Determining the spin of wavefunction

Question

We all know that by uncertainty principle, location of a wave-particle is perfectly determined when uncertainty of momentum becomes infinite. (I also heard that in reality, it is almost impossible to determine the location of a wave-particle perfectly and the location has some uncertainty.) So, what happens to spin? Spin can be superpositioned, so I am wondering what happens to spin.

Answer 1

Much like position is the operator $\hat x$ that has a continuous spectrum and whose predicted value has some uncertainty, the spin is the triplet of operators $(\hat J_x,\hat J_y,\hat J_z)$ that have a discrete spectrum (each of them) and they don't commute with other components, for example $$ [J_x,J_y] = i\hbar J_z $$ which means that two (or three components) of $\hat J$ cannot be measured at the same moment (unless the whole vector is zero) and we may derive the inequality: $$\Delta J_x \cdot \Delta J_y \geq \frac\hbar 2 \langle J_z \rangle$$ Moreover, the eigenvalues of $\hat J_z$ or any other components are discrete numbers from $-j\hbar$ through $+j\hbar$ where $j$ is an integer or half-integer multiple of $\hat J^2$ and the spacing is $\hbar$.

In the units of $\hbar$, each component is guaranteed to be measured as integer-valued or half-integer-valued but when one component is measured, the other components are completely uncertain.

Spin is a basic part of elementary quantum mechanics so see a textbook or e.g.

Spin at Wikipedia