3
$\begingroup$

I come across this article:

https://skullsinthestars.com/2016/03/29/1975-neutrons-go-right-round-baby-right-round/

I quote here a part of this article:

Spin 1/2 particles like the electron, proton and neutron, which constitute all of ordinary matter, must be rotated 2 times in order to look the same: a full 720°! This is incredibly bizarre, and there really isn’t anything in the world we experience that is comparable.

I am concerned about these two sentences quoted. Let me first state what I know:

  1. The Lie group $\text{SO}(3)$ has fundamental group $\Bbb Z_2$. Equivalently, we say that the universal cover of $\text{SO}(3)$ is a double cover.

  2. The belt trick is a demonstration of the fact that $\text{SO}(3)$ has universal cover being double cover. In terms of fundamental group, a 360° rotation in 3D space is not homotopic to the identity, while a 720° rotation is homotopic to the identity.

The quoted article talks about belt trick (the "rotate 720° to go back to original state" thing) as if it is related to spin-$\frac{1}{2}$ particles, that these particles are examples which 360° rotation will not give the original state, but 720° rotation will. They seem to imply that the very existence of spin-$\frac{1}{2}$ representation of $\mathfrak{so}(3)$ (those Pauli matrices used to describe spin-$\frac{1}{2}$) is connected to the fact that $\text{SO}(3)$ has double universal cover. Is this even true, that spin-$\frac{1}{2}$ representation is related to $\text{SO}(3)$ having double cover?

This answer by ACuriousMind seems to be related, but I am not sure if that directly answers my question.

Extra added question:

The article also states something like "spin-$2$ particle returns to its original state after 180° rotation". I believe this is also a false claim, is it not?

$\endgroup$
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/96045/2451 , physics.stackexchange.com/q/78536/2451 and links therein. $\endgroup$ – Qmechanic Nov 10 '19 at 15:08
  • 3
    $\begingroup$ Some statements in the paper you pointed out are wrong and misleading. An electron does not change its state after a rotation of $2\pi$. It is false that to come back in the initial state the electron needs $4\pi$! The author of that paper seems to have some confusion about the difference between state vectors and states, and seems not to know the notion of projective unitary representation. The physical symmetry group is always $SO(3)$. $SU(2)$ enters the game just because one prefers to deal with unitary reps instead of projective unitary reps. $\endgroup$ – Valter Moretti Nov 10 '19 at 15:10
  • $\begingroup$ @ValterMoretti Good to know. I believe me and my friends are confused by popular but wrong science for a long time but we had no idea. If you could, please also read my added question. $\endgroup$ – edm Nov 10 '19 at 15:40
  • $\begingroup$ What @ValterMoretti said is correct. However, there is a kind of loophole: we can do interference experiments in which neutrons pass through the interferometer one at a time, with a precession through angle $\theta$ induced on just one of the two paths (using a local magnetic field). Then we can observe how the interference pattern depends on $\theta$. The result: it's periodic with period $720^\circ$, not $360^\circ$. These experiments have been done. See see arxiv.org/abs/1601.07053 for a review. $\endgroup$ – Chiral Anomaly Nov 12 '19 at 2:35
  • $\begingroup$ @Chiral Anomaly Yes, that is an intersting phenomenon but it is disputable if it means that we need $4\pi$ to restore an electron in its original position. First of all they interpret the phenomenon as a precession as it will be true for a classical particle, furthermore what acts there it time evolution where the Hamiltonian is proportional to $\sigma_z$, which is true, it is also the generator of rotations. Finally the action is not on the whole system, since what they see is a relative interference of a single electron. However it is a very interesting phenomenon related rotations. $\endgroup$ – Valter Moretti Nov 12 '19 at 7:01
6
$\begingroup$

To clarify the misunderstanding arising from the text you have read, I should explain some general notions and result about the representation of $SO(3)$, viewed as the group of spatial rotations in the Hilbert space of a quantum system.

It is also fundamental in our discussion to bring in our mind that a state is a unit vector up a phase.

