Spin - any book recommendations?

[ansgar]

I think he means, you must have a "deeper" understanding of the physics than just the mathematical model involved.

It's very subjective, imo. Some people may find Newton's laws "deeper", while some may find the principle of least action "deeper". I don't see how one could judge either way...

Exactly, it is very subjective. Concepts of force, and fields etc, are very (or rather were) very debated entities back in those days.

I would say that deriving things from symmetries are probably the deepest thing one can do in physics, the simplicity and beauty of nature is simply manifested. And it is even more beautiful and deep if we can perform measurements and verify that the imposed symmetry of nature actually seems to be there, i.e. realized in nature.

If one can not understand what spin is, then we can not understand what electric charge is either...

Things like Angular momentum and Linear momentum are also related to symmetries, that is how one formally derive those quantities. So by saing that P = mv^2 / 2 is momentum is not a very deep answer, but if one says derives P from - assuming- that nature is invariant under linear translations, then one has a deeper connection between P and nature - i.e. one can argue that we have understood what P is - a manifestation of a underlying symmetry in nature. Now that is what i call philosophy!

 

[alxm]

I don't think there's a good answer to the OP's question really. If he's reading Griffiths, which is an introductory textbook, then there's not much you can do to 'explain' spin at that level.

It's a relativistic property (as pointed out), and can't really be 'explained' in a rigorous manner without getting into QM that's pretty far beyond the introductory level.

There's a good reason why most introductory textbooks simply assert it as a separate postulate.

 

[ansgar]

I don't think there's a good answer to the OP's question really. If he's reading Griffiths, which is an introductory textbook, then there's not much you can do to 'explain' spin at that level.

It's a relativistic property (as pointed out), and can't really be 'explained' in a rigorous manner without getting into QM that's pretty far beyond the introductory level.

There's a good reason why most introductory textbooks simply assert it as a separate postulate.

Yes it is simply postulated, but one can at least provide a clear presentation :)

 

[dreamspy]

Thats right, it's an introductory class. At the moment I'm mostly interested in beeing able to calculate simple questions to pass my test. Deeper understanding will be needed when I go to other QM classes.

 

[Fredrik]

Spin is not a relativistic property. The existence of the spin operators can be derived from the assumption that space is rotationally invariant. This is how the argument goes (details omitted because of time and because I only know some of them):

A symmetry is a bijection on the set of unit rays, that preserves probabilities, i.e. S is a symmetry if |\langle\psi'_1|\psi'_2\rangle|^2=|\langle\psi_1|\psi_2\rangle|^2 for all \psi_1,\psi_2\in \mathcal R and all \psi'_1,\psi'_2\in S(\mathcal R). The idea that space is rotationally invariant is incorporated into QM as the assumption that there's a group homomorphism from SO(3) into the group of symmetries.

Now the question is, does this mean that there also exists a group homorphism from SO(3) into the group of unitary linear operators on our Hilbert space H, i.e. is there a representation of the symmetry group on the Hilbert space of state vectors? Wigner showed that the answer is no, and also that it's "almost" yes. We can define a function from the set of continuous curves that start at the identity of SO(3), into the group of unitary linear operators on H. I'll write such operators as U(C), where C is the curve. Note that the other endpoint of C is a rotation matrix. It turns out that these curves can be divided into two equivalence classes, which I'll call 0 and 1, and that U(C) only depends on the endpoint and the equivalence class. So I'll write U([C],R) instead of U(C). It's also possible to prove that U(0,R)=-U(-1,R).

For every R, there's exactly one curve in 0 and one curve in 1 that goes from the identity to R. We can try to use only the 0 curves, hoping that this will enable us to drop the C's completely and just write U(R), but it turns out that we still won't get the simple relationship U(R')U(R)=U(R'R). The U(C) operators satisfy

U(C',R')U(C,R)=U(C'*C,R'R)

where C'*C is the curve that goes through C at twice the normal speed, and then through a translated version of C' from R to R'R at twice the speed. The problem is that C'*C might be in 1 even if C and C' are both in 0. This means that we can write

U(0,R')U(0,R)=U(s,R'R)=(-1)^s U(0,R'R)

where s=[C'*C], and this can't be simplified any further.

