rede96 said:
As I mentioned before I have no physics background and only basic Math. Sorry.
Quantum physics is, since its discovery, less intuitive relative to human being. Many physical phenomenon that take place in quantum level exhibits nature which seems to contradict our experience related to physical occurrences which are directly sensible, which we call classical physics. This may be the reason why the concepts and principles in quantum physics can be hard to visualize for beginners. In "Schroedinger picture" of quantum mechanics, the quantum state of a system is described by a
wavefunction, this object is introduced as a vector in linear algebra. Apart from wavefunction, we also use
operators to translate physical quantities like position, momentum, angular momentum, and energies into quantum mechanical language. With these two mathematical objects being essential in learning quantum mechanics, it's indeed difficult to comprehend a topic in this subject of physics without having any background in physics and math. To those without math and physics, quantum mechanics sounds just like a fairy tale.
As pointed above, the quantum state of a system is described by a wavefunction. When one considers system of one electron with spin and neglecting the spatial degree of freedom, the wavefunction of this system is a vector in 2D complex vector space, i.e. any arbitrary wavefunction ##|\psi\rangle## can be written as
$$
|\psi\rangle = c_1 |u_1\rangle + c_2 |u_2\rangle
$$
where ##|u_1\rangle## and ##|u_2\rangle## are basis vectors in this space and ##c_1## and ##c_2## are complex numbers. Born's rule states that the probability of finding this system in state ##|u_i\rangle## is given by ##|c_i|^2##. Now, considering an operator called
y component of spin ##S_y##, one can find the eigenvectors of this operator and having found these vectors denote them by ##|y;+\rangle## (spin up in y direction) and ##|y;-\rangle## (spin down in y direction). These vectors are usable as basis vectors for the space being considered. Therefore, any spin state of one electron system can be written as
$$
|\psi\rangle = c_{1y} |y;+\rangle+ c_{2y} |y;-\rangle
$$
This is why when you prepare the system in spin up along y-axis and measure the state using a device aligned at zero degree with respect to y axis, you got 100 % chance of measuring ##|y;+\rangle##. This is so because ##c_{1y} = 1## and ##c_{2y} = 0##.
Now spin is a vector operator, beside ##S_y## there are also ##S_x## and ##S_z## operators. When the measuring device is rotated 90 degree from the initial axis, it can represent either ##S_x## or ##S_z##. Take it to be ##S_z##, then the measurement will yield either ##|z;+\rangle## (spin up in z direction) or ##|z;-\rangle## (spin down in z direction). In turn, you can also use these last two vectors as a new basis and hence ##|\psi\rangle = c_{1z} |z;+\rangle+ c_{2z} |z;-\rangle##. Your initial state is ##|y;+\rangle##, it can be proven (using math of course) that the following equation holds
$$
|y;+\rangle = \frac{1}{\sqrt{2}}|z;+\rangle + \frac{i}{\sqrt{2}}|z;-\rangle
$$
You see from above that the probability of your device, after a rotation of 90 degrees, to detect spin up in the new (z) direction is ##|\frac{1}{\sqrt{2}}|^2 = 0.5## (50%).
If the device was rotated 60 degree from y axis, the calculation gets slightly more complicated because it involves applying another operator called rotation operator ##\exp (i\frac{S_z}{\hbar} \frac{\pi}{3})##. Has any of what I have made as examples made sense to you (assuming you still have no background in math and physics)?