# What are the I, X, Z gates in quantum gates? [closed]

Can someone please explan how the $$\rm I$$, $$\rm X$$ and $$\rm Z$$ gates work?

If $$\rm{I = X^2 = Z^2}$$, can you explain why this is the case or why it wouldn't work?

• What do you mean by "how do they work"? Can you be more specific? Also, what does their involution have anything to do with their ability to "work"? – keisuke.akira Aug 11 '20 at 7:47

$$I$$ is identity operator, which means that input state is not affected. In mathematical notation: $$I|\psi\rangle = |\psi\rangle$$.
$$X$$ operator is a negation. It changes 0 to 1 and conversely, i.e. $$X|0\rangle = |1\rangle$$ and $$X|1\rangle = |0\rangle$$. If it is applied on a qubit in arbitrary superposition, i.e. $$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$$, the state changes to $$X|\psi\rangle = \beta|0\rangle + \alpha|1\rangle$$.
$$Z$$ operator is a little bit more difficult to understand for beginners. The operator changes a phase of qubit. Consider for example state $$|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$. This state is an equally distributed superposition of state $$|0\rangle$$ and $$|1\rangle$$. When you apply $$Z$$ operator, you get a state $$|\psi\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$$. This is also an equally distributed superpositon of $$|0\rangle$$ and $$|1\rangle$$, however, the phase changed as you can see from minus sign before $$|1\rangle$$. If it is applied on a qubit in arbitrary superposition, i.e. $$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$$, the state changes to $$Z|\psi\rangle = \alpha|0\rangle - \beta|1\rangle$$. Again, only the phase changes.
An identity $$I = X^2 = Z^2$$ can be easily verified by direct multiplication of matrix representations of the operators: $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$$ and $$Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$
• A small addition: you may also find it intuitive to think of $X^{2} = Z^{2} = I$ like this: if $X$ flips the $|0\rangle$ to the $|1\rangle$ state and vice-versa, then $X^{2} = X*X$ just does the flip twice, which is equal to doing nothing (i.e. $I$ operation). The same reasoning applies to the $Z$ gate. – JSdJ Aug 11 '20 at 10:36