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Quantum Information Processing: Foundations - Part 2

10 min read

Previously, we had a brief introduction to the idea of geometrically visualizing a qubitqubit's quantum state in a 3D sphere called the Bloch Sphere ( Bloch, F., 1946 Bloch, F. (1946). Nuclear Induction. Physical Review, 460-474 ) — named after Felix Bloch, the Swiss-American physicist. Being able to represent an arbitrary qubitqubit state in space visually simplifies its complexity, and the Bloch Sphere does this. We will go a bit mathematical (with a touch of physics) in this part, primarily to bring into perspective how some expressions came about. Then, we'll pick some problems in ( Rieffel, E. G. & Polak, W. H., 2014 Rieffel, E. G. & Polak, W. H. (2014). Quantum Computing: A Gentle Introduction. MIT Press ) and work through solving them step-by-step to solidify our understanding better.

Quantum computing relies on mathematical (and, briefly in this article, physics) principles, but you can learn the essentials without being a math whiz. A working knowledge of high school math will equip you to understand the applications and fundamental ideas. Familiarity with the Python programming language will help you understand qiskit and/or cirq code.

Consider the Bloch sphere below:

Bloch Sphere
Figure 1: Bloch Sphere showing the major measurement basis and angles.

where:

  • 0\ket{0} represents the zz-axis;
  • +\ket{+} is the xx-axis;
  • i\ket{i} depicts the yy-axis;
  • ψ\ket{\psi} is an arbitrary state on the sphere;
  • θ\theta is the angle the state makes with zz-axis; and
  • ϕ\phi, the azimuthal angle, is the angle that the state's projection is making with the xx-axis.

This arbitrary state can be expressed, verbosely, using the state function:

ψ=eiγ(cosθ20+eiϕsinθ21)\ket{\psi} = e^{i\gamma}\left(\cos\frac{\theta}{2}\ket{0} + e^{i\phi}\sin\frac{\theta}{2}\ket{1} \right)

However, eiγe^{i\gamma}, regarded as a global phase, does not pose observable effects to the measurement obtained with or without its presence ( Glendinning, I., 2005 Glendinning, I. (2005). The Bloch Sphere ). This is because when a unitary operator, say UU, operates on ψ\ket{\psi}, its ketket-side remains unchanged whereas its brabra-side negates it (due to complex conjugation), which effectively eliminates eiγe^{i\gamma} ( UV Physics., 2023 UV Physics. (2023). Bloch sphere and Qubit Representation. https://www.youtube.com/watch?v=nVpj3_NOvRI.. Retrieved April 19, 2025 ):

eiγψUeiγψ=eiγeiγψUψ=ψUψ\braket{e^{i\gamma}\psi|U|e^{i\gamma}\psi} = e^{i\gamma}\cdot e^{-i\gamma}\braket{\psi|U|\psi} = \braket{\psi|U|\psi}

Effectively, the Bloch state function can be simplified to:

ψ=cosθ20+eiϕsinθ21\begin{equation} \ket{\psi} = \cos\frac{\theta}{2}\ket{0} + e^{i\phi}\sin\frac{\theta}{2}\ket{1} \end{equation}

How did they come about (1)? A curious mind would like to know. The following subsection unravels it!

Note:Physics Territory

We are delving into some concepts in Physics, such as Spin Angular Momentum. Reader's discretion is advised.

I will skip some details here for brevity. If you need a more detailed coverage and preliminaries, I recommend taking a look at ( UV Physics., 2023 UV Physics. (2023). Bloch sphere and Qubit Representation. https://www.youtube.com/watch?v=nVpj3_NOvRI.. Retrieved April 19, 2025 ; Zettili, N., 2009 Zettili, N. (2009). Quantum Mechanics: Concepts and Applications ).

Spherical representation of an arbitrary position vector in space
Figure 2: Spherical representation of an arbitrary position vector in space.

