# You may say i'm a idiots or what, but what is spin.

1. May 11, 2005

### ArielGenesis

you may say i'm a idiots or what, but what is spin.

2. May 11, 2005

### Tom Mattson

Staff Emeritus
Wouldn't dream of it.

In classical mechanics a constant applied torque (such that given by pushing on the unhinged end of a door) results in a constant increase in the "quantity of rotational motion" of the door. This quantity of rotational motion is called the angular momentum $L$. It is calculated by the vector product of the displacement $x$ and the momentum $p$: $L=x \times p$

In quantum mechanics we rarely talk about torques, but we still do talk about angular momentum. Only now it is expressed as an operator $\hat{L}$, and is given by $\hat{L}=\hat{x}\times\hat{p}$.

But in QM there is another type of angular momentum, one that is not a function of coordinates and momenta: It is the spin. It is still an angular momentum (because it has the same units as angular momentum), but it is an intrinsic angular momentum. And despite the name, it has nothing to do with classical "spinning" (as in a spinning top).

There is a lot more that could be said, but I think I'll leave it at that for now.

Last edited: May 11, 2005
3. May 11, 2005

### dextercioby

I'm not going to make that error again,so i'll ask before giving a book reccomandation.Question:What level of physical and mathematical knowledge do you have ?

Daniel.

4. May 11, 2005

### wangyi

spin is called angular momentum not only because it has the same unit as angular momentum, because the action, and the Planck constant are all the same unit as angular momentum. i think the reason we consider spin as angular momentum because it is conserved together with angular momentum, and it transforms as angular momentum in roation.

regards,
wangyi

5. May 11, 2005

### Tom Mattson

Staff Emeritus
I think the real reason we consider spin angular momentum is that $J=L+S$ is the generator of rotations. But when a guy asks, "what is spin?" I assume that he won't appreciate that yet.

6. May 12, 2005

### marlon

Suppose that you want some physical system to be invariant under certain operations. Then indeed, the 'pieces' that make up that system must all transform like irreducibe representations (IR) of the group that contains all operations under which the system must be invariant.

Now, what does this mean ? Well, lets look at QM and how the J-operator is connected to rotations. A QM-system is invariant under rotations if

1) the normalization of the wavefunction is preserved
2) the expectation value of any observable is preserved
3) if the Hamiltonian does NOT change under rotations.

In order to obey these commands, the wavefunctions (these are the 'parts' that make up the physical system, caracterized by the three above conditions) must transform in a certain way : w' = Uw...Where w is the original wavefunction and R denotes the rotation. U MUST BE UNITARY in order to obey the conditions (this is just the same as asking why time-evolution must be unitary).

In QM one can prove that if a wavefunction transforms like w' = Uw, then this U (which is a rotation) can be written in terms of the component of the L-operator along the rotational axis. But what does this component look like ? Well QM proves that we can write it in terms of its eigenvalues l just by calculating the expectation value of the L-operator in the appropriate base.
The same system works for J = L + S so if you take L to be zero, you are working with S, ie spin.

So what does this mean ? Answer : the IR representations are directly connected to the eigenvalues of the L-operator, which is also called the generator of the rotations. So l = 0, 1, 2,... all represent a different IR of the rotation-group.

Just as an addendum. If a system is invariant under rotations (which all are put in the symmetry group of that system), the parts which make up the system MUST transform as IR of that symmetry group. This is a very important rule that is quite logical, if you think about it. Let's take three operations out of the symmetry group and we call them A, B and C = A ° B...

Now we perform them on the system:
1) A(system) = system
2) B(system) = system
3) A°B(system) = A(B(system)) = A(system) = system
4) C(system) = system = A°B(system) = system

This means that if you perform either A°B or C, the result must be the same. This implies that the parts of the system must transform in the same way under either A°B and C. But this means that the parts actually obey the multiplication table of the group and by definition, this means that they must be representations of that group.

A spinor is a special kind of vector. I mean, it has the property that if you rotate it 360° you get the exact opposite (A ---> -A) of what you originally rotated. Rotate another 360° and you get where you started off in the first of the two rotations (A---> -A--->A).

