Decomposing the Hadamard gate (advanced)
As we know from our precedent article Introduction to quantum logic gates, conceptually quantum computing operations manipulates Φ and θ of the superposition to move points along the surface of the Bloch unit sphere. In the last article, even if we pointed out the position of the qubit after being applied the Hadamard gate, we did not precise the equivalent operations, in terms of rotations.
Say we want to reuse our Pauli-X gate introduced at the beginning of our previous article, then our Hadamard operation could be decomposed as below:
- rotate our up vector |0> of 180 degrees or π radians around the X-axis (Pauli-X gate)
- rotate our vector of -π/2 radians around the Y axis
If we had to rotate the vector of π radians around the Y axis, it would be an easy task: we would only need to apply the Pauli-Y matrice, which by definition, equates to a rotation around the Y-axis of the Bloch sphere by π radians (it maps |0> to i|1> and |1> to -i|0>). So how can we do?
Actually, more generally, the Pauli- X, Y and Z matrices are so-called because when they are exponentiated, they give rise to the rotation operators, which rotate the Bloch vector about the x, y and z axes, by a given angle θ. These rotation operators are defined as per below:
Now, if a operator A satisifies A2 = I, it can be shown that:
And since all the Pauli matrices satisfy X2 = Y2 = Z2 = I, the rotation operator can be expanded as:
If we apply the X gate first, then the intial |0> qubit is mapped to |1> and ends up in the south pole position (φ=0, θ=180) of the Bloch sphere.
To bring it it back to the equator position, we have to do a rotation of -π/2 radians around the y-axis. This is an easy task once we have the expression of a π/2 rotation along the y-axis, which is in our case Ry(π/2), as per above. We only need to take the hermitian conjugate of the previous matrice: this has for effect to rotate of an angle of opposite value.
Let's first calculate Ry(π/2), the π/2 rotation along the y-axis:
Y-1/2 is said to be the hermitian conjugate of Y1/2 and is the complex conjugate of Y1/2 transposed matrix.
Checking that Y-1/2X = H
We are just left at this point with calculating the product of these two matrices
Remark: we take the opportunity here to undertline that the Hadamard gate, like every observable in quantum mechanics, represent an hermitian operator: it is equal to its own conjugate transpose.
 Not all quantum devices can realize all quantum gates. In practice this is not an issue since any quantum gate can be constructed from a series of universal quantum gates provided the device can execute them. The downside is that the combination of gates takes longer to perform and hence introduces a higher error rate. For example on spin-qubit devices the CNOT gate is not directly available.