# Quantum Spin?

1. Oct 27, 2004

### Mk

What is quantum spin? I've read lots of things about it and describing it, I still don't quite understand it.

2. Oct 27, 2004

### misogynisticfeminist

From what I know, spin is not something which can be explained in entirely intuitive language. Its more of like a mathematical concept, but also part of physical reality.

3. Oct 27, 2004

### Mk

4 minutes! Must be a new record! Thanks for your information.

4. Oct 27, 2004

### dextercioby

The first lines of my QFT course contained these words:"Quantum field theory emerged from the unability of quantum mechanics to describe particles in a relativistic context".That is to say,quantum mechanics cannot provide a correct description of particles in a relativistic context.This thing is done in QFT

5. Oct 27, 2004

### dextercioby

,where spin emerges as the a natural RELATIVISTIC property of particles related to one of the fundamental requirements when "painting" a lagrangian action for the field,i.e. invarance to the Poincaré group (in the case of Standard Model QFT) described by means of the 10 parameters (3 for spatial boosts,3 for spatial rotations,and 4 for space-time unaccelerated translations).I would reccomend u to the good text found in Lewis H.Ryder's book:"Quantum field theory" Chapter 2 where u can find most of the answers.

6. Oct 27, 2004

### manesh

quantum spin is a quantum mechanical state. It has nothing to do with the spin in the real macroscopic Newtonian world. For instance electrons has 2 spin states (+1/2 and -1/2), that doesn't mean that the electron is rotating with different Angular Momentum directions!!! The angular momentum and the spin concept in quatum mechanics are abstract, you can't imagine the spin state of an electron.The name 'spin' has given to this quantum mechanical state during the development of the Quantum Mechanics..
cheers

7. Oct 27, 2004

### Coughlan

Spin is how you determine the quatum state of a particle. An interesting fact is that fermions or particles that have spins of even numbers like 1/2 3/2 etc, must rotate through 360 degress its quantum state is measurably different than when it started. If it rotates another 360 degrees its quantum state returns to where it started. So particles must 'rotate' a total of 720 degrees to make it back to the state they started in...spin is also denoted in units equal to .5(H-BAR) <-----don't know how ot do the equation thing. H-BAR = Planck's constant.

8. Oct 28, 2004

### Mk

Thanks Coughlan, misogynisticfeminist, and manesh, but isn't h-bar a unit? So you cannot visualize quantum spin?

9. Oct 28, 2004

### manesh

I just want to add the fact that quanum mechanical spin is different...there you can't assume an electron rotating unlike classical physics.....here the rotation means quantum mechanical transformation in Hilbert space...
c ya

10. Oct 28, 2004

### zefram_c

Not to contradict anything already said, but spin behaves in many ways as angular momentum. By that I mean that it is conserved as part of total angular momentum whenever an electron interacts with a photon or some other particle. Also its natural unit (Planck constant) carries dimension of angular momentum. Going back to the electron, it is indeed wrong to picture it as being in some rotation state - the way I think about it is that electrons (and other particles as well) have some internal degree of freedom that manifests itself like angular momentum in terms of quantization / projection.

11. Oct 28, 2004

### Fredrik

Staff Emeritus
In quantum mechanics, the states of a physical system are represented by rays in a complex Hilbert space. A ray is a subset of the Hilbert space that consists of vectors that "point in the same direction, but may have different lengths". If x is a vector and c a complex number, x and cx belong to the same ray.

Different observers would use different rays to represent the same state. E.g. if I'm using a wavefunction that represents a particle localized somewhere on my right (positive x axis), an observer that is rotated 180° will use a wavefunction that represents a particle localized somewhere on his left (negative x axis) to describe the same particle.

It is possible to define a function T that takes each ray that I would use to describe a system, to the ray that the rotated observer would use. If I say that the system is in the state represented by ray R, the rotated observer would say that the system is in the state represented by ray T(R). Let's call a function defined in this way a "symmetry transformation".