So spin arises due to symmetry of the physics, it does NOT imply that particles rotate around their axis.

marlon

7. May 12, 2005

### Edgardo

Hello Ariel Genesis,

maybe to give you an intuitive feeling for spin, I recommend reading about the Stern-Gerlach experiment
http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html#c5

In that experiment, a particle with spin 1/2 is shot through a inhomogeneous
magnetic field. What happens? You will notice that a force acts on that particle because the spin interacts with the magnetic field.

You can think of the spin as a tiny little bar magnet, that only has two orientations (for spin s=1/2 like free electrons) in a homogeneous magnetic field along z-direction.
http://hyperphysics.phy-astr.gsu.edu/hbase/spin.html

You see the red arrows? They represent the spin of the electron, but we are interested in the projection to the z-axis.
You have two choices: projection to the upper part of z-axis, so you get $+\frac{1}{2}\hbar$. In this case you talk about spin-up.

In the other case, where you project to the negative z-axis, you get
$+\frac{1}{2}\hbar$ as projection, and we talk about spin-down.

So when you hear someone talking about spin-up or spin-down, it is just the orientation of the spin (or projection).

There is an effect, called nuclear magnetic resonance, where you can flip the spin orientation.
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/nmr.html
In the left picture they use the analogy to the bar magnet.

8. May 12, 2005

### ArielGenesis

i think that it is the right question to ask, so : just a high school student.

edgardo, i once visit hyper physics is a good resource but, i just understand 'nothing' about it. and may be this should be a correct reply to marlon. i basically blank of those.

first of all thanks for all the effort
but it would be greate if it could be re-explain in an simpler form.

9. May 12, 2005

### dextercioby

I'm sorry,but for H-S,Tom's post and the material in a H-S textbook is the most u can ask.Simply put,the understanding of the concept of spin is highly nonintuitive and the mathematics required is way over the H-S level.

Daniel.

10. May 12, 2005

### nightcleaner

Hi
I'd like to request a reality check here. I'm not much in maths, but intuitive imagery of quantum process is a kind of hobby of mine.

Spin, as Wangli pointed out in another thread, has the same units as angular momentum. One is therefore tempted to imagine spin as determined by the motion of a tiny object spinning like a top. After all, the motions of a top can largely be determined using nothing but angular momentum.

However, spin in elementary particles does not behave in all ways like a spinning top. For one thing, spin (1/2) in particles can be in only one of two states, called up and down. All spin (1/2) particles are either spin up or spin down, and in some apparatus can be changed from up to down or down to up. The key difference between this behavior and that of a top should be obvious, but I have stumbled over that phrase in too many text books.

A top, or more formally, a gyroscope, can exhibit spin in any direction. It is quite possible to set one gyroscope turning at ninety degrees from another gyroscope. This is not the case for spin (1/2) particles. They can spin up, or they can spin down, but they cannot spin "sideways". In this way, they most definately do not behave like tops spinning under angular momentum.

Well, that is the best I have come up with after much study of spin using my limited maths. So what about it? Is this description simple enough to satisfy the non-mathematically inclined while also not offensive to quantum physics acolytes?

Just trying to help.....

nc

11. May 12, 2005

### wangyi

the J=L+S is the generator statement is equal to the statement that J is conserved on condition that the spacetime is invariant under rotation.

regards
wangyi

12. May 13, 2005

### gptejms

Referring to Marlon's post,could I say this:-
For every symmetry there's a conservation law,so if the system or parts of the system obey a certain symmetry(i.e. invariance under the transformations represented in a particular IR representation of SU(2)) that points towards a conserved quantity which is spin in this case:-the quantiative value of spin(s=1/2,1,..) depending on which particular representation of SU(2) it is(s=1/2 or s=1 etc.).

13. May 13, 2005

### dextercioby

I dunno,i have a hunch Marlon knows this part of QM.It's called "Noether's theorem for Quantum Mechanics".

Daniel.

14. May 14, 2005

### Neo7

Thanx Edgardo.I read it and it was really useful for me.