# Spin 1/2

Why people say that to understand spin is difficult because there in no a classical analogous...bla bla bla?
we got an OBSERVABLE, S associated with the physical quantity S, why all that confusion about the not classical analogous, isn't S the physical quantity associated with the observable the classical analogous?
I feel that I fully understand this because the relation between the physical quantity and its observable is well stablished in tannudji book in postulate 2, so everything get clear from that point on (the observable eigenvalues could be quantized!) Homework Helper
cire said:
Why people say that to understand spin is difficult because there in no a classical analogous...bla bla bla?

Who said anything about understanding the concept of spin veing difficult??It may be difficult,when u lack proper knowledge of mathematics.But then,what made u chose theoretical physics in general and qauntum mechanics in particular...?? :tongue2:

cire said:
we got an OBSERVABLE, S associated with the physical quantity S, why all that confusion about the not classical analogous, isn't S the physical quantity associated with the observable the classical analogous?

Nope.S without a hat doesn't exist.As for a handful of other operators,they exist at classical level.

cire said:
I feel that I fully understand this because the relation between the physical quantity and its observable is well stablished in tannudji book in postulate 2, so everything get clear from that point on (the observable eigenvalues could be quantized!)

Let's both agree on something:the physical quantity is something u measure.Energy,position,velocity,mass,momentum,angular momentum,and so on.An observable is something abstract right from the definiton.Think about Hamiltonian observables.They are real functions defined on the submanifold determined in the phase space by the solutions of the classical equations of motion.This,as stated,makes the notion "observable"quite intriguing:it makes you think:"observable":'hmmm...something that can be observed'.Wrong.U cannot observe/measure a hamiltonian,for example.So,the concept of "observable",both at classical and at quantum level requires some special attention.And definitely the necessary amount of mathematics.
The observables can be quantized,right??The eigenvalues are just numbers from generaly an arbitrary set.In fortunate cases,that set is discrete and u can call that (classical) observable as being quantized.

Daniel.

^_^
Spin doesn't have a classical analogue because it represents the angular momentum of a point particle. Electrons (for example) have no internal structure and so there is nothing there to spin and yet they have a magnetic dipole moment (Stern-Gerlech Experiment). This is silly to us because 400 years of classical physics says so and so we say there is no classical analogue.

Observable means you can measure a eigen-value for the operator. Measuring the Hamiltonian would give you a number in eV or J or whatever.

I think spin is hard to get because the whole fermion, boson, half integer, integer thing that everyone is taught in year twelve chemistry is some Quantum Field Theory result (theoretical, not empirical) that no one gets because QFT is stupid.

RedX
Spin has no classical counterpart because it comes from the fact that in quantum mechanics an electron for example is represented by a two component wave function called a spinor, so the orbital angular momentum is not the generator of infinitismal rotatations for you need something to rotate the components of the spinor and the coordinates.

humanino
(^_^) said:
I think spin is hard to get because the whole fermion, boson, half integer, integer thing that everyone is taught in year twelve chemistry is some Quantum Field Theory result (theoretical, not empirical) that no one gets because QFT is stupid.  Except that QFT ows a lot to experiment. It is very not fair to say that QFT emerged as a purely theoretical investigation. The success of QFT is due to its efficiency to describe reality, it was accepted because the phenomenology that one gets fits well observed behaviors.

Of course QFT has been axiomatized, and it was a necessary and useful task. But it was achieved a posteriori. Unlike Einstein's relativity, which emerged of a well-defined and unique empirical observation, translated into a sound new formalism, QFT originally tried at first to formulate a Lorentz-invariant quantization procedure by guessing a proper method. Many different versions have been proposed all along until a coherent view could be obtained, and in fact the process is still on its way.

