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Sakurai Ch.3 Pr.6

  1. Jan 9, 2008 #1
    [SOLVED] Sakurai Ch.3 Pr.6

    1. The problem statement, all variables and given/known data

    Let [tex] U = \text{e}^{i G_3 \alpha} \text{e}^{i G_2 \beta} \text{e}^{i G_3 \gamma}[/tex], where [tex] ( \alpha , \beta , \gamma ) [/tex] are the Eulerian angles. In order that [tex] U [/tex] represent a rotation [tex] ( \alpha , \beta , \gamma ) [/tex], what are the commutation rules satisfied by [tex] G_k [/tex]? Relate [tex] \mathbf{G} [/tex] to the angular momentum operators.

    2. Relevant equations

    3. The attempt at a solution

    I don't really know how to start here. In chapter 3.3 they represent rotation with Euler angles by 2x2 matrices but I don't think that's what I'm supposed to use. Instead I would guess that if [tex] U [/tex] is supposed to be a rotation operator then it has to be an element of either SO(3) or SU(3). Then from the fact that it belongs to one of these groups, one would hopefully get the desired commutation relations.

    Any thoughts?
  2. jcsd
  3. Jan 11, 2008 #2
    Ok, so I've gone over this problem like a hundred times and finally settled for an approach to the problem.

    From what I gather it is most simple to look at an infinitesimal rotation, we know that we can write an infinitesimal rotation as:

    [tex] \mathcal{D} ( d \theta ) = \left( 1 + i \mathbf{J} \cdot \hat{n} d \theta \right) [/tex]

    Where [tex] \mathbf{J} [/tex] is an Hermitian operator and [tex] \hat{n} [/tex] is the axis of rotation. Now if we look at the case of rotation by Euler angles, then each angle [tex] \alpha , \beta , \gamma [/tex] respectively "represents" a rotation about an axis. From this and the fact that rotations are closed under their binary operation we can write [tex] U [/tex] as a product of 3 infinitesimal rotations. Thus we write

    [tex] U ( dR )= \left( 1 + i G_3 d \alpha \right) \left( 1 + i G_2 d \beta \right) \left( 1 + i G_3 d \gamma \right) [/tex]

    Now we consider two rotations following upon each other, from the geometry of 3-D objects we know that 2 rotations do not necessarily commute. Therefore it seems appropriate to examine the following

    [tex] U ( dR ) U ( dR' ) - U ( dR' ) U ( dR ) [/tex]

    So that is what I am doing now. I'll get back with what I find as soon as I've finished the calculations. If anyone see any flaws in my reasoning, please point them out.
  4. Jan 17, 2008 #3
    Ok, I think I've finally come up with an acceptable solution. Here goes.

    We know that we can write an infinitesimal rotation [itex] R ( d \theta )[/itex] as

    [tex] R ( d \theta ) = 1 + i \mathbf{J} \cdot \hat{ \mathbf{n} } d \theta [/tex]. (1)

    We also know that if R is an operator generating equations it must fullfill the following relations

    [tex] R _i ( \epsilon ) R _j ( \epsilon ) - R _j ( \epsilon ) R _i ( \epsilon ) = R_k ( \epsilon ^2 ) - 1 [/tex], (2)

    where i,j,k=x,y,z and [itex] \epsilon [/itex] is an infinitesimal angle.
    Now remember that for Euler angles, [itex] \alpha , \beta , \gamma [/itex] represents rotations about the z,y and z axes respectively. Then it follows that in order for

    [tex] U ( \alpha , \beta , \gamma ) = \text{e}^{i G_3 \alpha} \text{e}^{i G_2 \beta} \text{e}^{i G_3 \gamma} [/tex] (3)

    to be an operator generating rotations it has to satisfy eq. (2). Evaluating an infinitesimal rotation [itex] \epsilon [/itex] about the y- and z-axes we find that (ignoring terms of higher order than [itex] \epsilon ^3 [/itex])

    [tex] U( 0 , \epsilon , \epsilon ) U( \epsilon , \epsilon , 0) - U( \epsilon , \epsilon , 0 ) U( 0 , \epsilon , \epsilon ) = 1 - [G_2,G_3] \epsilon ^2 - 1 + \mathcal{O}( \epsilon ^3 ) [/tex] (4)

    Now comparing (4) with (2) we notice that [itex] 1- [G_2,G_3] \epsilon ^2 [/itex] must represent a rotation about the x-axis with an infinitesimal angle [itex] \epsilon ^2 [/itex]. We get that

    [tex] [G_2,G_3] = -i G_1 [/tex],

    where [itex] G_1 [/itex] is a generator of rotations about the x-axis.
    Repeating the same argument about other axes we find the following commutation relations:

    [tex] [G_i, G_j] = - i \varepsilon _{ijk} G_k [/tex].

    Comparing these to the commutation relations for the angular momentum operators we find that

    [tex] G_i = - \frac{J_i}{\hbar} [/tex]
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