Recall that the cartesian coordinates (xx, yy, zz) of a position vector r\overrightarrow{r} relate to its polar coordinates (rr, θ\theta, ϕ\phi) in the following fashion ( Zettili, N., 2009 Zettili, N. (2009). Quantum Mechanics: Concepts and Applications ):

x=rsinθcosϕ,y=rsinθsinϕ,z=rcosθ\begin{equation} \tag{p1} x = r\sin\theta\cos\phi, \quad y=r\sin\theta\sin\phi, \quad z=r\cos\theta \end{equation}

Now, for a unitunit position vector n^\hat n (shown in the diagram above), these coordinates become:

x=sinθcosϕ,y=sinθsinϕ,z=cosθ\begin{equation} \tag{p2} x = \sin\theta\cos\phi, \quad y=\sin\theta\sin\phi, \quad z=\cos\theta \end{equation}

since r=1r=1 (hence the word unitunit).

Measuring the angular spin operator S\overrightarrow{S} in n^\hat n direction produces an operator, say A^\hat A which is:

A^=Sn^=Sxnx+Syny+Sznz\begin{equation} \tag{p3} \hat A = \overrightarrow{S} \cdot \hat n = S_x\cdot n_x + S_y\cdot n_y + S_z\cdot n_z \end{equation}

From ( Zettili, N., 2009 Zettili, N. (2009). Quantum Mechanics: Concepts and Applications ), the Pauli matrice — σx,σy,σz\sigma_x, \sigma_y, \sigma_z — are defined as:

σx=(0110),σy=(0ii0),σz=(1001)\begin{equation} \tag{p4} \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{equation}

For a spin-12\frac{1}{2}​ particle, the spin operators are 2x2 matrices which relate to Pauli matrices by 2\frac{\hbar}{2} so that:

Sx=2σx,Sy=2σy,Sz=2σz\begin{equation} \tag{p5} S_x = \frac{\hbar}{2}\cdot\sigma_x, \quad S_y = \frac{\hbar}{2}\cdot\sigma_y, \quad S_z = \frac{\hbar}{2}\cdot\sigma_z \end{equation}

where \hbar is the reduced Plank constant equalling h2π\frac{h}{2\pi}.

Substituting (p2), (p4) and (p5) into (p3), we have:

A^={2(0110)}sinθcosϕ+{2(0ii0)}sinθsinϕ+{2(1001)}cosθ\begin{equation} \notag \begin{aligned} \hat A = & \left\{\frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right\} \sin\theta\cos\phi \\ &+ \left\{\frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}\right\} \sin\theta\sin\phi \\ &+ \left\{\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\right\} \cos\theta \end{aligned} \end{equation} A^=2[(0sinθcosϕsinθcosϕ0)+(0isinθsinϕisinθsinϕ0)+(cosθ00cosθ)]\begin{equation} \notag \hat A = \frac{\hbar}{2} \left[ \begin{pmatrix} 0 & \sin\theta\cos\phi \\ \sin\theta\cos\phi & 0 \end{pmatrix} + \begin{pmatrix} 0 & -i\sin\theta\sin\phi \\ i\sin\theta\sin\phi & 0 \end{pmatrix} + \begin{pmatrix} \cos\theta & 0 \\ 0 & -\cos\theta \end{pmatrix} \right] \end{equation}

When the matrices are added, we have:

A^=2(cosθsinθcosϕisinθsinϕsinθcosϕ+isinθsinϕcosθ)=2(cosθsinθ(cosϕisinϕ)sinθ(cosϕ+isinϕ)cosθ)=2(cosθeiϕsinθeiϕsinθcosθ),Sinceeiϕ=cosϕ+isinϕ\begin{equation} \notag \begin{aligned} \hat A &= \frac{\hbar}{2} \begin{pmatrix} \cos\theta & \sin\theta\cos\phi - i\sin\theta\sin\phi \\ \sin\theta\cos\phi + i\sin\theta\sin\phi & -\cos\theta \end{pmatrix} \\ &= \frac{\hbar}{2} \begin{pmatrix} \cos\theta & \sin\theta(\cos\phi - i\sin\phi) \\ \sin\theta(\cos\phi + i\sin\phi) & -\cos\theta \end{pmatrix} \\ &= \frac{\hbar}{2} \begin{pmatrix} \cos\theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta \end{pmatrix}, \quad \text{Since} \,\, e^{i\phi} = \cos\phi + i\sin\phi \end{aligned} \end{equation}

