Generator of Rotations: J_x, J_y, J_z Relationship

[dEdt]

My textbook derives the relationship [J_x, J_y] = 2pi*iJ_z by considering the J's as the generators of the Euclidean rotation operator. Why does this result hold when J is the generator of other rotations, such as rotations in a quantum mechanical spin-state space?

 

[arkajad]

On the Lie algebra (infinitesimal) level there is no difference between integer and half-integer representations of the rotation group commutation relations. The difference comes in only when we want to integrate infinitesimal commutation relations to group operations. Spin representations can be thought of as two-valued (projective) representations of the Euclidean rotation group.

 

[Fredrik]

My textbook derives the relationship [J_x, J_y] = 2pi*iJ_z by considering the J's as the generators of the Euclidean rotation operator. Why does this result hold when J is the generator of other rotations, such as rotations in a quantum mechanical spin-state space?

If I understand the question correctly, it's more of an axiom than anything else. We want to somehow include the idea that space is isotropic, and the easiest way to do that is to postulate the existence of a representation of the rotation group or its Lie algebra, into the group of linear operators on the system's Hilbert space.

 

[tom.stoer]

One "axiom" of QM is that symmetries must be represented as hermitean operators in a Hilbert space. So if you want to have a qm system that is symmetric w.r.t. the rotation group SO(3) you have to translate the so(3) algebra of rotations in 3-space into hermitean operators (the angular momentum operators) acting on states in the Hilbert space (this is one examople, there are others, e.g. simple translation).
That's just one step during quantization of a classical system.

Whenever the result does not hold something went wrong during the quantization. Either there is no definition of the symmetry on the Hilbert space (weird, as that means that the translation somehow failed completely), or the algebra of the symmetry is violated (an anomaly, may happen e.g. in quantum field theory), or the symmetry is violated on certain states (spontaneous symmetry breaking).

 

[arkajad]

More exactly, we are talking about "presymmetry" here, because we may have symmetry operations even when the external forces (like an external potential) break the symmetry of the system under consideration. See, for instance, http://projecteuclid.org/DPubS?verb...=UI&handle=euclid.cmp/1103859880&page=record" by Y. Avishai and H. Ekstein

 

[tom.stoer]

I guess what you mean by pre-symmetry is that one can make disctinction like
a) introduce generators T^a
b) check if the system is invariant, i.e. if [H,T^a]
c) check if T^a|state> = 0

 

[arkajad]

You do not need c). There will always be states that are not symmetric, even if the system has a given symmetry. There is nothing wrong with such states. It is like that you can consider one trajectory in a perfectly symmetric Euclidean space in classical mechanics. But the distinction between a) and b) is relevant.
Somehow we associate momentum operators with the group of translations even when there is an external potential that evidently brakes the translational symmetry (soft symmetry breaking).

This may break when we have phase transitions and we have to change the Hilbert space in a drastic way so that translations (or rotations) are not even even implemented there (hard symmetry breaking). But that is another story.

 

[tom.stoer]

You are completely right; c) is irrelevant here.

Now: what do you mean by pre-symmetry?

 

[dEdt]

Thanks everyone for the answers.

Problem is, I didn't really understand them! I'm only a novice at QM, and I'm not familiar with Lie algebra or SO(3). Is there still a way to answer my question without using stuff I haven't seen, other than just saying it's a postulate (or a way of seeing how it relates to "isotropy of space)?

 

[tom.stoer]

I'll try to.

One "axiom" of QM is that symmetries must be represented as hermitean operators in a Hilbert space. So if you want to have a qm system that is symmetric w.r.t. rotations in 3-space you have to translate the 3 matrices generating these rotations i.e. generating rotation matrices acting on vectors in 3-space into hermitean operators acting on states in the Hilbert space.
That's just one step during quantization of a classical system.

Formally it looks like the following. You have 3-space with vectors $$\vec{r}$$ and rotation matrices $$R(\alpha_i)$$ acting on them, i.e.

$$\vec{r} \to \vec{r}^\prime = R(\alpha_i)\vec{r}$$

No instead of having this rotation symmetry acting in 3-space you want to have a qm system and study its behaviour w.r.t. rotations. In order to do that you translate the rotation specified by three angles into a unitary operator $$U(\alpha_i)$$. This operator does not act on the 3-vectors but on the state vectors, i.e.

$$|\psi\rangle \to |\psi^\prime\rangle = U(\alpha_i)|\psi\rangle$$

How are these two definitions related? Via the wave function. The wave function is defined as $$\psi(\vec{r}) = \langle\vec{r}|\psi\rangle$$

Now the unitary operator can act on the ket as follows

$$\psi(\vec{r}) \to \psi^\prime(\vec{r}) = \langle\vec{r}|\psi^\prime\rangle = \langle\vec{r}|U(\alpha_i)|\psi\rangle$$

In addition the operator can act on the bra

$$\langle\vec{r}|U(\alpha_i)|\psi\rangle = \langle R(\alpha_i)\vec{r}|\psi\rangle = \langle \vec{r}^\prime|\psi\rangle = \psi(\vec{r}^\prime)$$

So rotating 3-vectors and rotating state vectors is related.