As is well known, symmetries are represented by unitary operators (also antiunitary but when dealing with connected Lie groups they do not take place). The physical group of rotations $SO(3)$ acts on states, spin states in particular, and this action is expected to preserve the group structure.

As a consequence the map $$SO(3) \ni R \mapsto U_R\tag{1}$$ associating unitary operators $U_R$ to physical rotations $R\in SO(3)$ and representing the action in the space of the states of the rotations of the physical system must satisfy $$U_I= \alpha I\:, \quad U_RU_{R'}= \omega(R,R')U_{RR'}\tag{2}$$ where $\alpha, \omega(R,R')$ are phases: unit-modulus complex numbers (in principle they could also depend on the state in the argument of the unitary operators, but it is easy to prove that this dependence is actually trivial).

A map satisfying (2) is said to be a projective unitary representation.

This is the best you can obtain from basic assumptions of Quantum Physics when dealing with groups of physical symmetries.

Another important requirement is that the map (1) is continuous in a physical sense: for every pair of unit vectors $\psi, \phi$, the transition probability $$SO(3) \ni R \mapsto |\langle \psi|U_R \phi \rangle|^2$$ has to be continuous in the variable $R\in SO(3)$.

It is evident that unitary representations ($\alpha=\omega(R,R')=1$) are much more handy than projective unitary representations, so that mathematical physicists studied the problem of reducing projective unitary representations to properly unitary ones.

A celebrated theorem due to Bargmann established that, if the Lie algebra of the represented group $G$, $SO(3)$ in our case, has a certain cohomological property and the group is simply connected, then for a given continuous projective unitary representation of that group, it is always possible to redefine $U_g \to \chi_g U_g =: V_g$ (where $\chi_g$ is a suitable phase) in order that the new map $$G \ni g \mapsto V_g\tag{1'}$$ is a continuous unitary representation, $$V_e= I\:, \quad V_gV_{g'}= V_{gg'}\tag{2'}\:.$$

What happens for $G=SO(3)$?

Well, the cohomological condition is in fact fulfilled, but $SO(3)$ is not simply connected.

However one can pass to deal with $SU(2)$ which is the universal covering (double covering actually) of $SO(3)$. From definition of universal covering, on the one hand it has the same Lie algebra of $SO(3)$, so that the cohomological condition is safe, on the other hand it is simply connected.

The crucial observation is that, again because $SU(3)$ is the universal covering of $SO(3)$, there is a surjective Lie-group homomorphism which is a local isomorphism around the identity $$\Pi: SU(2) \to SO(3)\:.$$

It holds $ker(\Pi) = \{\pm I\}$, so that there are exactly two matrices of $SU(2)$ corresponding to a matrix of $SO(3)$ and these matrices differ only for their sign: $$\Pi^{-1}(R) = \{\pm A_R\}\:.\tag{3}$$

According to the theory outlined above and the general theory of irreducible representations (Schur's lemma) of compact topological groups (the Peter-Weyl theorem), it is possible to prove that, all irreducible continuous unitary representations of $SU(2)$ are

(i) finite-dimensional and

(ii) they include all possible irreducible continuous unitary projective representations of $SO(3)$. In the sense that if (1) and (2) holds for a given continuous projective unitary representation $U$ of $SO(3)$, it is then possible to change the phases $$V_R := \chi_R U_R$$ in other that either

(a) $SO(3) \ni R \mapsto V_R$ is an irreducible continuous unitary representation of $SO(3)$

or

(b) it still holds $V_{R}V_{R'} \neq V_{RR'}$ (so that $V$ remains a projective unitary rep. of $SO(3)$) but

$SU(2) \ni A \mapsto V_{\Pi(A)}$ is an irreducible continuous unitary representation of $SU(2)$

In the second case the representation of $SO(3)$ is still projective unitary, but we perfectly know how it is far from a properly unitary one: it is indeed possible to prove that $V_I=I$ and $V_RV_{R'} = \pm V_{RR'}$.

