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Rotations in SU(2) space

  1. Mar 30, 2015 #1
    Hi. I know that the [itex]\sigma [/itex] matrices are the generators of the rotations in su(2) space. They satisfy
    [tex] [\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k[/tex] It is conventional therefore to take [itex]J_i=\frac{1}{2}\sigma_i[/itex] such that [itex][J_i,J_j]=i\epsilon_{ijk}\sigma_k [/itex]. Isn't there a problem by taking these [itex] J_i [/itex] since [itex] \det J_i \neq 1[/itex]? (since we are talking about the special unitary group.)
    Also, how does one arrive at the general form of the rotation matrix [itex]e^{i\bf{\sigma} \theta\cdot \bf{\hat{n}}/2}[/itex]? the factor of 1/2 obviously comes from the definition of J above. Where does the [itex] \hat n [/itex] come from?
     
    Last edited: Mar 30, 2015
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  3. Mar 30, 2015 #2

    Orodruin

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    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.

    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).
     
  4. Mar 30, 2015 #3
    I see. Thank you


    Ok. And [itex] \hat n [/itex] 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?
     
  5. Mar 31, 2015 #4

    vanhees71

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    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.
     
  6. Mar 31, 2015 #5
    Thank you. How do you see the geometrical meaning of the expression [itex] \exp(-\mathrm{i} \vec{\sigma} \cdot \vec{\varphi}/2[/itex]? i.e. how do you see that it means a rotation at an angle [itex] \varphi [/itex] about the axis [itex] \hat{\varphi} [/itex]? I thought about showing that the rotation matrix doesn't change the axis vector [itex] \hat{\varphi}=\hat{n} [/itex], as the axis should not rotate, but I'm not sure how to do this.
     
  7. Apr 1, 2015 #6

    vanhees71

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    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).
     
  8. Apr 1, 2015 #7
    But how [itex] \hat{n} [/itex] and [itex] \varphi [/itex] are interpreted as the axis and the angle of the rotation?
     
  9. Apr 1, 2015 #8

    George Jones

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    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.
     
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