It seems to me that we could change all the current spin values of particles by multiplying them by two. Then we could describe Bosons as even spin particles and Fermions as odd spin particles. Is there some consequence or reason why we can't simply change 1/2 as the smallest unit of spin into 1?

It just seems prettier this way. In fact, there's almost an interesting relation to even and odd functions in that if you switch any two bosons around the wave function stays the same while if you switch any two fermions around the wave function becomes negative.

  • 2
    $\begingroup$ You absolutely could just define "Kainuispin" $S_{\text{Kainui}}$ by the equation $S_{\text{Kainui}} \equiv 2 S$. Nothing at all wrong with that. You could rewrite every equation which has ever been written in terms of $S_{\text{Kainui}}$. $\endgroup$ – DanielSank Sep 17 '14 at 3:33
  • $\begingroup$ @DanielSank that could probably be an answer. (Of course there is also room for an answer that explains the history of $\hbar$ coming to be the unit of spin, but I think you can write a perfectly valid answer without getting into that.) $\endgroup$ – David Z Sep 17 '14 at 3:42
  • $\begingroup$ You might be interested to read History of spin and statistics by Adam Martin $\endgroup$ – John Rennie Sep 17 '14 at 5:05
  • $\begingroup$ what about the different statistics exhibited by spin 1/2? and the Pauli exclusion? If all were integers all would have to follow bose-einstein statistics and no atoms could form. The spinors are crucial. $\endgroup$ – anna v Sep 17 '14 at 5:17
  • 2
    $\begingroup$ @annav: If you just make up a new variable defined as twice the usual spin, you would rewrite all known equations and everything would be fine. Whether "spin" is integer or half integer is meaningless without also staging its use in the theory. $\endgroup$ – DanielSank Sep 17 '14 at 5:45

The "spin" tells us how the wavefunction changes when we rotate space (or spacetime). Just because I double all charges by convention, the behaviour of the wavefunction will not be any different. What will happen is that the "doubling" or charges will lead to the "halving" of your definition of angles such that the physical results (which depends on angle multiplied by spin) remain the same.

Wrt. the observation on "odd" and "even" functions -- that is not an accident and it works quite like you think it does.

The crux of the matter is that a "full rotation" corresponds to $2 \pi$ so the phase picked up by a spin $\frac{1}{2}$ wavefunction is $e^{i \pi} = -1$.

Recall that (even in classical mechanics) "angular momentum" is the generator of rotations. So if I start using different units, eg: $\tau \equiv 2 \pi$ to represent a half rotation (instead of $\pi$) then the values of charge will halve to maintain the value of $e^{i q \theta}$

If you understand some representation theory, here goes:

  1. Representations of $SO(3)$ have integer charges. Since we're referring to the group of rotations, we call that charge as "angular momentum" or "spin". The representations correspond to scalars (spin 0), vectors (spin 1) and tensors (in general of spin 2 or higher).

  2. $SU(2)$ is a "double cover" of $SO(3)$ so representations of $SU(2)$ can have the "charges" as $SO(3)$ representations. Thus we also get half-integer spin. The new representations correspond to spinors.

  3. When we consider quantum relativistic physics (aka QFT), all physical fields/particles must form kosher reps of the Lorentz algebra, which happens to be $so(3,1) \sim so(4) = su(2) \oplus su(2)$. So (up to the "unitary trick") reps of the Lorentz group can be written as a tensor product of reps of the left and right-handed $SU(2)$ algebras. Based on #2 above, these continue to have integer or half-integer spin.

  • 2
    $\begingroup$ I think the exponent in $e^{-\pi}$ is missing a $i$ factor, (too small edit to do myself...) $\endgroup$ – ratchet freak Sep 17 '14 at 9:04
  • 1
    $\begingroup$ What if we switched from $\pi$ to $\tau$? $\endgroup$ – Erbureth Sep 17 '14 at 10:46
  • $\begingroup$ @Erbureth: Recall that (even in classical mechanics) "angular momentum" is the generator of rotations. So if I start using units of $\tau = 2 \pi$ to represent a half rotation (instead of $\pi$) then the values of charge will halve to maintain the value of $e^{i q \theta}$ $\endgroup$ – Siva Sep 17 '14 at 20:21
  • $\begingroup$ Thanks, I suppose I really just know too little and am currently biting off more than I can chew. I've only taken an undergraduate physical chemistry course in the subject and am trying my best to teach myself as much as possible, but I really don't have any direction or idea where to go from there. $\endgroup$ – Kainui Sep 18 '14 at 5:33
  • $\begingroup$ Ah, I see. I'm sorry about that. If you do a typical undergraduate course on classical mechanics (meant for physicists), you might learn enough about Lagrangians, Hamiltonians, symmetries and "generators" to get the flavour of my answer. If you follow that up with an undergraduate course on quantum mechanics, then hopefully my answer will make sense. $\endgroup$ – Siva Sep 18 '14 at 6:23

I would say it's because at its core quantum mechanics is the study of atomic transitions, and the knowledge of the internal states must be inferred. What we see from that perspective is that all angular momentum transitions are integer multiples of $\hbar$, and are quite frequently exactly $\hbar$. There are sets of related states in some atoms which are grouped in singlets, triplets, quintuplets, etc., and related states in other atoms which are grouped in doublets, quadruplets, sextuplets, etc. The fundamental bit of insight is that the odd-plet atoms may sometimes have zero angular momentum, while the even-plet atoms can never have zero angular momentum. Integer steps between half-integer values is a very clear way to achieve this; I think that dividing $\hbar$ again so that the smallest transition was "two spin quanta" would make the separation between these two systems much more confusing.

There's a lovely book by Tomonaga called The Story of Spin where he takes three competing rules for assigning quantum numbers to atomic angular momentum states from the early days of quantum mechanics, by Sommerfeld, Landé, and Pauli. Tomonaga presents all three models at face value. Each of them has its own internal consistency. We use Sommerfeld's model, where a state with multiplicity $n$ has angular momentum $j=\frac{n-1}{2}$. Landé used a supplementary quantum number $R=\frac n2$, which switches the places of integer and half-integer spin from the statistical standpoint, but by means of some different selection rules could recover the same physics. Deciding for myself which of the three models I preferred, and then convincing myself it wasn't just that I already had a lot invested in Sommerfeld's approach, was a very valuable intellectual exercise. None of the three has only integer quantum numbers for systems with spin; I suppose that the integer quantum numbers for systems with orbital angular momentum were already too well-established by the time that spin was discovered.

  • $\begingroup$ "quantum mechanics is the study of atomic transitions..." Eh? I do quantum mechanics every day and no atomic transitions in sight. $\endgroup$ – DanielSank Sep 17 '14 at 5:42
  • $\begingroup$ @DanielSank You've removed my emphasis --- or are you doing quantum mechanics with no transitions at all? $\endgroup$ – rob Sep 17 '14 at 13:41
  • $\begingroup$ Emphasis disappeared in my quote because I'm terrible at computers, not for any other reason. $\endgroup$ – DanielSank Sep 17 '14 at 16:47

Its just a consequence of garlac experiment the particles on the have possibly two eigen states up |z>and down ie half of particles go up nd rest half down |_z> that's y they are called spin half. And note that spin is nothing to do with rotation it is purely intrinsic and quantum mechanical


protected by Qmechanic Apr 6 '16 at 7:57

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.