Now, we need to obtain the eigenvector of this since its eigenvalue, λ\lambda, is ±2\pm\frac{\hbar}{2} ( UV Physics., 2023 UV Physics. (2023). Bloch sphere and Qubit Representation. https://www.youtube.com/watch?v=nVpj3_NOvRI.. Retrieved April 19, 2025 ; Zettili, N., 2009 Zettili, N. (2009). Quantum Mechanics: Concepts and Applications ) (this value remains the same irrespective of direction). To do this, we use the eigenvalue equation ( Zettili, N., 2009 Zettili, N. (2009). Quantum Mechanics: Concepts and Applications ):

A^ψ±=λψ±\hat A \psi_\pm = \lambda\psi_\pm

Since we are trying to obtain values for ψ±\psi_\pm, the unknown, we are at liberty to choose any symbol to represent it as long as it tallies with the dimension of A^\hat A. So,

2(cosθeiϕsinθeiϕsinθcosθ)(ab)=2(ab)\frac{\hbar}{2} \begin{pmatrix} \cos\theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta \end{pmatrix} \cdot \begin{pmatrix}a\\ b \end{pmatrix} = \frac{\hbar}{2}\begin{pmatrix}a\\ b \end{pmatrix} (cosθeiϕsinθeiϕsinθcosθ)(ab)=(ab)\begin{pmatrix} \cos\theta & e^{-i\phi}\sin\theta \\ e^{i\phi}\sin\theta & -\cos\theta \end{pmatrix} \cdot \begin{pmatrix}a\\ b \end{pmatrix} = \begin{pmatrix}a\\ b \end{pmatrix}

Multiplying out, we have:

acosθ+beiϕsinθ=aaeiϕsinθbcosθ=b\begin{align} \tag{p6} a\cdot\cos\theta + b\cdot e^{-i\phi}\sin\theta &= a\\ \tag{p7} a\cdot e^{i\phi}\sin\theta - b\cdot\cos\theta &=b \end{align}

To solve the simultaneous equations, we can express bb in terms of aa in (p6):

acosθ+beiϕsinθ=aa\cdot\cos\theta + b\cdot e^{-i\phi}\sin\theta = a b=a(1cosθ)eiϕsinθ=a(1cosθ)sinθeiϕ\begin{align} \notag b &= \frac{a(1-\cos\theta)}{e^{-i\phi}\sin\theta}\\ \notag &= \frac{a(1-\cos\theta)}{\sin\theta}e^{i\phi} \end{align}

From ( Wikipedia., 2025 Wikipedia. (2025). List of trigonometric identities. https://en.wikipedia.org/wiki/List_of_trigonometric_identities.. Retrieved April 20, 2025 ), 1cosθ=2sin2θ21-\cos\theta=2\sin^2\frac{\theta}{2} which is the half-angle formula for sine and sinθ=2sinθ2cosθ2\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2} (sine's double-angle formula). Therefore,

b=a(2sin2θ2)2sinθ2cosθ2eiϕ=asinθ2cosθ2eiϕ\begin{align} \notag b&= \frac{a(2\sin^2\frac{\theta}{2})}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}e^{i\phi}\\ \notag &= \frac{a\cdot\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}e^{i\phi} \end{align}

\therefore

ψ±=(aasinθ2cosθ2eiϕ)\psi_\pm = \begin{pmatrix}a\\ \frac{a\cdot\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}e^{i\phi} \end{pmatrix}

If we eliminate fractions, the recommended practices with eigenvector, we have:

ψ±=(acosθ2asinθ2eiϕ)\psi_\pm = \begin{pmatrix}a\cos\frac{\theta}{2}\\ a\cdot\sin\frac{\theta}{2}e^{i\phi} \end{pmatrix}

When we eliminate the common terms, aa, it becomes:

ψ±=(cosθ2sinθ2eiϕ)\psi_\pm = \begin{pmatrix}\cos\frac{\theta}{2}\\ \sin\frac{\theta}{2}e^{i\phi} \end{pmatrix}

Linearly transforming the vector, we have:

ψ=cosθ20+eiϕsinθ21\boxed{\psi = \cos\frac{\theta}{2}\ket{0} + e^{i\phi}\sin\frac{\theta}{2}\ket{1}}

which represents the qubitqubit state on the Bloch sphere.

