Operator with 3 degenerate orthonormal eigenstates

This means that the diagonal elements of ##\hat{A}## are the eigenvalues themselves, not just equal to them. In other words, the matrix is already diagonalized. However, the problem here is to show that any normalized linear combination of the three orthonormal vectors is also an eigenstate of ##\hat{A}## with eigenvalue ##a##. This means that any vector of the form ##| \psi \rangle = \alpha |1\rangle + \beta |2\rangle + \gamma |3\rangle##, where ##\alpha, \beta, \gamma## are complex numbers that satisfy ##|\alpha|^{2} + |\beta|^{2} + |\gamma|^{2
  • #1
Celso
33
1
Homework Statement
I've been given an operator ##\hat{A}## that has 3 orthonormal vectors as degenerate eigenstates corresponding to the same eigenvalue ##a##.

I know how the hamiltonian acts on these vectors and I want to use this information to check whether ##\hat{H}## and ##\hat{A}## commute or not.
Relevant Equations
## \hat{A} |1> a|1>, \hat{A} |2> a|2>, \hat{A} |3> a|3>##
##\hat{H} = \begin{bmatrix} \sigma & 0 & \sigma \\ 0 & \sigma & \delta \\ \sigma & \delta & \sigma \end{bmatrix} ##
With this information I concluded that the diagonal elements of ##\hat{A}## are equal to the eigenvalue ##a##, so ##\hat{A} = \begin{bmatrix} a & A_{12} & A_{13} \\ A_{21}& a & A_{23}\\A_{31} & A_{32} & a \end{bmatrix}## but I can't see how to go from this to the commuting relation, since I don't know the other terms.
 
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  • #2
Can you show that any normalized linear combination of the three orthonormal vectors is also an eigenstate of ##\hat A## with eigenvalue ##a##? If you can, what does this have to do with the question being asked here?
 
  • Informative
Likes Keith_McClary
  • #3
Celso said:
Relevant Equations:: ## \hat{A} |1> a|1>, \hat{A} |2> a|2>, \hat{A} |3> a|3>##

With this information I concluded that the diagonal elements of ##\hat{A}## are equal to the eigenvalue ##a##, so ##\hat{A} = \begin{bmatrix} a & A_{12} & A_{13} \\ A_{21}& a & A_{23}\\A_{31} & A_{32} & a \end{bmatrix}## but I can't see how to go from this to the commuting relation, since I don't know the other terms.
Note that your relevant equations say ##\hat{A} |n\rangle = a|n\rangle##, not ##\langle n \lvert \hat{A} \rvert n \rangle = a##.
 

Related to Operator with 3 degenerate orthonormal eigenstates

1. What is an operator with 3 degenerate orthonormal eigenstates?

An operator with 3 degenerate orthonormal eigenstates is a mathematical representation of a physical observable in quantum mechanics. It has three distinct eigenstates that have the same eigenvalue and are mutually orthogonal.

2. How is an operator with 3 degenerate orthonormal eigenstates represented?

An operator with 3 degenerate orthonormal eigenstates is typically represented by a matrix with three columns and three rows. The columns represent the eigenstates and the rows represent the coefficients of the eigenstates.

3. What is the significance of having three degenerate eigenstates?

Having three degenerate eigenstates means that the physical observable represented by the operator has three possible outcomes with the same probability. This is important in quantum mechanics as it allows for a greater range of possible outcomes and a more accurate representation of physical systems.

4. How are degenerate eigenstates different from non-degenerate eigenstates?

Degenerate eigenstates have the same eigenvalue and are mutually orthogonal, while non-degenerate eigenstates have different eigenvalues and are not necessarily orthogonal. This means that degenerate eigenstates have equal probabilities of being measured, while non-degenerate eigenstates have different probabilities.

5. Can an operator with 3 degenerate orthonormal eigenstates have more than 3 eigenstates?

Yes, an operator with 3 degenerate orthonormal eigenstates can have more than 3 eigenstates. However, the number of degenerate eigenstates will always be a multiple of 3. For example, an operator with 6 degenerate eigenstates will have 2 sets of 3 degenerate eigenstates. This is because degenerate eigenstates must have the same eigenvalue and be mutually orthogonal, so they must be grouped in sets of 3.

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