# Eigenvectors of Ly and associated energies

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1. Feb 2, 2017

### vbrasic

1. The problem statement, all variables and given/known data
Consider a particle with angular momentum $$l=1.$$ Write down the matrix representation for the operators $$L_x,\,L_y,\,L_z,$$for this particle. Let the Hamiltonian of this particle be $$H = aL\cdot L-gL_z,\,g>0.$$Find its energy values and eigenstates. At time $$t=0,$$we make a measurement of $$L_y$$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, $$(L_y)^2,$$ at time, $$t=0.$$In the above measurement, if we find that the particle has $$L_y=+1$$ in one measurement at time $$t=0,$$find an expression for the time-dependent wavefunction at any later time.

2. Relevant equations
Relevant equations are the ladder operators, $$L_{\pm}.$$

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

Because $$L^2,\,L_z$$ are simultaneously diagonalizable, they have the same eigenvectors, with eigenvalues corresponding to $$(1\,0\,0)\rightarrow 2a\hbar^2-g\hbar,\,(0\,1\,0)\rightarrow 2a\hbar^2,\,(0\,0\,1)\rightarrow 2a\hbar^2+g\hbar,$$such that the ground state corresponds to $$(1\,0\,0).$$That is the lowest energy. By symmetry I know that $$L_y$$ has eigenvalues, $$1,\,0,\,-1.$$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. Feb 3, 2017

### blue_leaf77

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
3. Feb 3, 2017

### PeroK

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 $L_y$.