle + |0 0 \rangle)
This is the final state which is measured. It is an entangled state. There are four possible outcomes; either q_1 is measured as | 0 \rangle and q_2 is measured as | 0 \rangle or q_1 is measured as | 0 \rangle and q_2 is measured as | 1 \rangle or q_1 is measured as | 1 \rangle and q_2 is measured as | 0 \rangle or q_1 is measured as | 1 \rangle and q_2 is measured as | 1 \rangle. From the 1/2 coefficients, the conclusion is that each outcome has a probability of 1/4. Part 2Design a circuit to convert the two-qubit input state |00\rangle into the (non-entangled) superposition two qubit output state comprising |01\rangle and |11\rangle with equal probability.There are various ways to achieve this. Once such circuit is shown in Figure 15.
Figure 15 A circuit to convert the two-qubit input state |00\rangle into the two qubit output state |01 \rangle or |11\rangle with equal probability
The figure shows two parallel horiontal lines. The upper one is labelled q1 and the lower one is labelled q2. Near the left end of the lower line is a large circle containing a large plus sign. Near the left end of the upper line is a square box containing a capital letter H.The circuit can be described as| q_1 q_2 \rangle_\text{final} = \widehat{\mathrm{H}}_1 \widehat{\mathrm{X}}_2 | 0 0 \rangle.Starting with input |00 \rangle the circuit applies a NOT gate to qubit q_2, resulting in |01\rangle, so we have| q_1 q_2 \rangle_\text{final} = \widehat{\mathrm{H}}_1 | 0 1 \rangle.A Hadamard gate is then applied to qubit q_1 to create a superposition for this qubit, \widehat{\mathrm{H}}_1 | 0 \rangle_1 = \frac{1}{\sqrt{2}}(|0\rangle_1 + |1 \rangle_1)This gives|q_1 q_2 \rangle_\text{final} = \frac{1}{\sqrt{2}} ( | 01 \rangle + | 11 \rangle )The output state is therefore a state whose non-entangled two qubit output state is a superposition of |01\rangle and |11\rangle with equal probability, as required.6 Real-world quantum computing
Quantum computing presents enormous technical challenges, which is why there many research groups and companies working on quantum computers and many competing technologies. Each of these technologies has its advantages and disadvantages.
Even though large systems are built of atoms and molecules that are well-described by quantum mechanics, large complicated systems such as computers or people don’t exhibit the quintessential quantum behaviour such as superposition or entanglement. 6.1 Schrödinger’s catA famous statement of this idea was given by Erwin Schrödinger in his Schrödinger’s cat thought experiment. In that thought experiment, the fate of a cat in an enclosed box depends on the disintegration of a single radioactive nucleus, which, when it decays, will trigger the release of poison and thereby the death of the cat. The nucleus is supposed to be in the superposition state| \text{nucleus} \rangle = a_0 | \text{metastable} \rangle + a_1 | \text{stable} \rangle
If the possible states for the cat are | \text{dead} \rangle and | \text{alive} \rangle, then the total state for the system of cat and nucleus is
| \text{cat + nucleus} \rangle = a_0 | \text{metastable} \rangle | \text{alive} \rangle + a_1 | \text{stable} \rangle | \text{dead} \rangle
The cat is entangled with the nucleus, since there exists a correlation between the state of the nucleus and the state of the cat (the state written down in Equation 17 is an entangled state). The point of the thought experiment is that it demonstrates that thinking of the nucleus + cat system as a superposition of those extreme states is absurd. Surely it’s absurd to believe that a cat is well-described by a state vector (| \text{alive} \rangle or | \text{dead} \rangle), and surely the cat is in the definite state ‘alive’until it definitely dies (or better, until the experiment is stopped before the cat is killed). The issue at hand is that a useful working quantum computer containing hundreds of thousands of qubits would be more akin to Schrödinger’s cat than a single quantum object, like a solitary microscopic superconducting circuit.6.2 A few examples of quantum technologiesThis section is to provide you with a starting point if you would like to learn more about quantum computer technologies. The difficulties of quantum