Give the set of all values θ\theta for which the following pairs of states are equivalent:

  • a. 1\ket{1} and 12(++eiθ)\frac{1}{\sqrt{2}} \left( \ket{+} + e^{i\theta}\ket{-} \right)
  • b. 12(i+eiθi)\frac{1}{\sqrt{2}} \left( \ket{i} + e^{i\theta}\ket{-i} \right) and 12(i+eiθi)\frac{1}{\sqrt{2}} \left( \ket{-i} + e^{-i\theta}\ket{i} \right)
  • c. 120321\frac{1}{2}\ket{0} - \frac{\sqrt{3}}{2}\ket{1} and eiθ(120321)e^{i\theta} \left( \frac{1}{2}\ket{0} - \frac{\sqrt{3}}{2}\ket{1} \right)

Solution

(a) Given:

1,12(++eiθ)\ket{1}, \quad \quad \frac{1}{\sqrt{2}} \left( \ket{+} + e^{i\theta}\ket{-} \right)

To solve this problem, we need to ensure that both states are in the same measurement basis. The first state is in the computational basis while the second is in Hadamard basis. It will be simpler to have both in the computational basis (or, if you want, Hadamard basis). We will transform the state in the Hadamard basis to computational basis in this solution.

From previous article, we know that:

+=12(0+1),=12(01)\ket{+} = \frac{1}{\sqrt{2}} (\ket{0} + \ket{1}), \quad \quad \ket{-} = \frac{1}{\sqrt{2}} (\ket{0} - \ket{1})

Therefore, substitute these into the second state:

12(++eiθ)=12(12(0+1)+eiθ12(01))=120+121+eiθ20eiθ21=1+eiθ20+1eiθ21\begin{align*} \frac{1}{\sqrt{2}} (\ket{+} + e^{i\theta}\ket{-}) &=\frac{1}{\sqrt{2}} \left( \frac{1}{\sqrt{2}} (\ket{0} + \ket{1}) + e^{i\theta} \frac{1}{\sqrt{2}} (\ket{0} - \ket{1})\right)\\ &=\frac{1}{2}\ket{0} + \frac{1}{2}\ket{1} + \frac{e^{i\theta}}{2}\ket{0}-\frac{e^{i\theta}}{2}\ket{1}\\ &= \frac{1 + e^{i\theta}}{2} \ket{0} + \frac{1 - e^{i\theta}}{2} \ket{1} \end{align*}

Now, we can compare both states since we want angles with which they are equivalent:

11+eiθ20+1eiθ21\ket{1} \equiv \frac{1 + e^{i\theta}}{2} \ket{0} + \frac{1 - e^{i\theta}}{2} \ket{1}

Here, we see that for both states to be equivalent, the coefficient of 0\ket{0}, in the left hand side, must be 00 and that of 1\ket{1} must be 11. So, going by the second option:

1=1eiθ22=1eiθeiθ=12=1\begin{align*} 1 &= \frac{1 - e^{i\theta}}{2}\\ 2 &= 1 - e^{i\theta}\\ e^{i\theta} &= 1-2 = -1 \end{align*}

From Euler's formula , eiθ=cosθ+isinθe^{i\theta}=\cos\theta + i\sin\theta.

\therefore

cosθ+isinθ=1\cos\theta + i\sin\theta = -1

Looking at this equation, we can identify that it's complex where cosθ\cos\theta and 1-1 are real while sinθ\sin\theta and 00 are imaginary. An important property of complex numbers is that:

For two complex numbers to be equal, their real and imaginary parts must be equal.

So,

cosθ=1,sinθ=0\cos\theta = -1, \quad \sin\theta = 0

Let's take a look at a sample sine and cosine graph.

