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Transcript
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
