Cannot interpret transformations on the bloch sphere as matrix multiplications

0

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.

Jimarious

Posted 2020-05-11T17:18:15.920

Reputation: 53

Answers

2

First of all if you take at look at how the $X$ gate works:

$X|0\rangle = |1\rangle$

Now applying a $Y$ you get

$Y|0\rangle = i|1\rangle$ and $Y|1\rangle = -i|0\rangle$, 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 $|1\rangle$ and $|0\rangle$ 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 $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, $Y|\psi\rangle = i\alpha|1\rangle -i\beta|0\rangle $, where $|i\alpha|^2 + |-i\beta|^2 = 1$, 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 $|0\rangle$ and $|1\rangle$, you are only shifting the phase of $1\rangle$, 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 $H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$, 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 $|0\rangle$ or $|1\rangle$.

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:

$|i\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{i}{\sqrt{2}}|1\rangle$

$|o\rangle\ = \frac{1}{\sqrt{2}}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle$

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

Sam Palmer

Posted 2020-05-11T17:18:15.920

Reputation: 561