1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Spin 1/2 Basis Change

  1. Oct 8, 2011 #1
    1. The problem statement, all variables and given/known data

    This is from my first-quarter graduate QM course. Part 4 of this problem asks me to compute the unitary operator U which transforms Sn into Sz, where Sn is the spin operator for spin 1/2 quantized along some arbitrary axis n = icos[itex]\phi[/itex]sinθ + jsin[itex]\phi[/itex]sinθ + zcosθ.

    2. Relevant equations

    From part 1, I solved Sn = \begin{pmatrix} cosθ&sinθe-i[itex]\phi[/itex]\\ sinθei[itex]\phi[/itex]&-cosθ \end{pmatrix}

    |n;+> = cos(θ/2)|+> + sin(θ/2)ei[itex]\phi[/itex]|->
    |n;-> = sin(θ/2)|+> - cos(θ/2)ei[itex]\phi[/itex]|->

    where |+> and |-> are the spin up and down states along the z-axis

    Also: U(adjoint)SnU = Sz

    3. The attempt at a solution

    U = Ʃ|n;±><±| = |n;+><+| + |n;-><-| = [cos(θ/2)|+> + sin(θ/2)ei[itex]\phi[/itex]|->]<+| + [sin(θ/2)|+> - cos(θ/2)ei[itex]\phi[/itex]|->]<-|

    I then calculated out <+|U|+>, <-|U|+>, <+|U|->, and <-|U|-> to get the 4 elements of the unitary operator matrix, and end up with:

    U = \begin{pmatrix}cos(θ/2)&sin(θ/2)ei[itex]\phi[/itex]\\sin(θ/2)&-cos(θ/2)ei[itex]\phi[/itex]\end{pmatrix}

    However I used that to transform Sn and could not recover Sz. U(adjoint) is just the transpose complex conjugate so then I should be using:

    U(adjoint) = \begin{pmatrix}cos(θ/2)&sin(θ/2)\\sin(θ/2)e-i[itex]\phi[/itex]&-cos(θ/2)e-i[itex]\phi[/itex]\end{pmatrix}

    I'm not quite sure where to go from here.

    Edit: I also can't figure out how to post matrices, if someone can help with that also. Thanks!

    Edit 2: Thanks vela for the fix.
    Last edited: Oct 8, 2011
  2. jcsd
  3. Oct 8, 2011 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Fixed your LaTeX.
  4. Oct 9, 2011 #3
    Okay so I did out the matrix math a little further, found some more double angle identities and so forth, and if I set [itex]\phi[/itex] to 0 after multiplying through it comes out to σ3 which is what I want for Sz.

    I think that this is because the unitary operator has to keep the inner product and its therefore an orthogonal transformation, meaning U must be orthogonal. If that's the case then it has to have real elements and the only way for that is if phi = 0, which makes sense in that the the azimuthal angle needs to be 0 in order to be orthogonal to the z-axis.

    Is that correct?
  5. Oct 9, 2011 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your matrix for Sn is missing an overall factor of 1/2.

    If you multiply the second eigenvector by e-iΦ, you'll have
    \vert + \rangle_n &= \begin{pmatrix} \cos(\theta/2) \\ e^{i\phi}\sin(\theta/2) \end{pmatrix} \\
    \vert - \rangle_n &= \begin{pmatrix} e^{-i\phi}\sin(\theta/2) \\ -\cos(\theta/2)\end{pmatrix}
    \end{align*}(It's not really necessary to do this. I'm just doing it for aesthetic reasons.)

    If you recall your basic linear algebra, you know that the matrix U that diagonalizes Sn has the eigenvectors of Sn as its columns, so you have[tex]U=\begin{pmatrix}
    \cos(\theta/2) & e^{-i\phi}\sin(\theta/2) \\
    e^{i\phi}\sin(\theta/2) & -\cos(\theta/2)
    You can verify that this matrix is in fact unitary for any value of the angles and that it will diagonalize Sn. Requiring Φ=0 isn't acceptable since n can point in any direction.

    For some reason, you ended up with the transpose of this matrix.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook