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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?

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    $\begingroup$ 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"? $\endgroup$ – keisuke.akira Aug 11 '20 at 7:47
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$I$ 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 $$ Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}. $$

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    $\begingroup$ 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. $\endgroup$ – JSdJ Aug 11 '20 at 10:36

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