This is why we take the symmetry group to be SU(2) instead of SO(3). SO(3) is homeomorphic to a 3-sphere with opposite points identified, and SU(2) is homeomorphic to a 3-sphere. This difference means that there will only be one equivalence class of curves in SU(2) instead of two, and that enables us to write U(r')U(r)=U(r'r), where r and r' are members of SU(2) that correspond to rotation operators, and r and -r correspond to the same rotation operator.

The R's and the r's can be expressed as functions of three parameters (e.g. Euler angles), and therefore both U(R) and U(r) can be expanded in Taylor series:

U(r)=1+\frac i 2\omega_{ij}M_{ij}+\cdots[/itex]<br /> <br /> where second-order terms have been omitted and repeated indices are summed over, and \omega is the first-order term in a series expansion of r. The fact that \omega is symmetric implies that M can be defined as symmetric. The spin operators can now be defined as<br /> <br /> S_i=\epsilon_{ijk}M_{jk}<br /> <br /> where the epsilon is the Levi-Civita symbol. The commutation relations<br /> <br /> [S_i,S_j]=i\epsilon_{ijk}S_k<br /> <br /> can be derived from the fact operators must transform as<br /> <br /> X\rightarrow U(r)^\dagger X U(r)<br /> <br /> in order to make expectation values invariant.<br /> <br /> As you guys can see, this stuff is pretty complicated. I think that&#039;s the reason why a proper discussion of spin is usually delayed until an advanced class where no one is going to be bothered by the fact that we consider the restricted Poincaré group instead of just SO(3) or SU(2). SO(3) is a subgroup of the restricted Poincaré group, so when we do the relativistic version of this, we get everything I said above and then some. That includes the concepts concepts &quot;momentum&quot;, &quot;energy&quot; and &quot;mass&quot;, and we end up with a proper definition of a quantum mechanical &quot;particle&quot;.

 

[Truecrimson]

If your aim is to learn by examples and problems, you might want to take a look at Zettili.

 

[dextercioby]

why should I waste my valuable time to preach from the R-QM textbooks? I have already said that intrinsic angular momentum (called spin) is a manifestation due to Lorentz Symmetry: Which we have pretty good proofs that this symmetry is realized in nature.

All particles have intrinsic angular momentum; fermions have 1/2, i.e. transform under the irreducible representation of the Lorentz group. Scalars have 0, i.e. transforms as singlets. Vectors are particles that transforms as four-vectors, tensor-2 particles are particles that transforms as rank-2 tensors etc etc.

So compact: Intrinsic Angular momentum / spin is a symmetry manifestation.

Analogy: electric charge is a symmetry manifestation of the gauge transformations in electromagnetism.

Physics is about symmetries, once you know the symmetries in nature, we can deduce the dymanics.

As far as I know, you don't need to include the special relativity into quantum mechanics in order to to "derive" the concept of spin. You can do that very well with a Newtonian "background".

 

[dextercioby]

Spin is not a relativistic property. The existence of the spin operators can be derived from the assumption that space is rotationally invariant.

I think you're wrong. Spin cannot be "derived" (it's rather <introduced>) without the assumption of <relativity>, be it special/Minkowski or Galilean.

 

[Fredrik]

I think you're wrong. Spin cannot be "derived" (it's rather <introduced>) without the assumption of <relativity>, be it special/Minkowski or Galilean.

You can take the existence of the spin operators to be an axiom if you prefer, but that doesn't make what I said wrong. What is it that you think I'm wrong about?

 

[dextercioby]

I didn't mention <axioms> in any place. I just asserted that there's no discussion of spin, unless one brings in relativity in quantum mechanics. So i'd say that spin is a relativistic property (flat spacetime), just like mass.

 

[Fredrik]

OK, I misunderstood what you were trying to say. I get it now. You can certainly derive the existence of spin from the axiom I used. If we replace the rotation group in my argument with the Galilei group (i.e. if we consider the principles of QM combined with Galilean relativity), we get a lot more, including momentum, energy and mass operators, and the Schrödinger equation.