Cosine and Sine graph for various angles
Figure 3: Cosine and Sine graph for various angles plotted together

We can clearly deduce that:

cosθ=1whenθ={[π,π],[3π,3π],[5π,5π],...}\cos\theta = -1 \quad \text{when} \quad \theta=\{[-\pi,\pi],[-3\pi,3\pi],[-5\pi,5\pi],...\}

Generally, if nn is any integer, then:

cosθ=1whenθ=(2n+1)π=π+2πn\cos\theta = -1 \quad \text{when} \quad \theta=(2n+1)\pi = \pi + 2\pi n

For sin\sin,

sinθ=0whenθ={[0],[π,π],[2π,2π],[3π,3π],[4π,4π],[5π,5π]...}\sin\theta = 0 \quad \text{when} \quad \theta=\{[0],[-\pi,\pi],[-2\pi,2\pi],[-3\pi,3\pi],[-4\pi,4\pi],[-5\pi,5\pi]...\}

Generally,

sinθ=0whenθ=0+πnorθ=πn\sin\theta = 0 \quad \text{when} \quad \theta=0+\pi n \,\, \text{or} \,\, \theta=\pi n

To account for both cases, π+2πn\pi + 2\pi n is chosen as this satisfies sinθ\sin\theta too (albeit forfeiting some values). Hence, both states will be equivalent when:

θ=π+2πn\theta= \pi + 2\pi n

(b) We want to find the values of θ\theta for which the following pair of states are strictly equal:

12(i+eiθi)and12(i+eiθi)\frac{1}{\sqrt{2}} \left( \ket{i} + e^{i\theta}\ket{-i} \right) \quad \text{and} \quad \frac{1}{\sqrt{2}} \left( \ket{-i} + e^{-i\theta}\ket{i} \right)

To do this, we transform both states to the computational basis — in 0\ket{0} and 1\ket{1}. We recall that:

i=12(0+i1),i=12(0i1)\begin{equation} \tag{1} \ket{i} = \frac{1}{\sqrt{2}}(\ket{0} + i\ket{1}), \quad\quad \ket{-i} = \frac{1}{\sqrt{2}}(\ket{0} - i\ket{1}) \end{equation}

Substituting (1) into the state expressions, we get:

12(i+eiθi)=12(12(0+i1)+eiθ12(0i1))=12[(1+eiθ)0+i(1eiθ)1]\frac{1}{\sqrt{2}}\left(\ket{i} + e^{i\theta}\ket{-i}\right) = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}(\ket{0} + i\ket{1}) + e^{i\theta}\frac{1}{\sqrt{2}}(\ket{0} - i\ket{1})\right) = \frac{1}{2}\left[(1 + e^{i\theta})\ket{0} + i(1 - e^{i\theta})\ket{1}\right] 12(i+eiθi)=12(12(0i1)+eiθ12(0+i1))=12[(1+eiθ)0+i(1+eiθ)1]\frac{1}{\sqrt{2}}\left(\ket{-i} + e^{-i\theta}\ket{i}\right) = \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}(\ket{0} - i\ket{1}) + e^{-i\theta}\frac{1}{\sqrt{2}}(\ket{0} + i\ket{1})\right) = \frac{1}{2}\left[(1 + e^{-i\theta})\ket{0} + i(-1 + e^{-i\theta})\ket{1}\right]

For the two states to be strictly equal, their coefficients in the computational basis must be equal:

12[(1+eiθ)0+i(1eiθ)1]=12[(1+eiθ)0+i(1+eiθ)1]\frac{1}{2}\left[(1 + e^{i\theta})\ket{0} + i(1 - e^{i\theta})\ket{1}\right] = \frac{1}{2}\left[(1 + e^{-i\theta})\ket{0} + i(-1 + e^{-i\theta})\ket{1}\right]

Multiplying by 2:

(1+eiθ)0+i(1eiθ)1=(1+eiθ)0+i(1+eiθ)1(1 + e^{i\theta})\ket{0} + i(1 - e^{i\theta})\ket{1} = (1 + e^{-i\theta})\ket{0} + i(-1 + e^{-i\theta})\ket{1}