computing means that the number of systems that have been proposed as quantum computers is almost as large as the number of tasks in which quantum computers can (in principle) outperform classical computers. In this section there is a brief introduction to three, the transmon qubit, NMR and cold atom technology. It is not possible to expect this section to keep up-to-date as advances are being made all the time and often announcements appear in the news. You can watch out for these and follow up your interests if you would like to. Some links are provided at the end to get you started.The transmon qubitThe transmon qubit is based on a miniaturised superconducting circuit, built from a capacitor and a non-linear inductor called a Josephson junction. This qubit is at the heart of the IBM quantum computers, which makes it one of the most advanced platforms. Within superconducting circuits, quantum logic gates may be implemented by applying an AC voltage to the qubits. The qubits in the superconductor involve the charge carriers which respond to the applied potential.The NMR qubitNMR was used as an early test of quantum computing because the technology is well-developed as a result of medical research and its use for medical purposes (MRI scanners). In NMR quantum computing, each qubit is realised as a collection of particles called an ensemble. The ensemble is all the molecules in a sample. The qubits are molecular sites in the target molecule. Therefore, the sample contains many copies of the qubits. To increase the number of qubits, the molecule must become more complex, but this is difficult because as the size of the molecule increases, environmental effects mean that each molecule is likely to be different. This leads to the conclusion that NMR quantum computing is limited to about 20 qubits and it is unlikely that this number of qubits can be increased. Using cold-atom technologyCold-atom technology (which includes cold ions) is able to take advantage of techniques developed to build atomic clocks. The appeal of cold-atom systems is the degree of control that can be achieved. For example, from the gas phase, individual atoms can be trapped at specific sites in a vacuum chamber, and then addressed by individual laser beams for the purposes of implementing gates or measurements. The weakness of the cold-atom approach is that, so far, its complexity scales poorly with the number of qubits. LinksHere are some links to websites that explain current activities in quantum computing from different companies.IBMhttps://www.ibm.com/quantum/technologyAmazonhttps://aws.amazon.com/what-is/quantum-computingQuantum Insiderhttps://thequantuminsider.com/2023/06/06/types-of-quantum-computersXanadu AI and From a state of light to state of the artGoogle Quantum AI and What our quantum computing milestone means or Microsofthttps://quantum.microsoft.com/en-us/explore/concepts/topological-qubits7 SummaryIn this course you have learnt the fundamentals of quantum computing.The key points are as follows.Quantum computers may be able solve problems more quickly than classical computers if problem solving algorithms which have exponential run-times on a classical computer can be written to have polynomial run-times on a quantum computer.For a given square matrix, \mathrm{A}, it is possible to solve the equation \mathrm{A}\mathbf{v}= \lambda \mathbf{v} where \mathbf{v} are column vectors known as eigenvectors and \lambda is a scalar called an eigenvalue. In quantum mechanics, an operator is a mathematical entity which converts one function into another function. Given an operator \mathrm{\widehat{A}}, the eigenvalue equation for that operator is \mathrm{\widehat{A}}f(x)= \lambda f(x). Here, the eigenvalue \lambda may be a complex number, and f(x) is function known as an eigenfunction. There may be more than one eigenvalue and corresponding eigenfunction associated with each eigenvalue equation.A general spin state | A \rangle (known as a ket) can be written as a linear combination of a spin-up state | \uparrow \rangle and a spin-down state | \downarrow \rangle (known as basis vectors), thus | A \rangle = a_1 | \uparrow \rangle + a_2 | \downarrow \rangle where a_1 and a_2 are complex numbers. For an atom in any spin