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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.

Qmechanic
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Kshitij Patil
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2 Answers2

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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.

Gary Godfrey
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  • Spinors are not half-integer "spin vectors" (rather they are representations of the Dirac algebra) and I don't see what "the length $(2s+1)$ of the vector that a real world particle rotates like" is. – gented Oct 30 '15 at 15:30
  • He said he's 15. Linear transformations on kets? Fine answer maybe, but likely not so helpful to the asker... – anon01 Oct 30 '15 at 15:40
  • @Gennaro: Spinor refers to how something transforms. The even length vectors I talked about transform like spinors, and are the carrier space that your representations (ie: matrices) of the Dirac algebra act on. Under conjugation, an operator can also transform like a spinor or vector. For example, the Pauli spin matrices $\sigma_x, \sigma_y, \sigma_z$ transform like a 3-vector (spin=1), but they are 2x2 matrices which operate on 2-vectors (s=1/2 spinor). To stand for an electron, you write a little 2-vector that you rotate with a 2x2 representation of the rotation group. – Gary Godfrey Oct 30 '15 at 19:19
  • @ann0909: I hear you, and tried (perhaps failing) to be simple. Kets may have been an unnecessary word. Spin is an abstract concept that labels a property of a particle; it's not something like color, balls, and string that we are familiar with. Please add an answer to this question. It is an important one to make clear. – Gary Godfrey Oct 30 '15 at 19:32
  • @GaryGodfrey That was exactly my point: just because you may (accidentally, in some simple cases) perform the same operations on different things (that are in general not defined otherwise) this doesn't mean that one is allowed to define the two different things according to the way they (accidentally) happen to transform. In particular I still don't see how a 2-vector is a $s=1/2$ spinor (it may be a representation thereof, which is different). – gented Oct 30 '15 at 19:45
  • @Gennaro: There may be a difference in what we each mean by the word "representation". I mean, if $J_x,J_y,J_z$ are some abstract operators, then the 2x2 matrices $\sigma_x,\sigma_y,\sigma_z$ are a representation of the abstract operators because they obey the same commutation relations. There are other reps by matrices of dim (2s+1, 2s+1). In my meaning, a 2-vector does not "represent" a $s=1/2$ spinor. Instead, the 2-vector "transforms like" a $s=1/2$ spinor (ie: transforms by the 2x2 matrix rep of the rotation group $SU(2,C)$. You call something a spinor if it transforms like a spinor. – Gary Godfrey Oct 30 '15 at 21:20
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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.

MildBubby
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