[Q]Degeneracy and Commutability


by good_phy
Tags: commutability, qdegeneracy
good_phy
good_phy is offline
#1
Oct27-08, 09:03 PM
P: 43
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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vanesch
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#2
Oct28-08, 07:22 AM
Emeritus
Sci Advisor
PF Gold
P: 6,238
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.


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