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Cannot interpret transformations on the bloch sphere as matrix multiplications

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.

One Answer

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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