# Cannot interpret transformations on the bloch sphere as matrix multiplications

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:

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

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