But arguing this way against what spin is, where it comes from, ... actually requires a few technical notions. The covering of $$SO(3)$$ is $$SU(2)$$ In fact, there is a discussion which I am aware of about [thread=52153]spin[/thread] In the form dispayed there, for the representation of a vector, is :
$$(V)_{\alpha\dot{\beta}}=\left( \begin{array}{cc}V_+,V_1^*\\V_1,V_-\end{array} \right) = \frac{1}{\sqrt{2}} \left( \begin{array}{cc}V_0+V_1,V_2-\imath V_3\\V_2+\imath V_3,V_0-V_1\end{array} \right) =V_a (\sigma^a)_{\alpha\dot{\beta}}$$
and this representation has some convenience, and is no less legitimate than the more conventional one. Besides it makes it apparent that a spinor is a sort of square root for the vector. Under rotations, the aforementioned representation for a vector transforms as $$V'=UVU^\dagger$$ with $$U$$ unitary, whereas the spinor transform as $$\Psi'=U\Psi$$ Oh !? Where did we quantize anything so far ? Nowhere ! Actually, there are classical solutions to the Dirac equation. And when dealing with a circulary polarized light beam, do I need the whole QED formalism ? Actually, no. I have a classical solution to the Maxwell equation for the propagation of a light wave. This is helicity $$\pm1$$ of the beam, this is the classical spin of a massless vector particle.

Homework Helper
humanino said:  Except that QFT ows a lot to experiment. It is very not fair to say that QFT emerged as a purely theoretical investigation. The success of QFT is due to its efficiency to describe reality, it was accepted because the phenomenology that one gets fits well observed behaviors.

Of course QFT has been axiomatized, and it was a necessary and useful task. But it was achieved a posteriori. Unlike Einstein's relativity, which emerged of a well-defined and unique empirical observation, translated into a sound new formalism, QFT originally tried at first to formulate a Lorentz-invariant quantization procedure by guessing a proper method. Many different versions have been proposed all along until a coherent view could be obtained, and in fact the process is still on its way.

Agree,here.

humanino said:
But arguing this way against what spin is, where it comes from, ... actually requires a few technical notions. The covering of $$SO(3)$$ is $$SU(2)$$ In fact, there is a discussion which I am aware of about [thread=52153]spin[/thread] In the form dispayed there, for the representation of a vector, is :
$$(V)_{\alpha\dot{\beta}}=\left( \begin{array}{cc}V_+,V_1^*\\V_1,V_-\end{array} \right) = \frac{1}{\sqrt{2}} \left( \begin{array}{cc}V_0+V_1,V_2-\imath V_3\\V_2+\imath V_3,V_0-V_1\end{array} \right) =V_a (\sigma^a)_{\alpha\dot{\beta}}$$
and this representation has some convenience, and is no less legitimate than the more conventional one. Besides it makes it apparent that a spinor is a sort of square root for the vector. Under rotations, the aforementioned representation for a vector transforms as $$V'=UVU^\dagger$$ with $$U$$ unitary, whereas the spinor transform as $$\Psi'=U\Psi$$ Oh !? Where did we quantize anything so far ? Nowhere ! Actually, there are classical solutions to the Dirac equation.

Yes,u're right,but you have to admit that,even though the Dirac field is built ONLY BY RELATIVISTIC METHODS,it means s*** at classical (relativistic and nonquatum) level.The classical Dirac field is just an ordinary field and describes no particle,just as the classical em field.One u quantize it,u can speak of particles associated with the field.

humanino said:
And when dealing with a circulary polarized light beam, do I need the whole QED formalism ?

Obviously not.
humanino said:
Actually, no. I have a classical solution to the Maxwell equation for the propagation of a light wave. This is helicity $$\pm1$$ of the beam,this is the classical spin of a massless vector particle.

Your analogy works only for the em and gravitational field,as thery are the ONLY QUANTUM MASSLESS FIELDS WITH CLASSICAL ANALOGUS.Try to see whether the neutrino,gravitino,and other massless fields from susy can have polarization at classical level.They don't and,because they "live" only in the quantum world,for them "helicity" could not be put in connection with "polarization".And i believe the 2 Q fields (em+gravity) have classical correspondent,because they're infinite range,appearing also at macroscopical/classical level.

Daniel.

^_^
I don't actually think QFT is stupid, just complicated. I'm the stupid one who finds it difficult was the upshot of my remark. Also I ment that you can't get spin-statistics out of non-relativistic quantum but it is predicted by QFT. That fermions and bosons have the spins they do is a purely empirical result until you do QFT, where it is predicted by theory.

Don't get me wrong, I'm not some heathen who thinks QFT is crap and wrong simply because I don't get half of it. I'm not going to go out and write "The Final Theory" or "Natures simeltanious 4-dimiensional harmonic Time Cube" because tensors are hard to get.

thank you guys for your comments I will post get in touch when I finish my ED test next thursday
thanks 