Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

My first question is fairly basic, but I would like to clarify my understanding. The second question is to turn this into something worth answering.

Consider a relativistic electron, described by a spinor wave function $\psi(\vec x ,\sigma)$ and the Dirac equation. The conventional wisdom is that rotating everything by 360 degrees will map the spinor to its negative $\psi \mapsto -\psi$. However, it appears to me that this statement is "obviously false", because a rotation by 360 is, when viewed as an element of the group $SO(3)$, exactly equal to the identity map and cannot map anything to its negative.

Thus, to make sense of the behavior of spin under "rotation", I have to conclude the following

The rotation group $SO(3)$ does not act on the configuration (Hilbert) space of electrons. Only its double cover $SU(2)$ acts on the space of electrons.

Is this interpretation correct?

So, essentially, there is a symmetry group $SU(2)$ which acts on "physics", but its action on the spatial degrees of freedom is just that of $SO(3)$.

What other groups, even larger than $SU(2)$, are there that (could) act on "physics" and are an extension of $SO(3)$? Is it possible to classify all possibilities, in particular the ones that are not direct products?

Of course, gauge freedoms will give rise to direct products like $SO(3) \times U(1)$ (acting on space $\times$ electromagnetic potential), but I would consider these to be trivial extensions.

share|improve this question
    
Thanks always wanted to know this –  JPM Jan 9 '12 at 17:32
add comment

2 Answers

The group $SO(3)$ is the physical transformation group of spatial rotations. It's also contained in the bigger Lorentz group, which is the isomorphism group of the spacetime metric in special relativity. And you transform object like vectors $\vec x$ as in $$\vec x \mapsto \vec x'= R\cdot\vec x,$$

where $R\in SO(3)$ is a matrix.

In quantum mechanics you are interested in expectation values of observables such as $$\langle X\rangle_\psi:=\frac{\langle\psi|X|\psi\rangle}{\langle \psi|\psi\rangle}.$$ This object is invariant under $|\psi\rangle\mapsto c\ |\psi\rangle$, where $c$ is some nonzero complex number. The state denoted with $\psi$ is a whole ray of vectors. Now you want to rotate the state in Hilbert space

$$|\psi\rangle\mapsto T_R |\psi\rangle,$$

where $T_R$ is a transformation corresponding to the rotation with fundamental representation $R$. Putting everything together, you see that if you represent transformations (rotations,...) in a Hilbert space of quantum mechanics, you have to represent them ($SO(3),...$) only up to a phase. Therefore you now have more options. Your factor $-1$ is such a phase.

You can read the second chapter of Weinbergs book on QFT to get into some more details. The result is the possibility for $SU(2)$ representations, i.e. spinors. Roughly speaking you could say that you still use the $SO(3)$-angles for your transformation, since both have three parameters. But as you say in your first question, $SU(2)$ is actively acting on the two vector components of the fermion. Note that if you have coordinate dependent objects $\psi(\vec x)$, then the argument $\vec x$ still transforms conventionally, while the field $\psi$ as such has some transformation law.

Of course, there are many more groups relevant for Quantum Mechanics and other physical theories (like gauge groups, as you mentioned). But concerning your second question, as far as "rotations only" go, $SU(2)$ is already the universal covering group. These are the simple connected Lie-Groups, which lie on top of these upward chains you're searching for.

As a side note, this double covering can also be done for other dimensions of some $S$pecial groups.

share|improve this answer
1  
+1. Also, another way to think of the double valued nature of the spinor rotations is to consider not just the rotation of an object, but to maintain a view of its relation with its environment. This has been called the "orientation-entanglement" relation and was discussed in Misner Thorne and Wheeler (the relevant bit is reproduced here) –  twistor59 Jan 9 '12 at 19:57
    
Reading around the links you gave, I think you want to say that $SO(3)$ acting on the wave function is a projective representation, i.e. up to a phase. Then, it is a "known fact" that every projective representation of $SO(3)$ arises from its universal cover $SU(2)$? –  Greg Graviton Jan 11 '12 at 17:17
    
@GregGraviton: Is this a question? Well, yes, what you say is pretty much this paragraph. As a side note, understanding $SU(2)$ is also an essential part of understanding $SU(3)$, for example. (Small remark: I guess all of this "projective business" should not be mixed up with that). –  NiftyKitty95 Jan 11 '12 at 18:25
    
@NickKidman: Ok. Concerning accepting an answer, I'm happy with the information you provided, but I'd like to reword it heavily. Shall I edit your answer or should I make my my own and accept that one? –  Greg Graviton Jan 13 '12 at 13:58
    
@Greg Graviton: Whatever, I don't care. –  NiftyKitty95 Jan 13 '12 at 14:16
add comment
up vote 4 down vote accepted

Actually, the rotation group $SO(3)$ does act "on physics", even in the presence of spin.

The thing is that the wave function $\psi(\vec x, \sigma)$ is a redudant description of a physical state. A wave function with a different overall phase $c\psi(\vec x, \sigma)$ describes exactly the same physical state. After all, the only quantities of interest are only the expectation values of observables

$$\langle X\rangle_\psi:=\frac{\langle\psi|X|\psi\rangle}{\langle \psi|\psi\rangle}.$$

and these are invariant under a rotation $R$

$$\langle X\rangle_{R\psi} = \langle X\rangle_{\psi} .$$

Mathematically, we can say that the action of the rotation group on physical states is a projective representation, i.e. it acts on lines $\lbrace\lambda\psi(\vec x,\sigma), \lambda\in\mathbb C\rbrace$ (one-dimensional subspaces) in a Hilbert space, but not on the individual vectors. However, as you can read in the wikipedia page above, every projective representation of a Lie group like $SO(3)$ can usually be obtained from a linear representation of its universal covering group like $SU(2)$. (Linear representation just means that the group acts on individual vectors.)

To summarize, the rotation group $SO(3)$ acts on ordinary quantum mechanics, too, but for practical calculations, it's useful to generalize it to $SU(2)$ instead.

There is even a bit hair splitting as to whether you consider wave functions as physically relevant quantities and add an additional symmetry ("up to phase"), or whether you take the quotient "wave function up to phase" as physically relevant quantities and work in the quotient space.


As for the second question, I think it's possible to classify all Lie groups $G$ with a homomorphism $G \to SO(3)$ via group cohomology, but I'm not familiar enough with this topic to give an answer.

share|improve this answer
2  
"The thing is ..." I like how you use casual jargon even if you're basically writing a note to yourself :) –  NiftyKitty95 Jan 22 '12 at 1:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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