Equating the coefficients of 0\ket{0}:

1+eiθ=1+eiθ1 + e^{i\theta} = 1 + e^{-i\theta} eiθ=eiθe^{i\theta} = e^{-i\theta}

Using Euler's formula eix=cosx+isinxe^{ix} = \cos x + i\sin x:

cosθ+isinθ=cosθisinθisinθ=isinθ2isinθ=0sinθ=0\begin{align*} \cos\theta + i\sin\theta &= \cos\theta - i\sin\theta\\ i\sin\theta &= -i\sin\theta\\ 2i\sin\theta &= 0\\ \sin\theta &= 0 \end{align*}

From the previous solution, we know that:

sinθ=0whenθ=0+πnorθ=πn\sin\theta = 0 \quad \text{when} \quad \theta=0+\pi n \,\, \text{or} \,\, \theta=\pi n

Now to 1\ket{1}, let's equate its coefficients:

i(1eiθ)=i(1+eiθ)1eiθ=1+eiθ1+1=eiθ+eiθ2=eiθ+eiθ2=(cosθ+isinθ)+(cosθisinθ)2=2cosθcosθ=1\begin{align*} i(1 - e^{i\theta}) &= i(-1 + e^{-i\theta})\\ 1 - e^{i\theta} &= -1 + e^{-i\theta}\\ 1 + 1 &= e^{i\theta} + e^{-i\theta}\\ 2 &= e^{i\theta} + e^{-i\theta}\\ 2 &= (\cos\theta + i\sin\theta) + (\cos\theta - i\sin\theta) \\ 2 &= 2\cos\theta\\ \cos\theta &= 1 \end{align*}

For the two states to be strictly equal, both conditions must be met simultaneously:

sinθ=0andcosθ=1\sin\theta = 0 \quad \text{and} \quad \cos\theta = 1

The values of θ\theta for which sinθ=0\sin\theta = 0 are θ=nπ\theta = n\pi,
and the values of θ\theta for which cosθ=1\cos\theta = 1 are θ=2πn\theta = 2\pi n, where nn is an integer (nZn \in \mathbb{Z}).

For both conditions to be true, θ\theta must be an integer multiple of 2π2\pi.

θ=2πn,where nZ\theta = 2\pi n, \quad \text{where } n \in \mathbb{Z}

If we restrict nn to non-negative integers (K{0,1,2,...}K \in \{0, 1, 2, ...\}), the solution is:

θ=2πn,where nN\theta = 2\pi n, \quad \text{where } n \in \mathbb{N}

(c) Given:

ψ1=120321,ψ2=eiθ(120321)\ket{\psi_1} = \frac{1}{2}\ket{0} - \frac{\sqrt{3}}{2}\ket{1}, \quad \ket{\psi_2} = e^{i\theta} \left( \frac{1}{2}\ket{0} - \frac{\sqrt{3}}{2}\ket{1} \right)

Directly, we can deduce that:

ψ2=eiθψ1\ket{\psi_2} = e^{i\theta}\ket{\psi_1}

This is regarded as global phase equivalence ( Viamontes, G. F., Markov, I. L., & Hayes, J. P., 2007 Viamontes, G. F., Markov, I. L., & Hayes, J. P. (2007). Checking equivalence of quantum circuits and states ). This is the case since both states only differ in phase. It means that for all values θ\theta, both states remain the same.

What are θ\theta and ϕ\phi for each of the states +\ket{+}, \ket{-}, i\ket{i}, and i\ket{-i}?

Solution

Before going into specifics, let's lay out some foundations.