There's a theorem that was proven by Eugene Wigner, that says that it's possible to define a function U from the set of symmetry transformations into the set of operators on the Hilbert space, such that each U(T) is either linear and unitary or antilinear and antiunitary.

It is natural to assume that there exists a symmetry transformation for each member of the rotation group SO(3). When we do, Wigner's theorem leads us to the concept of spin.

A rotation can be parametrized by three numbers (e.g. Euler angles). Let's use the greek letter θ to represent these parameters, and let X(θ) be the rotation operator on $$\mathbb R^3$$ that corresponds to the parameters θ. Let T(X(θ)) be the symmetry transformation that corresponds to that particular rotation. Wigner's theorem tells us that there must exist a linear and unitary operator U(T(X(θ))), which I will write simply as U(θ) from now on, that tells us what vector a rotated observer would use to represent the same state. If I say that the state is represented by a ray that contains the vector x, the rotated observer would say that the state is represented by a ray that contains the vector U(θ)x.

The spin operators show up when we Taylor expand U(θ). The result of a Taylor expansion can be written as

$$U(\theta)=1-\frac{i}{2}\sum_{i,j}\omega_{ij}J^{ij}+\dots$$

where the J's are hermitean operators, the omegas are real numbers (that depend on θ), and the second and higher order terms have been omitted. J and omega are both anti-symmetric:

$$\omega_{ij}=-\omega_{ji}$$
$$J^{ij}=-J^{ji}$$

The zeroth order term is the identity operator on the Hilbert space if we choose to use parameters that have the property that X(0) is the identity operator on $$\mathbb R^3$$.

Since J is anti-symmetric, it only has three independent components. These can be identified with the spin operators:

$$J^1=J^{23}$$
$$J^2=J^{31}$$
$$J^3=J^{12}$$

It can be shown that the properties of the rotation group SO(3) imply that these three operators must satisfy the commutation relations

$$[J^i,J^j]=i\sum_k\epsilon_{ijk}J^k$$

(If you're not used to the epsilon notation, you may not recognize that these are the commutation relations for spin operators that you've seen lots of times before, but they are).

It should be clear from the above that spin is a quantum mechanical property that isn't easily explained in plain english. It should also be clear that spin is a property that is very closely related to rotations. One more thing that indicates this is that the orbital angular momentum operators defined by

$$\vec L=\vec x\times\vec p$$

satisfies the same commutation relations as the spin operators.

12. Oct 28, 2004

### dextercioby

To be very strict,actually to be correct,the description of states in the standard version of QM (i.e. the version with vectors and operators (due mostly to Dirac)) is achived not by rays in a complex Hilbert space,but by UNITARY RAYS IN THE COMPLEX HILBERT SPACE OF STATES ASSUMED SEPARABLE (THE SPACE).It's basically the core of the first principle.The definition of an unitary ray of a vector $$I\psi>$$ in such a space is
$$\displaystyle{{c I\psi>,c\in C,I\psi>\in H,\underline{IcI^2 = 1}}}$$
That's why it's called "unitary".Coz it preserves "lenght",defined by means of the sesquiliniar application form H to C called "scalar product".So the affirmation "A ray is a subset of the Hilbert space that consists of vectors that <<point in the same direction, but may have different lengths>>" is basically correct,but it doesn't apply to QM.

13. Oct 28, 2004

### Fredrik

Staff Emeritus
I think the term is "unit ray", not "unitary ray". At least that's what Steven Weinberg calls it in "The quantum theory of fields", volume 1. It doesn't really matter if you say that "states are represented by rays" or "states are represented by unit rays", since there's exactly one "ray" for each "unit ray" and vice versa.

14. Oct 28, 2004

### dextercioby

unnecessary details...

In my version of the book it's simply "ray" (Steven Weinberg "The quantum theory of fields",C.U.P.volume 1,p.49-50).In the QM course i took back home it was called:"raza unitara".I should have probably translated it "unit ray",instead of "unitary ray".In other QM books (in English),it's either "ray" or no other name.This is irrelevant.I'm glad if you got the picture right.

15. Oct 28, 2004

### manesh

this is the apt answer to the initial question...well done..