[Q]Degeneracy and Commutability

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The discussion centers on the theorem regarding the commutation of operators and the implications of degeneracy in quantum mechanics. Specifically, it highlights that while two operators, such as the identity operator and the projector Px, commute ([1, Px] = 0), eigenstates of one operator are not necessarily common eigenstates of the other. The example provided illustrates that the vector (1,2,0) is an eigenstate of the identity operator but not of the projector Px, which only has specific eigenstates. This emphasizes the nuanced relationship between degeneracy and commutativity in quantum systems.

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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.
 
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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.
 

Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?
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