What is the general form of the rotation matrix in SU(2) space?

[Splinter1]

Hi. I know that the \sigma matrices are the generators of the rotations in su(2) space. They satisfy
[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k It is conventional therefore to take J_i=\frac{1}{2}\sigma_i such that [J_i,J_j]=i\epsilon_{ijk}\sigma_k. Isn't there a problem by taking these J_i since \det J_i \neq 1? (since we are talking about the special unitary group.)
Also, how does one arrive at the general form of the rotation matrix e^{i\bf{\sigma} \theta\cdot \bf{\hat{n}}/2}? the factor of 1/2 obviously comes from the definition of J above. Where does the \hat n come from?

 

[Orodruin]

Hi. I know that the \sigma matrices are the generators of the rotations in su(2) space. They satisfy
[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k It is conventional therefore to take J_i=\frac{1}{2}\sigma_i such that [J_i,J_j]=i\epsilon_{ijk}\sigma_k. Isn't there a problem by taking these J_i since \det J_i \neq 1? (since we are talking about the special unitary group.)

No there is no problem here, su(2) is the Lie algebra of the group SU(2) and is a linear vector space. The determinant of the generators is not relevant (you will also notice that the determinants of the $\sigma$s is -1.

Also, how does one arrive at the general form of the rotation matrix e^{i\bf{\sigma} \theta\cdot \bf{\hat{n}}/2}? the factor of 1/2 obviously comes from the definition of J above. Where does the \hat n come from?

Any SU(2) matrix can be written as an exponentiation of an element of the Lie group su(2). The $\theta$ and the $\hat n$ simply parametrise the three-dimensional Lie algebra su(2).

 

[Splinter1]

No there is no problem here, su(2) is the Lie algebra of the group SU(2) and is a linear vector space. The determinant of the generators is not relevant (you will also notice that the determinants of the $\sigma$s is -1.

I see. Thank you

Any SU(2) matrix can be written as an exponentiation of an element of the Lie group su(2). The $\theta$ and the $\hat n$ simply parametrise the three-dimensional Lie algebra su(2).

Ok. And \hat n parametrizes it since it can include any (normalized) 3d vector, thus describing any linear combination, and we can choose it to be normalized since a multiplicative factor in the exponent doesn't matter. Is that correct?

 

[vanhees71]

The general spin-1/2 rotation matrix reads
$$D(\vec{\varphi})=\exp(-\mathrm{i} \vec{\sigma} \cdot \vec{\varphi}/2),$$
where the magnitude of $\vec{\varphi}$ denotes the rotation angle and its direction $\hat{n}=\vec{\varphi}/|\vec{\varphi}|$ in the sense of the right-hand rule.

Since
$$\mathrm{det} D=\exp[\mathrm{Tr}(\ln D)]=\exp(-\mathrm{i} \mathrm{Tr} \vec{\varphi} \cdot \vec{\sigma}/2) \stackrel{!}{=} 1,$$
it follows that
$$\mathrm{Tr} \sigma_j=0, \quad j \in \{1,2,3 ,\}.$$
Further $D^{\dagger}=D^{-1}$ implies that
$$\sigma_j^{\dagger}=\sigma_j, \quad j \in \{1,2,3 \}.$$
Thus the generators of spin-1/2 rotations are the Hermitean traceless $\mathbb{C}^{2 \times 2}$ matrices, which build a vector space and together with the commutator a Lie algebra.

 

[Splinter1]

The general spin-1/2 rotation matrix reads
$$D(\vec{\varphi})=\exp(-\mathrm{i} \vec{\sigma} \cdot \vec{\varphi}/2),$$
where the magnitude of $\vec{\varphi}$ denotes the rotation angle and its direction $\hat{n}=\vec{\varphi}/|\vec{\varphi}|$ in the sense of the right-hand rule.

Since
$$\mathrm{det} D=\exp[\mathrm{Tr}(\ln D)]=\exp(-\mathrm{i} \mathrm{Tr} \vec{\varphi} \cdot \vec{\sigma}/2) \stackrel{!}{=} 1,$$
it follows that
$$\mathrm{Tr} \sigma_j=0, \quad j \in \{1,2,3 ,\}.$$
Further $D^{\dagger}=D^{-1}$ implies that
$$\sigma_j^{\dagger}=\sigma_j, \quad j \in \{1,2,3 \}.$$
Thus the generators of spin-1/2 rotations are the Hermitean traceless $\mathbb{C}^{2 \times 2}$ matrices, which build a vector space and together with the commutator a Lie algebra.

Thank you. How do you see the geometrical meaning of the expression \exp(-\mathrm{i} \vec{\sigma} \cdot \vec{\varphi}/2? i.e. how do you see that it means a rotation at an angle \varphi about the axis \hat{\varphi}? I thought about showing that the rotation matrix doesn't change the axis vector \hat{\varphi}=\hat{n}, as the axis should not rotate, but I'm not sure how to do this.

 

[vanhees71]

You can map the three vectors to
$X=\vec{x} \cdot \vec{\sigma}.$
Then the rotation is given by
$X'=D(\vec{\varphi}) X D^{-1}(\vec{\varphi}).$
This is the socalled adjoint representation, which is the fundamental representation of rotations in terms of SO(3).

 

[Splinter1]

You can map the three vectors to
$X=\vec{x} \cdot \vec{\sigma}.$
Then the rotation is given by
$X'=D(\vec{\varphi}) X D^{-1}(\vec{\varphi}).$
This is the socalled adjoint representation, which is the fundamental representation of rotations in terms of SO(3).

But how \hat{n} and \varphi are interpreted as the axis and the angle of the rotation?

 

[George Jones]

But how \hat{n} and \varphi are interpreted as the axis and the angle of the rotation?

Take an arbitrary vector and decompose it into a part parallel (or ant-parallel) to n, and a part orthogonal to n. The part parallel to n remains invariant (because it is along the axis of rotation), and the part orthogonal to n rotates by angle psi in a plane orthogonal to n.