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Homework Help: Normalized Eigenvectors of a Hermitian operator

  1. Oct 29, 2012 #1
    Hi all

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

    Given is a Hermitian Operator H
    [tex]H= \begin{pmatrix}
    a & b \\
    b & -a

    where as [tex] a=rcos \phi , b=rsin \phi[/tex]

    I shall find the Eigen values aswell as the Eigenvectors. Furthermore I shall show that the normalized quantum states are:

    [tex] \mid + \rangle =\begin{pmatrix}
    \frac{cos \phi}{2} \\
    \frac{sin \phi}{2}


    [tex] \mid - \rangle =\begin{pmatrix}
    \frac{-sin \phi}{2} \\
    \frac{cos \phi}{2}

    I know that: [tex] \frac {tan \phi}{2} = \frac {1-cos \phi}{sin \phi}[/tex]

    2. Relevant equations


    3. The attempt at a solution

    I calculated the Eigen values vie the determinant (=0) (let's call them lambda). I think I did that right and the solutions are:
    [tex] \lambda_1=r ,\lambda_2=-r[/tex]

    Furthermore I calculated two Eigenvectors. I have something like an inner product

    from H times a vector, so I just swapped the two entries and put a minus in front of one. My two Eigenvectors are:

    [tex] \vec v_1 =\begin{pmatrix}
    -rsin \phi \\
    rcos \phi - r

    \vec v_2 =\begin{pmatrix}
    -rsin \phi \\
    rcos \phi + r

    I calculated the norm which is
    [tex] ||v_1||^2 = 2r^2(1-cos \phi)[/tex]

    But now I'm stuck. I don't get the solution I should get. Did I do something wrong?

    Thanks for your help.
  2. jcsd
  3. Oct 29, 2012 #2


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    Gold Member

    Don't know if it helps any, but that should read [tex]tan \frac {\phi}{2} = \frac {1-cos \phi}{sin \phi}[/tex]
  4. Oct 30, 2012 #3
    Something looks wrong with your eigenvectors. Try working on them again / show your work.
  5. Oct 30, 2012 #4
    I'm in a hurry. Thanks for your help guys. I'll post tomorrow how I calculated my eigenvectors.


    Thanks for that info, it wasn't clear on the sheet where that /2 belongs to. The vectors + and - which are given are also wrong then, it should be sin(phi/2) etc I guess then.
  6. Oct 30, 2012 #5
    Yup...the original eigenvectors you were supposed to prove seemed kind of dubious as they arent normalised.
  7. Oct 31, 2012 #6
    Well, my attempt was totally wrong. But I'm stuck now.

    I try to find the Eigenvector for the Eigenvalue r

    I have:

    [tex] \begin{pmatrix}
    rcos \phi & rsin \phi \\
    rsin \phi & -rcos \phi
    \end{pmatrix} \begin{pmatrix}
    x \\
    0 \\

    When I try to solve this I come to:

    [tex] xrcos \phi -xr +yrsin \phi = 0 \\
    xrsin \phi -yrcos \phi -yr= 0[/tex]

    Furthermore I come to:

    [tex] x= y \frac{sin \phi}{1-cos\phi}[/tex]

    aswell as
    [tex] y= x\frac{sin \phi}{cos\phi +1}[/tex]

    My problem is the following: I don't know what my x and y are and I can't solve it properly. I'm stuck. Did I do it right thus far? And what to do next?

    Thanks for your help
  8. Oct 31, 2012 #7
    Well that's it! In general, there is no unique eigenvector - if |+> is an eigenvector of the matrix, then clearly any multiple of |+> too is a possible eigenvector.

    So, our generalised eigenvector (for eigenvalue r) is
    \left(\begin{matrix} \frac{sin \phi}{1-cos\phi} \\ 1 \end{matrix}\right)
    \equiv \left(\begin{matrix} sin \phi \\ 1-cos\phi \end{matrix}\right)
    \equiv \left(\begin{matrix} 2 sin \frac{\phi}{2} cos \frac{\phi}{2} \\ 2 sin^{2}\frac{\phi}{2} \end{matrix}\right)
    \equiv \left(\begin{matrix} cos \frac{\phi}{2} \\ sin \frac{\phi}{2} \end{matrix}\right)
    where we recognise that the final eigenvector is normalised.
    (you could of course simply normalise the first eigenvector and arrive at the same result)
  9. Oct 31, 2012 #8
    So you just chose your y as one because you say any multiplied vector to the eigenvector is one eigenvector aswell?

    Thanks for the help everyone.

    The task was to show that the normalized eigenvectors are the given ones. I did it just with the equation

    H*ev=lambda*ev (ev=eigenvector)

    but I wanted to know the other way around, because this looks more elegant to me^^.
  10. Oct 31, 2012 #9
    Yup, it doesn't matter which multiple of the eigenvector I choose to use. After normalising, I will still end up with the same normalised eigenvector.
  11. Oct 31, 2012 #10
    Ok, I think I understand it a bit better now. Thanks for your help again
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