state, | A \rangle, the probability of the outcome of a measurement indicating spin-up is | a_1 |^2 and for spin-down is |a_2|^2. Since these are the only possible outcomes the corresponding probabilities must sum to one, therefore |a_1|^2+|a_2|^2=1.Matrices can be used as an alternative representation of spin states to simplify calculations. | \uparrow \rangle and | \downarrow \rangle are represented by the following column vectors:| \uparrow \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \text{and} \quad | \downarrow \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix} Any vector | A \rangle in spin space may be written as a linear combination of | \uparrow \rangle and | \downarrow \rangle. This means that | A \rangle becomes:| A \rangle = a_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix} + a_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} In this way any spin state of a spin-½ particle can be represented as a two-element matrix, which is called a spinor.For two electrons, the two-particle spin state can have an overall spin function which is either symmetric or antisymmetric to exchange of electrons. There is a set of triplet states and a singlet state, as follows:\begin{split} | 1, 1 \rangle & = | \uparrow \uparrow \rangle , \\
| 1, 0 \rangle & = \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle ), \\
| 1, -1 \rangle & = | \downarrow \downarrow \rangle, \\
| 0, 0 \rangle & = \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle ). \end{split}where the first arrow in each ket refers to particle 1 and the second to particle 2. The triplet states are symmetric and the singlet state is antisymmetric under particle exchange. Two-particle states which cannot be factorised (i.e. | 0,0 \rangle and | 1, 0 \rangle) are known as entangled states and exhibit entanglement. The other states
(i.e. | 1,1 \rangle and | 1, -1 \rangle) are not entangled.Quantum computing is based on units of information called qubits (quantum bits, and pronounced kew-bits), which obey the laws of quantum mechanics. A qubit is the quantum analogue of a classical bit. The classical bit values 0 and 1 are replaced by the orthonormal basis states of the quantum-mechanical qubit |0 \rangle and | 1 \rangle. The basis states are given the name logical states , since they correspond to the classical bits upon which the logic gates operate. The key difference between qubits and classical bits is that qubits can exist in a superposition of the |0 \rangle and |1 \ranglestates, which means qubits can be prepared in the superposition state | \psi \rangle = a_0 | 0 \rangle + a_1 | 1 \rangle.The quantum NOT gate is denoted by the operator, \widehat{\mathrm{X}} and, in the basis of the logical qubits |0 \rangle and |1 \rangle, is represented by the matrix:\widehat{\mathrm{X}} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.which is also the Pauli-X operator, \widehat{\sigma}_x. If the input to a quantum NOT gate is a_0 |0 \rangle + a_1 | 1 \rangle then the output is a_0 |1 \rangle + a_1 | 0 \rangle. The Hadamard gate is a single-qubit gate defined by the matrix:\widehat{\mathrm{H}} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}.If the input to a Hadamard gate is a_0 |0 \rangle + a_1 | 1 \rangle then the output is \frac{1}{\sqrt{2}} ( a_0 + a_1) | 0 \rangle + \frac{1}{\sqrt{2}} ( a_0 - a_1) | 1 \rangle . A Hadamard gate allows the transformation of the logical qubit state into a superposition state.A straightforward way to define the two-qubit states is to build the two-qubit basis states from product states | \Psi \rangle = | q_1 \rangle | q_2 \rangle = | q_1 q_2 \rangle. There are four possible product states of the usual single-qubit basis states: |00 \rangle, |01 \rangle, |10 \rangle, | 11 \rangle. A general two-qubit state, therefore can be expressed in terms of the product states as | \Psi \rangle = a_{00} |00 \rangle + a_{01} |01 \rangle + a_{10} |10 \rangle + a_{11} |11 \rangle, where | \Psi \rangle is normalised in the usual way:| a_{00}|^2 + | a_{01}|^2 + | a_{10}|^2 + | a_{11}|^2 =1The quantum CNOT gate acts on two qubits, a control qubit and a target qubit. It is represented by an operator \widehat{\mathrm{CX}}_{C,T} which acts on a target qubit | \phi_T \rangle depending on the