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Quantum physics problem- Bra-Ket notation and operators

  1. Apr 24, 2012 #1
    Hello all,

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

    I’m trying to derive a result from a book on quantum mechanics but I have trouble with bra-ket notation and operators…
    Let’s say we have a photon moving along the cartesian z-axis. It is polarized and its state is
    Psi(theta) = cos (theta) x1 + sin(theta) x1
    Here, x1 and x2 are the base vectors.
    The book states that a rotation about z axis is represented by an operator U, which has the matrix (respective to x1 and x2 base):
    cos(fi) sin(fi)
    -sin(fi) cos(fi)
    It is the next step I have trouble with, the book states that by applying a rotation to psi(theta) you will get psi(theta+fi).
    When I use simple matrix multiplication of U and psi, I don’t get this result but rather Psi(fi-theta)…
    I did manage to produce the correct result when I used hermetian conjugate od U… Why is this so?


    2. Relevant equations

    What is the correct procedure and why? What am I doing wrong?

    3. The attempt at a solution

    Simple matrix multiplication...
     
  2. jcsd
  3. Apr 25, 2012 #2

    diazona

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    Homework Helper

  4. Apr 25, 2012 #3
    Thanks for your replay.

    I did actually do quite extensive search on the web, but I did not find what I was looking for.

    Maybe I should explain...

    The matrix (for the base [itex]\chi_{1}[/itex] and [itex]\chi_{2}[/itex]) of the rotation operator is (according to the book):
    [itex]U_{\phi, \textbf{k}}[/itex] =
    cos [itex]\phi[/itex] sin [itex]\phi[/itex]
    -sin [itex]\phi[/itex] cos [itex]\phi[/itex]

    the state of the photon is

    [itex]\Psi_{\theta}[/itex] = cos [itex]\theta[/itex] [itex]\chi_{1}[/itex] + sin [itex]\theta[/itex] [itex]\chi_{2}[/itex]

    The book also states that when rotating this state by the rotation matrix above, you will get

    [itex]\Psi_{\theta + \phi}[/itex]

    But this is not what I get unless I transpose the rotation matrix.

    If I do that, the matrix looks exactly like the one in the wikipedia article.

    So is the book wrong?

    The most confusing part of it all is when I look at the lecture notes. According to those, the professor first calculated the effect of the rotation on the base, like this:

    [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] | [itex]\chi_{1}[/itex] > = cos [itex]\phi[/itex] | [itex]\chi_{1}[/itex] > + sin [itex]\phi[/itex] | [itex]\chi_{2}[/itex] >
    and
    [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] | [itex]\chi_{2}[/itex] > = -sin [itex]\phi[/itex] | [itex]\chi_{1}[/itex] > + cos [itex]\phi[/itex] | [itex]\chi_{2}[/itex] >

    then he writes

    [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] | [itex]\Psi_{\theta}[/itex] > = [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] (cos [itex]\theta[/itex] | [itex]\chi_{1}[/itex] > + sin [itex]\theta[/itex] | [itex]\chi_{2}[/itex] > ) =

    then he simply rearanges:

    = cos [itex]\theta[/itex] [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] | [itex]\chi_{1}[/itex] > + sin [itex]\theta[/itex] [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] | [itex]\chi_{2}[/itex] > )

    and then he substitues the result [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] | [itex]\chi_{1}[/itex] > and [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] | [itex]\chi_{2}[/itex] > from above and that gives the result from the book.

    My question is:

    1) why is he using this procedure instead of simple matrix multiplication?
    2) why is Matrix multiplication wrong?
    3) why does the rotation matrix in the book look exactly like the transposed version of the one in wikipedia article (and any other article I could find)
    4) How did the professor calculate the effect of the operator on the base (i.e. [itex]\widehat{U_{\phi, \textbf{k}}}[/itex] | [itex]\chi_{1}[/itex] >)

    I've been tormented by this for a week now, any input is appreciated...
     
  5. Apr 25, 2012 #4
    This is most probably a problem of active and passive coordinate transformations.

    Basically, if you rotate your coordinate axes one way, then the new coordinates of a vector are obtained by rotating the old coordinates in the opposite direction. This is a constant pain in the rear that creeps up every time you do coordinate transformations.

    This explains why the rotation matrix you give in the first post has the opposite sign of theta than the one on the Wiki page linked in the second post.

    http://en.wikipedia.org/wiki/Active_and_passive_transformation

    http://arxiv.org/abs/1106.4446 section 1.5 (very brief)
     
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