1.8 Commutator
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Definition 1.8.1 Let T 1 , T 2 : V −→ V be two linear operators on V . The commutator between T 1 and T 2 is defined by
[T 1 , T 2 ] := T 1 T 2 − T 2 T 1 .
If [T 1 , T 2 ] ≡ 0, then T 1 commutes with T 2 .
Note first that [T 1 , T 2 ] is a function from V into V . Since T 1 and T 2 are linear then
also is [T 1 , T 2 ]. Thus we can view the commutator as an operation on the vector space
Hom(V, V ) of all linear operators on V , that is, [, ] is a function [, ] : Hom(V, V ) ×
Hom(V, V ) −→ Hom(V, V ) such that, for every pair (T 1 , T 2 ) of linear operators
assigns the linear operator [T 1 , T 2 ].
Analogously, we can define the anti-commutator between two operators.
Definition 1.8.2 Let T 1 , T 2 : V −→ V be two linear operators on V . The anticommutator between T 1 and T 2 is defined as
{T 1 , T 2 } := T 1 T 2 + T 2 T 1 .
If [T 1 , T 2 ] ≡ 0, then T 1 commutes with T 2 . When {T 1 , T 2 } ≡ 0 we say that T 1 anticommutes with T 2 .
As in the case of commutators, anti-commutators can be also viewed as an
operation on the vector space Hom(V, V ): {, } : Hom(V, V ) × Hom(V, V ) −→
Hom(V, V ) such that, for every pair (T 1 , T 2 ) of linear operators assigns the linear
operator {T 1 , T 2 }.
Theorem 1.8.1 (Simultaneous diagonalization theorem) Assume that T 1 , T 2 : V −→
V are two Hermitian operators on V . Then [T 1 , T 2 ] ≡ 0 if and only if there exists an
orthonormal basis B such that both T 1 and T 2 are diagonal with respect to B. If there
exists such basis, the operators T 1 and T 2 are called simultaneously diagonalizable.
In other words, two Hermitian operators commute themselves if and only if they
are simultaneously diagonalizable.
Exercise 1.8.1 (a) Show that X and Y are not simultaneously diagonalizable.
(b) Show the same to Y and Z e for Z and X .
(c) Adopting the notation of Definition 1.7.5, show that {ρ i , ρ j } ≡ 0, for each
i = j, i, j = 1, 2, 3.
An interesting case is when two operators commute and anti-commute themselves
at the same time and one of them is invertible. Let us see what happens. Assume that
T 1 , T 2 : V −→ V are linear operators such that [T 1 , T 2 ] ≡ 0, {T 1 , T 2 } ≡ 0 and T 1 is
invertible. We then have
T 1 T 2 = T 2 T 1
(1.1)
15
Definition 1.8.1 Let T 1 , T 2 : V −→ V be two linear operators on V . The commutator between T 1 and T 2 is defined by
[T 1 , T 2 ] := T 1 T 2 − T 2 T 1 .
If [T 1 , T 2 ] ≡ 0, then T 1 commutes with T 2 .
Note first that [T 1 , T 2 ] is a function from V into V . Since T 1 and T 2 are linear then
also is [T 1 , T 2 ]. Thus we can view the commutator as an operation on the vector space
Hom(V, V ) of all linear operators on V , that is, [, ] is a function [, ] : Hom(V, V ) ×
Hom(V, V ) −→ Hom(V, V ) such that, for every pair (T 1 , T 2 ) of linear operators
assigns the linear operator [T 1 , T 2 ].
Analogously, we can define the anti-commutator between two operators.
Definition 1.8.2 Let T 1 , T 2 : V −→ V be two linear operators on V . The anticommutator between T 1 and T 2 is defined as
{T 1 , T 2 } := T 1 T 2 + T 2 T 1 .
If [T 1 , T 2 ] ≡ 0, then T 1 commutes with T 2 . When {T 1 , T 2 } ≡ 0 we say that T 1 anticommutes with T 2 .
As in the case of commutators, anti-commutators can be also viewed as an
operation on the vector space Hom(V, V ): {, } : Hom(V, V ) × Hom(V, V ) −→
Hom(V, V ) such that, for every pair (T 1 , T 2 ) of linear operators assigns the linear
operator {T 1 , T 2 }.
Theorem 1.8.1 (Simultaneous diagonalization theorem) Assume that T 1 , T 2 : V −→
V are two Hermitian operators on V . Then [T 1 , T 2 ] ≡ 0 if and only if there exists an
orthonormal basis B such that both T 1 and T 2 are diagonal with respect to B. If there
exists such basis, the operators T 1 and T 2 are called simultaneously diagonalizable.
In other words, two Hermitian operators commute themselves if and only if they
are simultaneously diagonalizable.
Exercise 1.8.1 (a) Show that X and Y are not simultaneously diagonalizable.
(b) Show the same to Y and Z e for Z and X .
(c) Adopting the notation of Definition 1.7.5, show that {ρ i , ρ j } ≡ 0, for each
i = j, i, j = 1, 2, 3.
An interesting case is when two operators commute and anti-commute themselves
at the same time and one of them is invertible. Let us see what happens. Assume that
T 1 , T 2 : V −→ V are linear operators such that [T 1 , T 2 ] ≡ 0, {T 1 , T 2 } ≡ 0 and T 1 is
invertible. We then have
T 1 T 2 = T 2 T 1
(1.1)