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Eigenvectors of Ly and associated energies

  1. Feb 2, 2017 #1
    1. The problem statement, all variables and given/known data
    Consider a particle with angular momentum [tex]l=1.[/tex] Write down the matrix representation for the operators [tex]L_x,\,L_y,\,L_z,[/tex]for this particle. Let the Hamiltonian of this particle be [tex]H = aL\cdot L-gL_z,\,g>0.[/tex]Find its energy values and eigenstates. At time [tex]t=0,[/tex]we make a measurement of [tex]L_y[/tex]in the ground state. Find the possible values we will find from such a measurement, and the probability for each of these possible results. Hence find the expectation value of, [tex](L_y)^2,[/tex] at time, [tex]t=0.[/tex]In the above measurement, if we find that the particle has [tex]L_y=+1[/tex] in one measurement at time [tex]t=0,[/tex]find an expression for the time-dependent wavefunction at any later time.

    2. Relevant equations
    Relevant equations are the ladder operators, [tex]L_{\pm}.[/tex]

    3. The attempt at a solution
    The first part is straightforward enough. We rewrite [tex]L_{x,y},[/tex] in terms of ladder operators to find the 3x3 matrices in the basis, [tex]|1,1>=(1\,0\,0),\,|1,0>=(0\,1\,0),\,|1,-1>=(0\,0\,1),[/tex]and observing how the basis transforms. Doing this I got,[tex]L_x=\frac{\hbar}{\sqrt{2}}\begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0 \end{pmatrix},\,L_y=\frac{\hbar}{\sqrt{2}}\begin{pmatrix} 0&-i&0\\ i&0&-i\\ 0&-&0 \end{pmatrix}.[/tex]

    Because [tex]L^2,\,L_z[/tex] are simultaneously diagonalizable, they have the same eigenvectors, with eigenvalues corresponding to [tex](1\,0\,0)\rightarrow 2a\hbar^2-g\hbar,\,(0\,1\,0)\rightarrow 2a\hbar^2,\,(0\,0\,1)\rightarrow 2a\hbar^2+g\hbar,[/tex]such that the ground state corresponds to [tex](1\,0\,0).[/tex]That is the lowest energy. By symmetry I know that [tex]L_y[/tex] has eigenvalues, [tex]1,\,0,\,-1.[/tex]However, I'm having some trouble finding the eigenvectors in order to take the inner product. As well, what would be the associated energies to find time evolution?
     
  2. jcsd
  3. Feb 3, 2017 #2

    blue_leaf77

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    Consider this
    $$
    L_z|z\rangle = m\hbar|z\rangle \\
    e^{-i\phi L_x/\hbar} L_z e^{-i\phi L_x/\hbar} e^{-i\phi L_x/\hbar}|z\rangle = m\hbar e^{-i\phi L_x/\hbar} |z\rangle
    $$
    If you set ##\phi = 90^o##, you will get ##Ly## from the first three operators in LHS. So, if you apply rotation operator around x axis by 90 degree, ##\exp(-i\frac{\pi L_x}{2\hbar})##, to the eigenvectors of ##L_z## you can actually get the eigenvectors of ##L_y##.
     
    Last edited: Feb 3, 2017
  4. Feb 3, 2017 #3

    PeroK

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    If you don't understand the clever way to do this above, then it's just an eigenvector problem. You know ##L_y## expressed as a matrix (in the z-basis) and you know its eigenvalues in advance (so that saves you having to work those out). You just have to look for the eigenvectors (expressed in the z-basis).

    Note that it's relatively easy to show that ##\langle L_y \rangle = 0##, but this doesn't help you with ##L_y^2##. It tells you ##L_y = \pm \hbar## are measured equally likely but it doesn't tell you how frequently ##L_y = 0## is measured. So, I don't see an obvious short cut: you'll have to express the z-eigenstate ##|1 \ 1 \rangle## in terms of the eigenstates of ##L_y##.
     
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