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Rotation Operator/Orthogonal matrices in quantum mechanics

  1. Dec 31, 2008 #1
    1. The problem statement, all variables and given/known data

    Write all orthogonal matrices in the form [tex]e^{i\phi \frac{\hat{n}\bullet\vec{L}}{\hbar}}[/tex].

    2. Relevant equations
    3. The attempt at a solution

    I couldn't understand the question. An orthogonal matrix [itex]R[/itex] satisfies

    [tex]R^{T}R = RR^{T} = I[/tex]

    and rotation matrices in 3 dimensions are orthogonal. Further,

    [tex]e^{i\phi \frac{\hat{n}\bullet\vec{L}}{\hbar}}[/tex]

    describes a finite rotation of [itex]\phi[/itex] about the axis [itex]\hat{n}[/itex]. What do I have to show here?

    Thanks in advance.
  2. jcsd
  3. Dec 31, 2008 #2
    Hello Vivek :smile:

    All you have to do is to show that any orthogonal matrix can be interpreted as a rotation matrix. While you have stated that all rotation matrices are orthogonal, you have to show that all orthogonal matrices can be seen as some sort of rotation matrix. Now, if you can do this, you have proven the problem :smile:

    Happy New Year!
  4. Dec 31, 2008 #3
    Thanks for your reply Domnu, but I understand that the exponential form is indeed a rotation matrix so while I was able to paraphrase the problem in the form you stated it, it wasn't obvious to me how we can say that

    for every orthogonal matrix [itex]R[/itex], we can find [itex]\hat{n}[/itex] and [itex]\phi[/itex] such that

    [tex]R = D(\phi, \hat{n})[/tex]

    where D denotes the exponential form.

    How do I show this for arbitrary [itex]\hat{n}[/itex]? I know that a finite rotation can be broken up in terms of Euler angles, but even so, how do I prove the existence rigorously?

  5. Dec 31, 2008 #4
    Right, so we can try to do something of the following form: let [tex]A[/tex] be our matrix such that

    [tex]A = \begin{bmatrix} a & b & c\\ d & e & f \\ g & h & j\end{bmatrix}[/tex]​

    Now, each of the columns are orthogonal to each other, and we need that the determinant of [tex]A[/tex] be 1. In addition, we know that the sums of squares of each row and column are all 1, so if we look at each column vector put in space and form a box out of the three vectors, we find that this box is a "unit box" whose side lengths are all 1 (which nicely proves the determinant idea), and that it can be rotated somehow. In other words, this is just taking the x,y,z unit vectors and rotating them somehow (to preserve orthonormality). Now, getting this orthogonal matrix is simple: just apply the 3D rotation matrix (which you can get from Eulerian angles, etc.) to the identity matrix.

    Does this help? :smile:
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