What is the meaning of spin of an particle is $1/2$ and $2$ or something? On which factor does these spin no. depend?

Question

I have read a book. The writer had written that if the spin of an particle is $\frac{1}{2}$, then we have to rotate it at $720$ degree. Imagine that there are two balls joined. Then we have to rotate the two balls first through $180$ degrees then we will reach at the joining point of two balls. Then again rotate it through $180$ degrees.Then we will reach at end point of second ball. Like this we have to rotate it through $720$ degrees. The writer has also written that if spin is $0$ then it will be like a sphere. I can't understand this all. I am fifteen years old. So my questions can be silly with respect to other members. Please tell me about these.

Answer 1

The spin indicates the length $(=2s+1)$ of the vector that a real world particle rotates like. They do not all rotate like pencils (3-vectors). Your questions are not silly!

Part of Quantum mechanics involves 1) making a correspondence between a symbol (a |ket>) that you write on piece of paper and an object in the real world, and 2) making a correspondence between linear transformations done to the |ket> and the actual physical transformations you do to the object in the real world. Rotations are one of the transformations you can do to an object in the real world. For example, a pencil in the real world corresponds to 3-vector on a piece of paper which you rotate with a 3x3 matrix. As an example, here is the matrix which rotates the 3-vector about the z-axis being applied to a pencil pointing in the x-direction: $$ R(\theta_z)\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} =\begin{bmatrix} cos(\theta_z) & -sin(\theta_z) & 0 \\ sin(\theta_z) & cos(\theta_z) & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} $$ The product of any two rotations is another rotation. Every rotation has an inverse (ie: do it by a negative angle). There is also an identity transformation (ie: rotate by 0 degrees). Therefore, rotations form a group. Now, it happens, there are matrices with dimension other than 3 that satisfy the same group multiplication law (ie: which two rotations multiplied together make which single rotation). For example, here is the 2x2 matrix that rotates a 2-vector about the z-axis being applied to a $s_z=- {1\over 2}$ state of spin 1/2 particle: $$ R(\theta_z)\begin{bmatrix} 1 \\ 0 \end{bmatrix} =\begin{bmatrix} e^{-i {{\theta_z}\over 2}} & 0 \\ 0 & e^{i {{\theta_z}\over 2}} \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} $$ Notice that this matrix does not operate on a vector in x,y,z space. Instead it operates on an abstract 2-vector with components labelled by $s_z=- {1\over 2}, {1\over 2}$ that stands for the particle. The dimension of the vector (and the matrix) is labelled by the spin s, where the dimension=$2s+1$, and the vector component labels are $s_z=-s,-s+1,...0,...s-1,s$. Hence, the 2-vector is $s=1/2$ and the 3-vector is $s=1$. Examples of particles that transform under rotation as different dimension vectors are:

s=0.....(1-vector).....pion

s=1/2...(2-vector).....electron, muon, tau, neutrinos, proton, neutron

s=1.....(3-vector).....photon, rho meson

s=3/2...(4-vector).....delta baryon

s=2.....(5-vector).....graviton

and so on. There is a representation of rotation for every integer and half-integer s, and EVERY object in the real world transforms under rotation as some s … there are no exceptions.

If you further study the SU(2) rotation group you will find the 3 generators of the group correspond to angular momentum. The half-integer spin vectors are called spinors. It requires an angle of $4\pi$ radians to be put into the spinor rotation matrices to get the identity (as exemplified by the 2x2 matrix above) whereas it only takes a rotation of $2\pi$ radians to get an integer particle back to where it started.

Answer 2

The spin of a particle is a number that describes its angular momentum. The earth orbits the sun, making years- that is angular orbital momentum. The earth spins on its own axis, making days- that is angular rotational momentum

The spin of a particle is analogous to the latter of those two. Not exactly alike due to the quantum nature of spin, but ´same idea´. The differences arise partly from the nature of spin being measured within a 2d vector space(which will require additional reading to really make sense.)

The other part of your question, ´on what factor does these spin no. depend´ can be answered by arguing the opposite question:¨Upon the spin no. do what factors depend upon,¨ as that is the nature of fermions and bosons- the spin determines whether they are a fermion or boson and from this the properties of the particle arise. This is a tautological statement, but without going terribly in depth it can just be assumed.

Bosons(which have integer spins) are force carriers while Fermions(which have 1/2 integer spin) are the constituents of ordinary mass. If their spins were reversed, the properties would be very different from what we have observed.

To learn more about why that actually is, reading about the Pauli Exclusion principle and becoming familiar with quantum numbers will help. I think that with this information you will be able to go out and find the rest of the desired answers while actually gaining a familiarity with the basics of particle physics.