Benefit of phase shift in quantum computing

3

1

I am new to quantum computing.

I compare Pauli-X gate and Pauli-Y gate as equivalent to NOT gate in classical computers. Though I am not very sure when to use Pauli-X and Pauli-Y gates as the result only differ in global phase.

As per my understanding phase shift gates (like Pauli-Z gate) doesn't change the amplitude of the qubit but only changes the relative phase of the qubit on the bloch sphere.

I am unable to understand the benefit of phase change (global - like in Pauli-Y and relative - like in Pauli-Z gate) in quantum computer? I assume it is only the amplitude which determines the observed state during measurement of qubit.

Abhinav

Posted 2020-10-07T17:58:57.197

Reputation: 209

Answers

4

Note that:

$$ X |0\rangle = |1\rangle \hspace{1 cm} X|1\rangle = |0\rangle $$

but

$$ Y |0\rangle = i|1\rangle \hspace{1 cm} Y|1\rangle = -i|0\rangle $$

So $X \neq iY$. In fact, the set $\{I, X, Y, Z\}$ is an orthogonal basis set for $2 \times 2$ matrices. They are not just some factor off from each other. They are independent from one another.


As for overall phase on quantum gate, $U$ v.s $e^{i\theta}U$ , these two are indistinguishable as long as you don't do any control operation. That is, controlled-$U$ is not the same as controlled-$e^{i\theta }U$ . This is because the phase of the target qubit can be kicked back to the controlled-qubit and it is essentially how quantum phase estimation algorithm works.

KAJ226

Posted 2020-10-07T17:58:57.197

Reputation: 6 322

2

You can consider X and Z gates as "inversion" in computational basis and circular and Hadamard bases, respectively.

Lets start with X. It holds that $$ X|0\rangle = |1\rangle\,\,\,\,\,\,\ X|1\rangle = |0\rangle, $$ so X is analogical to classical negation, i.e. it converts 0 to 1 and conversely.

Instead of computational basis $\{|0\rangle, |1\rangle\}$, you can express qubits as combination of members of Hadamard basis $\{|+\rangle, |-\rangle\}$, where $$ |+\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle) \\ |-\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle) $$

You can verify that $$ Z|+\rangle = |-\rangle\,\,\,\,\,\,\ Z|-\rangle = |+\rangle, $$

Circular basis is composed of $\{|\uparrow\rangle, |\downarrow\rangle\}$ (note that I was not able to find proper symbol for circular arrows), where $$ |\uparrow\rangle = \frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle) \\ |\downarrow\rangle = \frac{1}{\sqrt{2}}(|0\rangle-i|1\rangle) $$

You can again verify that $$ Z|\uparrow\rangle = |\downarrow\rangle\,\,\,\,\,\,\ Z|\downarrow\rangle = |\uparrow\rangle, $$

All Pauli gates also define rotation around axes x, y and z. Consider $A \in \{X,Y,Z\}$ then rotation by angle $\theta$ around axis $a \in \{x,y,z\}$ is defined as $$ R_a(\theta) = \mathrm{e}^{-i\frac{\theta}{2}A} $$ Note that the exponential is so-called matrix exponential.

Martin Vesely

Posted 2020-10-07T17:58:57.197

Reputation: 7 763