[Q]Degeneracy and Commutability

[good_phy]

Hi

Liboff said that with two operator [A,B] = 0, If some eigenstate of operator A are degenerate, they are not necessarily completely common eigenstate of B.

How is it proved or where can i find proof of this theorem?

Please answer me.

 

[vanesch]

Simple example: consider in C^3, the unity operator, 1, and the projector on the first coordinate Px (Px(x,y,z) = (x,0,0) )

Clearly they commute: [1,Px] = 0 (as the unity operator commutes with all).
now, consider the vector (1,2,0). This is an eigenstate of 1 (all vectors are eigenstates of 1). However, it is not an eigenstate of Px. Only vectors of the kind (x,0,0) are eigenstates of Px with eigenvalue 1, and vectors of the kind (0,y,z) are eigenstates of Px with eigenvalue 0. But (1,2,0) is neither.

 

[Entropia]

Hello,

Thank you for your question. The concept of degeneracy and commutability is an important one in quantum mechanics. In short, degeneracy refers to the situation where multiple eigenstates of a single operator have the same eigenvalue. Commutability refers to the relationship between two operators, and specifically whether or not they commute, meaning that their order of operation does not affect the outcome.

Liboff's statement is referring to the fact that even if two operators, A and B, commute (meaning [A,B] = 0), it does not necessarily mean that all eigenstates of A are also eigenstates of B. This can be proven mathematically using the commutator relationship [A,B] = AB-BA = 0. If two operators commute, it means that their product is symmetric, meaning that AB = BA. However, this does not necessarily mean that all eigenstates of A are also eigenstates of B. This can be seen by considering an example where A and B are the position and momentum operators, respectively. While they commute, their eigenstates are not completely common.

There are many resources available where you can find a proof of this theorem. I would recommend consulting a quantum mechanics textbook or searching for academic articles on the topic. Additionally, there are many online resources and forums where you can discuss and learn more about this concept. I hope this helps!