# Eigenvectors of Ly and associated energies

• vbrasic
In summary: Fortunately, you know how to do this as I just mentioned above.In summary, we consider a particle with angular momentum l=1 and write down the matrix representation for the operators L_x, L_y, and L_z. The Hamiltonian of this particle is then given as H = aL*L - gL_z, where g>0. We then use the ladder operators L+ and L- to rewrite L_x and L_y in terms of these operators and find the 3x3 matrices in the basis |1,1>, |1,0>, and |1,-1>. The eigenvalues for L^2 and L_z are then determined to be 2aℏ^2 - gℏ,
vbrasic

## Homework Statement

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.

## Homework Equations

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

## 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?

vbrasic said:
However, I'm having some trouble finding the eigenvectors in order to take the inner product. As we
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:
vbrasic said:
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?

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##.

## 1. What is the significance of eigenvectors in the study of Ly and associated energies?

Eigenvectors play a crucial role in the study of Ly and associated energies as they represent the directions in which a linear transformation (such as the Ly operator) has simple behavior. In other words, eigenvectors are the special vectors that are unchanged in direction when the Ly operator is applied to them, making them essential in understanding the behavior of Ly and its associated energies.

## 2. How are eigenvectors of Ly and associated energies determined?

Eigenvectors of Ly and associated energies can be determined by solving the characteristic equation of the Ly operator. This equation involves finding the values of the associated energies (eigenvalues) that satisfy the equation and then finding the corresponding eigenvectors by plugging in these eigenvalues into the original equation.

## 3. Can eigenvectors of Ly and associated energies be complex numbers?

Yes, eigenvectors of Ly and associated energies can be complex numbers. This is because the characteristic equation of the Ly operator can have complex eigenvalues, which in turn leads to complex eigenvectors. These complex eigenvectors can still provide valuable insight into the behavior of Ly and its associated energies.

## 4. How are eigenvectors of Ly and associated energies used in practical applications?

Eigenvectors of Ly and associated energies have various applications in fields such as physics, engineering, and data analysis. They are used to decompose complex systems into simpler components, identify dominant patterns and behaviors, and make predictions about the future behavior of a system. For example, in quantum mechanics, eigenvectors of the Hamiltonian operator (which is related to the Ly operator) represent the possible states of a quantum system, and their associated energies represent the possible energy levels of the system.

## 5. Can eigenvectors of Ly and associated energies change over time?

No, eigenvectors of Ly and associated energies remain constant over time. This is because eigenvectors are defined as the vectors that are unchanged in direction when the Ly operator is applied to them. Therefore, even if the associated energies of the system change, the eigenvectors will remain the same unless the Ly operator itself changes. However, the coefficients of the eigenvectors (i.e. the amount of each eigenvector present in a given state) can change over time as the system evolves.

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