Commutator Properties: [A,B]C+B[A,C]=[A,B](C+B)?

  • Context: Undergrad 
  • Thread starter Thread starter Kyle Nemeth
  • Start date Start date
  • Tags Tags
    Commutator Properties
Click For Summary
SUMMARY

The discussion centers on the commutator property defined as [A,BC] = [A,B]C + B[A,C]. It is established that if B equals C, then the expression [A,B]C + B[A,C] simplifies to [A,B](C+B). The commutator is explicitly defined as [A,B] = AB - BA. The participants clarify that since the operators A, B, and C do not commute, the order of operations cannot be switched, leading to the conclusion that the left-hand side does not equal the right-hand side without additional conditions.

PREREQUISITES
  • Understanding of commutators in linear algebra
  • Familiarity with operator theory
  • Knowledge of linear mappings
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of commutators in quantum mechanics
  • Learn about non-commutative algebra
  • Explore linear operator theory in functional analysis
  • Investigate the implications of operator commutation in physics
USEFUL FOR

Mathematicians, physicists, and students studying linear algebra and operator theory, particularly those interested in the properties of commutators and their applications in quantum mechanics.

Kyle Nemeth
Messages
25
Reaction score
2
Given the property,

[A,BC] = [A,B]C+B[A,C],

is it true that, if B=C, then

[A,B]C+B[A,C]=[A,B]C+B[A,B]=[A,B](C+B)?

I apologize if I have posted in the wrong forum.
 
Physics news on Phys.org
Kyle Nemeth said:
Given the property,

[A,BC] = [A,B]C+B[A,C],

is it true that, if B=C, then

[A,B]C+B[A,C]=[A,B]C+B[A,B]=[A,B](C+B)?

I apologize if I have posted in the wrong forum.
The commutator (in the assumed context above) is defined as ##[A,B]=AB-BA##.
Now you have ##[A,B]C+B[A,C]=[A,B]B+B[A,B]=AB^2-B^2A## on the left and a factor ##2## on the right.
The (presumably) operators (or linear mappings) ##A,B,C## do not commutate, so you cannot switch orders.
 
  • Like
Likes   Reactions: Kyle Nemeth
Thank you for your help. I understand now.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
722
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 1 ·
Replies
1
Views
633
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 80 ·
3
Replies
80
Views
8K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K