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I study physics and am attending a course on quantum field theory. It is hard for me to draw connections from there to the old conventional theories.

In quantum field theory spin originates from the Dirac equation in the case of fermions.

I don't remember where it comes from in quantum mechanics. I just remember that there was the Stern Gerlach experiment where you shoot neutral spin 1/2 Ag atoms in. Is there also a electrically neutral elementary particle? If this is the case, how would this experiment look like in quantum field theory? Of course I ask this for the lowest order, otherwise we would have to calculate an infinite number of Feynman graphs, won't we?

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The Higgs and $Z^0$ bosons, photon, gluon, and all three neutrinos are all elementary particles and all are electrically neutral. But you might be confusing spin and electric charge. –  Mike Jun 10 '13 at 18:13
    
Possible duplicates: physics.stackexchange.com/q/1/2451 , physics.stackexchange.com/q/822/2451 , and links therein. –  Qmechanic Jun 10 '13 at 18:39

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Fundamentally, the spin originates from the fact that we want our quantum fields to transform well-behaved under Lorentz transformations.

Mathematically, one can start to construct the representations of the Lorentz group as follows: The generators $M^{\mu \nu}$ can be expressed in terms of the generators of rotations $J^{i}$ and those of boosts $K^{i}$. They fulfill $$ [J^{i}, J^{j}] = i \epsilon_{ijk} J^k, \, [K^i, K^j] = -i \epsilon_{ijk} J^k,\, [J^i, K^j] = i \epsilon_{ijk} K^k.$$ From them one can construct the operators $M^i = \frac{1}{\sqrt{2}} (J^i + i K^i)$ and $N^i = \frac{1}{\sqrt{2}} (J^i - i K^i)$. They fulfill $$[M^i, N^j] = 0,\, [M^i, M^j] = i \epsilon_{ijk} M^k \, [N^i, N^j] = i \epsilon_{ijk} N^k$$ These are just the relations for angular momentum that you should know from your QM introductory course. Group theoretically this means that every representation of the Lorentz group can be characterized by two integer of half-integer numbers $(m, n)$. If you construct the transformations explicitly you will find

  • $(m = 0, n = 0)$ is a scalar, i.e. does not change under LT.
  • $(m = 1/2, n = 0)$ is a left-handed Weyl spinor
  • $(m = 0, n = 1/2)$ is a right-handed Weyl spinor
  • $(m = 1/2, n = 1/2)$ is a vector

A Dirac spinor is a combination of a right and a left handed Weyl spinor.

Actually, one can now use these objects and try to find Lorentz invariant terms in order to construct a Lagrangian. From that construction (that is too lengthy for this post) one finds that the Dirac equation is the only sensible equation of motion for a Dirac spinor simply from the Lorentz group's properties! Similarly one finds the Klein-Gordon equation for scalars and so forth. (One can even construct higher spin objects than vectors, but those have no physical application except maybe in supergravity theories).

So, as you can see now, spin is fundamentally a property of the Lorentz group. It is only natural that we find particles with non-zero spin in our Lorentz invariant world.

Sidenote: Since we found the Dirac and Klein-Gordon equations from Lorentz invariance alone, and their low-energy limit is the Schroedinger equation, we get a 'derivation' of the Schroedinger equation, too. Most of the time the SE is simply postulated and worked with: this is where it comes from!

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Thank you for your answer, that gave me some deeper insight into the derivations we have done so far. What did confuse me was the fact, that quantum mechanics is non relativistic. This means that the Lorentz invariance is not given and if i am right, then there is no such a thing as spin in pure quantum mechanics as it comes from Lorentz invariance. Did we just 'invent' it because it is needed to use the pauli exclusion principle for the H-Atom? Shouldn't there be no spin in QM? –  physicsGuy Jun 11 '13 at 12:42
    
There is spin in pure quantum mechanics. However, Pauli put it there from a purely phenomenological perspective - he saw the experiment and came up with a mathematical description that could explain the data. –  Neuneck Jun 11 '13 at 12:45
    
Thank you that explains it :-) –  physicsGuy Jun 11 '13 at 12:47

The spin is the intrinsic angular momentum of an object, usually a particle, as measured in its rest frame. Large objects may rotate around their axis. Even smaller objects may rotate around an axis. Quantum mechanics implies that there's nothing that would prevent otherwise pointlike particles from rotating around an axis, too.

In quantum mechanics, the angular momentum around an axis is a multiple of $\hbar/2$; this may be shown from the single-valuedness of the wave function under 720-degree rotations (360-degree rotations are allowed to change the wave function in a modest way, namely to flip its sign).

The precise amount of spin of a given particle may be determined by a deeper theory or an experiment. It's just a fact that the electron – or any other lepton, including neutrinos, or quark – has the spin of $J=\hbar/2$, the smallest allowed nonzero value.

Other elementary particles have different spins. The Higgs boson has $J=0$, the photons, gluons, and W/Z-bosons have $J=1\hbar$ while the graviton has $J=2\hbar$.

There's no direct relationship between the spin and the electric charge. For example, neutrinos and Z-bosons have a vanishing electric charge but a nonzero spin; the goldstone bosons eaten by the W-bosons have a vanishing spin but a nonzero charge.

There is nothing such as "a way how experiment looks in a theory". Experiments are independent of theories; they're always the same; they are performed in order to test the validity of theories. Some theories pass, some theories fail because they disagree with the results of the experiments. For certain (slow-speed etc.) experiments, it's OK to use a more primitive theory such as non-relativistic quantum mechanics; for more general setups, one needs a more complete theory (quantum field theory) because the simpler theories would produce inadequate predictions. Every theory must define the set of experiments for which it should work, and if it disagrees with an experiment in this set, it has to be abandoned.

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