Cases (a) and (b) are distinguished by a parameter labeling the representations: it is nothing but the $j=0,1/2,1,3/2, \ldots$ in the expression $j(j+1)$ of the eigenvalue of $J^2$ which physicists known very well.

(a) is proper of integer spin representations and (b) is proper of semi-integer spin representation.

We are ready to come back to the misunderstanding about $4\pi$.

Consider a spin-$1/2$ representation. In this case the Hilbert space is $\mathbb{C}^2$ and from the point of view of $SU(2)$, the representation is $SU(2) \ni A \mapsto A= V_{\Pi(A)}$.

From the point of view of $SO(3)$ the representation is strictly projective. The fundamental observation is this one. Consider the group of rotations $R_\theta$ of $SO(3)$ around an axis ${\bf n}$. This is a one-parameter subgroup of $SO(3)$ which is the image through $\Pi$ of an analogous one-parameter subgroup of $SU(2)$ denoted by $A_\theta$. It turns out that this subgroup of $SU(2)$ is completely covered by taking $\theta \in [0, 4\pi]$ with in particular $$A_0=I, \quad A_{2\pi}= -I,\quad A_{4\pi}=I.$$ This is coherent with (3), since $\Pi(\pm I) = I$.

If $\psi \in \mathbb{C}^2$, we have that $$V_{\pi(A_{2\pi})}\psi = A_{2\pi}\psi = -\psi \tag{4}$$ whereas $$V_{\pi(A_{4\pi})}\psi = A_{4\pi}\psi = \psi\tag{5}$$

(4) and (5) lead to the misunderstanding that a rotation of $2\pi$ changes the state of a spin-$1/2$ particle and that a rotation of $4\pi$ is necessary to restore the initial state. This is false because a state is a unit vector up to a phase so that $\psi$ and $-\psi$ in particular define the same state.

Furthermore, we must not forget that the physical group we are representing is not $SU(2)$ but $SO(3)$ (even if the resulting representation is projective unitary). From the point of view of physics, $A_{2\pi}$ and $A_{4\pi}= -A_{2\pi}$ define the same rotation $\Pi(A_{2\pi})= \Pi(A_{4\pi}) =R_{2\pi}= I$.

Just a final remark. The partition into (a) and (b) of the continuous projective unitary representations of $SO(3)$ has non-negligible physical consequences. It gives rise to a superselection rule, but this is not the place where discussing it.

Regarding your last extra question: according to what I wrote above, the considered representation is now a proper unitary representation of $SO(3)$. I do not know how it is made, but it is possible that a rotation of $SO(3)$ around an axis of $\pi$ produces $e^{ia}I$. These matrices can be found in the literature and it is easy to check.

$\endgroup$
  • $\begingroup$ I'm curious as to why you make your continuity argument on $\vert \langle \psi\vert U_r\phi\rangle\vert^2$ and not on $\langle \psi\vert U_r\phi\rangle$ alone. $\endgroup$ – ZeroTheHero Nov 10 '19 at 18:39
  • $\begingroup$ Because the latter cannot physically measured as the two vectors and $U_r$ itself are affected by arbitrary phases, whereas the former has a direct physical intepretation in terms of probabilities ( statistical frequencies). $\endgroup$ – Valter Moretti Nov 10 '19 at 18:41
  • $\begingroup$ Right... I need to think about formulating my question in a sharper way. I'll think about this more. $\endgroup$ – ZeroTheHero Nov 10 '19 at 18:45
  • 1
    $\begingroup$ There is a huge literature on these issues. A summary can be found in my last two books...An extensive textbook on projective unitary representations is Varadarajan's book on the geometry of quantum theory $\endgroup$ – Valter Moretti Nov 10 '19 at 18:46
  • $\begingroup$ I understand your argument - it's quite clear - and I'm not disputing the projective stuff; I'm just surprised that continuity of $\vert U_r\phi\rangle$ isn't enough... will definitely read more. Thanks. $\endgroup$ – ZeroTheHero Nov 10 '19 at 19:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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