1
$\begingroup$

How to find a ket $|\psi \rangle$ that illustrates how changing X to iX in the definition of the CNOT gate makes an important difference because of what happens when CNOT is applied to $|\psi \rangle$?

The definition of the CNOT gate is: $$CNOT = |0\rangle\langle 0| \otimes I + |1\rangle\langle 1| \otimes X $$

What is the physical meaning of this?

$\endgroup$

3 Answers 3

2
$\begingroup$

Assuming your question is "What $| \psi_i \rangle $ would have different resulting $| \psi_f \rangle $ having undergone a $C-X$ gate vs. a $C-iX$ gate?"

Consider $| \psi_i \rangle = \frac{1}{\sqrt2} \Big( |00 \rangle + |11 \rangle \Big)$

$$C-X | \psi_i \rangle = \frac{1}{\sqrt2} \Big( |00 \rangle + |10 \rangle \Big)$$ $$C-iX | \psi_i \rangle = \frac{1}{\sqrt2} \Big( |00 \rangle + i|10 \rangle \Big)$$

Obviously these two resulting states are different since there is a relative phase. To make the difference more clear you can try to consider measurements in the Y-basis. Just as a note a $C-U$ gate is a "Controlled $U$ gate", and often times you will find $CNOT$ written as $C-X$ to be consistent with $C-H$ (Controlled Hadamard), $C-Z$ (Controlled Z gate), etc.

The Physical meaning of this is that adding a global phase to a single Qubit Gate on a multi Qubit system results in a change in the multi Qubit system. Notice how if we only had a single Qubit, there would be no way of knowing the difference between $X$ and $iX$, because it would be a global phase; however in a 2 Qubit system, that phase shows itself as a relative phase which indeed is measurable.

$\endgroup$
1
$\begingroup$

These two gates only differ by a local unitary, namely $$ \begin{pmatrix}1&0\\ 0&i\end{pmatrix} $$ on the control qubit. Thus, there is no important difference between them, at least in terms of their entangling power and and non-local properties - they are exactly equivalent in that respect. But of course, it depends what you deem an important difference, which you would have to clarify in your question.

$\endgroup$
0
$\begingroup$

Gates $X$ and $iX$ differs in global phase which is $\pi/2$ (since $i = e^{i\frac{\pi}{2}}$). As two states (gates) which differ in global phase only cannot be distinguished, there is no difference between $X$ and $iX$.

However, the difference appears in controlled versions of the gates. Controlled $iX$ is composed of controlled $X$, i.e. CNOT, and controlled global phase which is in this case described by gate $U1(\pi/2) \otimes I$ where $$ U1= \begin{pmatrix} 1 & 0 \\ 0 & e^{i\frac{\pi}{2}} \end{pmatrix} $$ So $C-iX = \text{CNOT} (U1(\pi/2) \otimes I)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.