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Commutative Operators and Common Basis David Chen October 22, 2012 1 Non-degenerate case Given two operators A and B that satisfy the commutation relation, i.e. [A, B] = AB − BA = 0, and given a non-degenerate eigenstate of A, |ai, we have the following relation A (B |ai) = BA |ai = Ba |ai = a (B |ai) (1) (2) (3) In other words, B |ai is an eigenstate of A with eigenvalue a. Since |ai is nondegenerate, B |ai has to be proportional to |ai, or B |ai = b |ai (4) Therefore, |ai is simultaneously an eigenstate of A and B, and it is usually labeled by their eigenvalues, i.e. |a, bi 2 Degenerate case For simplicity, let’s assume a twofold degeneracy, but it can be easily generalized. A a(1) = a a(1) (5) (2) (2) A a = a a (6) 1 (1) (2) a a In the way as above, B and B are eigenstates of A with eigenvalue a. (2) (1) are degenerate, we can only state that and a However, because a (7) B a(1) = b11 a(1) + b12 a(2) (2) (1) (2) B a = b21 a + b22 a (8) We can B in the sub-space associated to a, which is spanned by diagonalize (1) a , a(2) , resulting 1 B b(1) = b1 b(1) (9) B b(2) = b2 b(2) Finally, we can express b(i) in terms of a(1) and a(2) (i) X (j) (i) (j) b = a |b a (10) j=1,2 (1) b and b(2) are eigenstates of B by definition, Eq.(9), and we (i)can see from Eq.(10) that they are also eigenstates of A. As before, the state b is usually denoted as |a, bi 3 Example The Hamiltonian operator of a free particle is p̂2 (11) Ĥ = 2m There are two degenerate eigenstates for this operator, and the common eigenvalue is p2 /2m. On the other hand, it is clear that the momentum operator p̂ commutes with Ĥ. Therefore, there is a common basis for both operators, which is {|E, pi , |E, −pi} where E = p2 /2m. References [1] Gasiorowics, Quantum Physics [2] Sakurai, Modern Quantum Mechanics 1 Assuming that a(1) and a(2) are orthogonal. What happens in the general case? 2