I Why is this 3D operator with degeneracies only giving me 2 eigenstates

Click For Summary
The discussion centers on the eigenvalue problem of a 3D operator represented by a specific matrix, revealing eigenvalues of 1 (with degeneracy) and -1. The non-degenerate eigenstate corresponding to the eigenvalue -1 is found to be a specific vector, while the degenerate eigenvalue 1 yields a two-parameter family of eigenstates. This leads to a realization that degeneracy does not imply a fixed number of distinct eigenstates but rather a subspace of solutions. The participants clarify that selecting different values for the parameter results in linearly independent vectors, and orthogonality can be enforced to obtain an orthogonal basis within the degenerate subspace. Ultimately, the interpretation of degeneracy and eigenstates is confirmed, emphasizing the nature of eigenvalue solutions in quantum mechanics.
struggling_student
Messages
9
Reaction score
1
The matrix representation of a certain operator in a certain basis is

$$\begin{bmatrix} 1 & 0 & 0 \\0 & 0 & -i \\ 0 & i & 0
\end{bmatrix} .$$

The eigenvalue problem leads to this equation

$$0=det\begin{bmatrix} 1-\lambda & 0 & 0 \\0 & -\lambda & -i \\ 0 & i & -\lambda
\end{bmatrix} =-(\lambda-1)^2(\lambda+1).$$

So the eigenvalues are ##1##, ##1## and ##-1##. There is a degenacy and this means that there are two distinct states with the same eigenvalue but I would like to find all the eigenstates, generically denoted as ##\begin{bmatrix} a \\b \\ c \end{bmatrix}##.

First, the non-degenerate ##\lambda=-1## leads to the following equation

$$0=\begin{bmatrix} 2 & 0 & 0 \\0 & 1 & -i \\ 0 & i & 1 \end{bmatrix} \begin{bmatrix} a \\b \\ c \end{bmatrix},$$

$$\therefore 2a=0 \ and \ b=ic.$$

So the nondegenerate eigenstate is ##\begin{bmatrix} 0 \\ic \\ c \end{bmatrix}=\begin{bmatrix} 0 \\i \\ 1 \end{bmatrix}## because without the loss of generality we can set ##c:=1##.

Now, for the degerate ##\lambda=1## the equation is

$$0=\begin{bmatrix} 0 & 0 & 0 \\0 & -1 & -i \\ 0 & i & -1 \end{bmatrix} \begin{bmatrix} a \\b \\ c \end{bmatrix},$$

##\therefore b=-ic##, any value for ##a## will do so the degerate state is ##\begin{bmatrix} a \\-ic \\ c \end{bmatrix}##. How do I interpret this?

I expected to find two different vectors but instead I've got a two-parameter result. What is the degerate state? Is what I wrote not a valid operator? Am I mistaken in thinking that a degerate eigenvalue corresponds to exactly two distinct eigenstates?
 
Physics news on Phys.org
If you put in two different values for ##a## you get two linearly independent vectors.
 
  • Like
Likes vanhees71 and struggling_student
Your result makes sense. The eigenvector calculation needs to give you all the vectors in the subspace (two dimensional here) with the degenerate eigenvalue. However, if you want to have a set of orthogonal vectors, then after you pick the first value of ##a## you determine the second vector by enforcing orthogonality.
 
  • Like
Likes vanhees71 and struggling_student
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...