We can recall that a single qubitqubit state in the computational basis is of the form:

ψ=α0+β1\ket{\psi} = \alpha\ket{0} + \beta\ket{1}

It has this Bloch sphere state function:

ψ=cosθ20+eiϕsinθ21\ket{\psi} = \cos\frac{\theta}{2}\ket{0} + e^{i\phi}\sin\frac{\theta}{2}\ket{1}

Equating both, we have:

α0+β1=cosθ20+eiϕsinθ21\alpha\ket{0} + \beta\ket{1} = \cos\frac{\theta}{2}\ket{0} + e^{i\phi}\sin\frac{\theta}{2}\ket{1}

For both sides to be equal, it means:

cosθ2=α,andeiϕsinθ2=β\begin{equation} \tag{f1} \cos\frac{\theta}{2} = \alpha, \quad \text{and} \quad e^{i\phi}\sin\frac{\theta}{2}=\beta \end{equation}

where θ[0,π]\theta \in [0, \pi] and ϕ[0,2π]\phi \in [0, 2\pi].

Now to the specifics.

(i) +\ket{+} can be expressed in computational basis as:

+=120+121\ket{+} = \frac{1}{\sqrt{2}}\ket{0} + \frac{1}{\sqrt{2}}\ket{1}

Here, α=β=12\alpha=\beta=\frac{1}{\sqrt{2}}. Substituting into (f1f1), we have:

cosθ2=12θ=2cos1(12)θ=2π4θ=π2\begin{align*} \cos\frac{\theta}{2} &= \frac{1}{\sqrt{2}}\\ \theta &= 2\cdot\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\\ \theta &= 2\cdot \frac{\pi}{4}\\ \theta &= \frac{\pi}{2} \end{align*}

To solve for ϕ\phi:

eiϕsinθ2=12(cosϕ+isinϕ)sinπ22=12(cosϕ+isinϕ)sinπ4=12(cosϕ+isinϕ)12=12cosϕ+isinϕ=1\begin{align*} e^{i\phi}\sin\frac{\theta}{2} &= \frac{1}{\sqrt{2}}\\ (\cos\phi + i\sin\phi)\sin\frac{\frac{\pi}{2}}{2} &= \frac{1}{\sqrt{2}}\\ (\cos\phi + i\sin\phi)\sin\frac{\pi}{4} &= \frac{1}{\sqrt{2}}\\ (\cos\phi + i\sin\phi)\frac{1}{\sqrt{2}} &= \frac{1}{\sqrt{2}}\\ \cos\phi + i\sin\phi &= 1 \end{align*}

We can now equate both real and imaginary parts:

cosϕ=1andsinϕ=0ϕ=cos1(1)andϕ=sin1(0)ϕ=0andϕ=0\begin{align*} \cos\phi = 1 \quad &\text{and} \quad \sin\phi = 0\\ \phi = \cos^{-1}(1) \quad &\text{and} \quad \phi = \sin^{-1}(0)\\ \phi = 0 \quad &\text{and} \quad \phi = 0 \end{align*}

\therefore

+\ket{+} will be represented on the Bloch sphere with θ=π2,ϕ=0\theta=\frac{\pi}{2},\phi = 0.

With this, I will leave those of \ket{-}, i\ket{i}, and i\ket{-i} as exercise. It'll be fun!

Note:Hint

In all, θ=π2\theta=\frac{\pi}{2} while ϕ\phi varies.

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References

  1. Bloch, F. (1946). Nuclear Induction. Physical Review. https://doi.org/10.1103/PhysRev.70.460
  2. Glendinning, I. (2005). The Bloch Sphere. [Lecture notes]. Pdx University. https://web.cecs.pdx.edu/~mperkows/june2007/bloch-sphere.pdf
  3. Rieffel, E. G. & Polak, W. H. (2014). Quantum Computing: A Gentle Introduction. MIT Press. https://books.google.com/books?id=CQ3YoAEACAAJ
  4. UV Physics. (2023). Bloch sphere and Qubit Representation. https://www.youtube.com/watch?v=nVpj3_NOvRI.. Retrieved April 19, 2025
  5. Viamontes, G. F., Markov, I. L., & Hayes, J. P. (2007). Checking equivalence of quantum circuits and states..
  6. Wikipedia. (2025). List of trigonometric identities. https://en.wikipedia.org/wiki/List_of_trigonometric_identities.. Retrieved April 20, 2025
  7. Zettili, N. (2009). Quantum Mechanics: Concepts and Applications..