- #1

vbrasic

- 73

- 3

## Homework Statement

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.

## Homework Equations

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

## 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?