state of a control qubit | \psi_C \rangle. If the input to a CNOT gate is \frac{1}{\sqrt{2}} ( | 00 \rangle \pm | 10 \rangle ) then the output is \frac{1}{\sqrt{2}} ( | 00 \rangle \pm | 11 \rangle ). The CNOT gate can entangle a pair of disentangled qubits and can also disentangle a pair of entangled qubits. Single-qubit gates and two-qubit gates can be combined in a structured sequence to give a quantum circuit that executes a particular algorithm. Measurements can also be made in quantum circuits when one of the eigenstates of the measurement operator will be obtained with a certain probability. Real-world quantum computing has been impemented on platforms including the transmon qubit (as in the IBM quantum computers), the NMR qubit (based on technology used in MRI scanners), and using ultra-cold atom technology. <Number>8</Number> QuizAnswer the following questions in order to test your understanding of the key ideas that you have been learning about.Question 1Which of the following statements about eigenvalues, eigenstates, eigenvectors and eigenfunctions are true? An eigenfunction is special type of function that remains essentially unchanged (except for a scaling factor) when acted upon by a given linear operator.TrueAn eigenstate in quantum mechanics is a special quantum state that remains unchanged, except for a multiplicative factor, when a specific quantum operator acts on it.TrueAn eigenvalue is a special scalar associated with a linear transformation of a square matrix. It represents how much a given vector is scaled when that matrix is applied to it.TrueAn eigenstate is a state for which the outcome of a measurement of a certain observable (like energy, position, or momentum) will always yield a specific, definite value.TrueAn eigenvector of a square matrix is a nonzero vector that gets scaled by a certain value when the matrix is applied to it. TrueThere is always only one eigenvalue and corresponding eigenfunction associated with each eigenvalue equationFalseThe first five statements are all true. The last one is false: there may be more than one eigenvalue and corresponding eigenfunction associated with each eigenvalue equation.Question 2What are the eigenvalues and eigenvectors of the following matrix?\begin{bmatrix} 7 & 4 \\ 2 & 5 \end{bmatrix}For eigenvalue \lambda_1 = 7 the eigenvector is \mathbf{v}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} and for eigenvalue \lambda_2 = 5 the eigenvector is \mathbf{v}_2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix}For eigenvalue \lambda_1 = 9 the eigenvector is \mathbf{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix} and for eigenvalue \lambda_2 = 3 the eigenvector is \mathbf{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}For eigenvalue \lambda_1 = 4 the eigenvector is \mathbf{v}_1 = \begin{bmatrix} 2 \\ 2 \end{bmatrix} and for eigenvalue \lambda_2 = 2 the eigenvector is \mathbf{v}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}For eigenvalue \lambda_1 = 1 the eigenvector is \mathbf{v}_1 = \begin{bmatrix} 2 \\ 3 \end{bmatrix} and for eigenvalue \lambda_2 = 8 the eigenvector is \mathbf{v}_2 = \begin{bmatrix} -3 \\ 1 \end{bmatrix}Following the prescription described in the course: a=7, b=4, c=2 and d=5. So we first need to solve the quadratic equation \lambda^2 - (7+5) \lambda + ((7 \times 5) - (4 \times 2)) = 0which is simply \lambda^2 - 12 \lambda + 27 = 0This can be written as (\lambda - 9) (\lambda - 3) = 0So it has solutions \lambda_1 = 9 and \lambda_2 = 3. These are the two eigenvalues.We now write the two eigenvector equations:\begin{split}
(7 - \lambda) x + 4y & = 0 \\
2x + (5 - \lambda)y & = 0
\end{split}For eigenvalue \lambda_1 = 9 these reduce to
\begin{split}
-2x + 4y & = 0 \\
2x - 4y & = 0
\end{split}Both equations imply that x=2y, so x=2 and y=1 and the first eigenvector is \mathbf{v}_1 = \begin{bmatrix} 2 \\ 1 \end{bmatrix}For eigenvalue \lambda_2 = 3 these reduce to
\begin{split}
4x + 4y & = 0 \\
2x + 2y & = 0
