Rotations in Bloch sphere

4 min readJan 14, 2023

In this post, I will try to explain about the rotational operators. They are extremely significant in quantum computation. Why are they so significant? Well, any Qubit can be represented as a point on Bloch sphere. We can transform any Qubit to a target Qubit using just three rotations. Any arbitrary single qubit unitary operator can be written as

U = exp(iθ)Rz(α)Ry(β)Rz(γ)

How to understand this intuitively? Let’s start that exercise by ignoring the global phase factor exp(iθ). So, our U becomes the following

U = Rz(α)Ry(β)Rz(γ)

If we apply this unitary operator on a Qubit, it will first get rotated by γ about Z axis, followed by a rotation of β about Y axis and then a rotation of α about Z axis.

The above image is a cross section of Bloch sphere(It is convenient to think of bloch sphere as earth and grey lines as latitudes. However rotation about Y axis can’t be thought as traversing longitudes as all of them pass through poles). Rotation about Z axis is equivalent to traversing grey line. Rotation about Y axis is equivalent to traversing the black line. So, applying U is equivalent to traversing grey line followed by black line followed by grey line. From the Qubit represented by blue circle, we can always imagine a path to any other point on the sphere just traversing grey, black and grey line.

Isn’t two rotations enough for any Qubit to get transformed to any Qubit on circle? No. The Qubit represented by yellow circle is a counter example. We can’t traverse grey line from that as it lies on pole of bloch sphere. So, we have to traverse black line to travel vertically along the sphere. To reach any point not on circle joining poles, we need a rotation about Z axis again. That is why we need three rotations.

Let us now discuss more about the rotation operators Rx(θ), Ry(θ), Rz(θ) and their derivations. These are expressed in terms of Pauli Matrices given by

Rotation about X axis is given by the following formula

Rx(θ) = cos(θ/2)I — i sin(θ/2)X

Rotation about Y axis is given by the following formula

Ry(θ) = cos(θ/2)I — i sin(θ/2)Y

Rotation about Z axis is given by the following formula

Rz(θ) = cos(θ/2)I — i sin(θ/2)Z

It took me an embarassingly long time to derive these equations. Derivations are done in the following sections

Derivation of Rx(θ)

Let’s assume our initial state was making an angle of θ with Z axis and we intend to rotate it by θ’ about X axis as shown in the figure

Initial vector as shown in blue is

i = cos(θ/2) |0> + i sin(θ/2) |1>

After rotation, the rotated vector as shown in green is

f = cos(θ/2+θ’/2)|0> + i (sin(θ/2 + θ’/2)

Expanding

f = cos(θ/2)cos(θ’/2)|0> — sin(θ/2)sin(θ’/2)|0>
+ i sin(θ/2)cos(θ’/2) |1> + i sin(θ’/2) cos(θ/2) |1>

taking common terms

f = cos(θ’/2) * (cos(θ/2) |0> + i sin(θ/2))
— sin(θ’/2) * (sin(θ/2) |0> -i cos(θ/2) |1>)

term in first brackets is initial vector i and term in second brackets is i times Rx of state vector i

Hence

Rx = cos(θ/2) I — i sin(θ/2) X

Derivation of Ry(θ)

Use the same approach as the one used for Rx(θ) but set the initial vector i to be a point on XZ plane. Let it be cos(θ/2) |0> + sin(θ/2) |1>

If we rotate the vector by θ’ about Y axis, the final vector would be

f = cos(θ/2 + θ’/2) |0> + sin(θ/2 + θ’/2)|1>

Expanding f, we get

f = cos(θ/2)cos(θ’/2) |0> — sin(θ/2)sin(θ’/2)|0> +
sin(θ/2)cos(θ’/2) |1> + cos(θ/2)sin(θ’/2) |1>

Taking cos(θ’/2) and sin(θ’/2) common, we get

f = cos(θ’/2)(cos(θ/2) |0> + sin(θ/2)|1>) -sin(θ’/2)(sin(θ/2)|0> — cos(θ/2)|1>)

This can be written as

f = [cos(θ’/2)I -sin(θ’/2)(iY)]*i

Derivation of Rz(θ)

Let’s assume our initial state was making an angle of θ with Z axis, ϕ with Y axis and we intend to rotate it by ϕ’ about Z axis as shown in the figure