Quantum Computing Asked by Jimarious on July 27, 2021
I understand that X,Y and Z gates are rotations around the axes with the respective letters, but I cannot understand how can Y gate multiply the amplitude of 0 with unreal number and have it landing on the bloch sphere, or how can S gate add a phase of 90 degrees.
First of all if you take at look at how the $X$ gate works:
$X|0rangle = |1rangle$
Now applying a $Y$ you get
$Y|0rangle = i|1rangle$ and $Y|1rangle = -i|0rangle$, so you can see that you are flipping the state of the qubit, i.e. an X rotation with a phase rotation (you can also see this from the commutor relation $[X,Z] = XZ - ZX =2iY $). In the case of the pure states $|1rangle$ and $|0rangle$ you can see that it ends up in another pure state, and as such the phases, $i$ and $-i$ applied by the $Y$ gates can be treated a global phase and in these cases 'ignored' when taking a measurement, you will always be measuring with probability $1$ the state that you are in.
Now in the more general case consider a state $|psirangle = alpha|0rangle + beta|1rangle$, $Y|psirangle = ialpha|1rangle -ibeta|0rangle $, where $|ialpha|^2 + |-ibeta|^2 = 1$, when measuring these states the factor of $i$, where $|i^2|=1$ can be ignored. However we should always keep track of phases as in mixed states they can't be ignored as they impact the probability of measurement.
Again when applying a phase gate to $|0rangle$ and $|1rangle$, you are only shifting the phase of $1rangle$, but this doesn't change the probability of measuring the state.
So what about the $H$ gate, this is a combination of $Z$ and $Y$ rotations, and takes $H|0rangle = frac{1}{sqrt{2}}(|0rangle + |1rangle)$, in this case we can't ignore the phase $frac{1}{sqrt{2}}$, because $|frac{1}{sqrt{2}}|^2 = frac{1}{2}$, and changes the measurement probability such that it is 50/50 measuring either $|0rangle$ or $|1rangle$.
As a side, to visualise the poles of the $Y$ axis, in (into the screen) and out (out of the screen), they are given by:
$|irangle = frac{1}{sqrt{2}}|0rangle + frac{i}{sqrt{2}}|1rangle$
$|orangle = frac{1}{sqrt{2}}|0rangle - frac{i}{sqrt{2}}|1rangle$
so on the Bloch Sphere applying a $Y$ gate to either of these poles flips between them.
I would recommend watching Prof Shor explain this better than me https://courses.edx.org/courses/course-v1:MITx+8.370.1x+1T2018/courseware/Week2/lectures_u1_3/?child=first
Answered by Sam Palmer on July 27, 2021
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