\end{split}Both equations imply that x=-y, so x=-1 and y=1 and the second eigenvector is \mathbf{v}_2 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}Question 3If a general spin state is written as |a \rangle = a_1 | \uparrow \rangle + a_2 | \downarrow \rangle, which of the following statements is true?The probability of a measurement indicating spin-up is a_1The probability of a measurement indicating spin-down is |a_2|^2The probability of a measurement indicating spin-down is |a_1|^2The probability of a measurement indicating spin-down is a_1 - a_2The probability of a measurement indicating spin-up is |a_1 + a_2|^2The probability of the outcome of a measurement indicating spin-up is |a_1|^2 and for spin-down is |a_2|^2. Question 4Match the following two-particle spin states with the correct descriptions. symmetric not entangled statesymmetric entangled stateantisymmetric entangled stateThe triplet states (i.e. |1,1\rangle and |1,0\rangle and |1,-1\rangle) are symmetric and the singlet state (i.e. |0,0\rangle) is antisymmetric under particle exchange. Two-particle states which cannot be factorised (i.e. |0,0\rangle and |1,0\rangle) are known as entangled states. The other states (i.e. |1,1\rangle and |1,-1\rangle) are not entangled.Question 5The quantum NOT gate is represented by which of the following matrices?\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}The quantum NOT gate is represented by the Pauli-X operator, \widehat{\mathrm{X}} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}Question 6The Hadamard gate is represented by which of the following matrices?\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\frac{1}{\sqrt{2}} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\frac{1}{\sqrt{2}} \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix}The Hadamard gate is represented by \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}Question 7Match the following quantum gates with the correct result. Flips the state of a qubitEntangles a pair of disentangled qubitsTransforms a qubit into a superposition stateA NOT gate flips the state of a qubit. A CNOT gate can entangle a pair of disentangled qubitts. A Hadamard gate can transform a qubit into a superposition state.Question 8If an input qubit | 0 \rangle is passed to a Hadamard gate, and the output from the Hadamard gate is then passed as input to another Hadamard gate, what will be the output from the second Hadamard gate?\frac{1}{\sqrt{2}}(|0 \rangle + |1 \rangle)|0 \rangle\frac{1}{\sqrt{2}}(|0 \rangle - |1 \rangle)\frac{1}{2}(|0 \rangle + |1 \rangle)\frac{1}{2}(|0 \rangle - |1 \rangle)|1 \rangleThe circuit can be written as \widehat{\mathrm{H}} \widehat{\mathrm{H}} |0 \rangle. The action of the first Hadamard gate produces a superposition state:\widehat{\mathrm{H}}|0 \rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}| 1 \rangleThen passing this through the second Hadamard gate we have\widehat{\mathrm{H}}\widehat{\mathrm{H}}|0 \rangle = \frac{1}{\sqrt{2}}\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) |0\rangle + \frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) | 1 \rangle\widehat{\mathrm{H}}\widehat{\mathrm{H}}|0 \rangle = \frac{1}{\sqrt{2}}\left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) |0\rangle = \frac{1}{\sqrt{2}} \left( \frac{2}{\sqrt{2}} \right) | 0 \rangle = | 0 \rangleThe action of the second Hadamard gate is therefore to restore the original input qubit.AcknowledgementsThis free course was written by Christine Leach, Silvia Bergamini, Calum MacCormick and Andrew Norton and first published in June 2025.OpenLearn editor: Dale HarryExcept for third party materials and otherwise stated (see terms and conditions), this content is made available under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 Licence.The material acknowledged below is Proprietary and used under licence (not subject to Creative Commons Licence). Grateful acknowledgement is made to the following sources for permission to reproduce material in this free course: ImagesCourse image: bpawesome/istock/Getty ImagesFigure 1: Radek Grzybowski/UnsplashEvery effort has been made to contact copyright owners. If any have been inadvertently overlooked, the publishers will be pleased to make the necessary arrangements at the first opportunity.Don’t miss outIf reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover our free learning resources and qualifications by visiting The Open University – www.open.edu/openlearn/free-courses.Discussion20250114002025021001Discussion