Prove Commutator Exponentials Algebra

  • Thread starter Thread starter chill_factor
  • Start date Start date
  • Tags Tags
    Algebra Commutator
Click For Summary
SUMMARY

The discussion centers on proving the identity for operators A and B: e^A B e^-A = B + [A,B] + (1/2!) * [A,[A,B]] + (1/3!) * [A,[A,[A,B]]] + ... This formula involves the Baker-Campbell-Hausdorff theorem, which is crucial for understanding operator exponentials in quantum mechanics. The user expresses difficulty in generalizing the proof beyond the special case where [A,B] is a constant and commutes with A and B. The discussion highlights the need for a solid grasp of commutators and operator algebra.

PREREQUISITES
  • Understanding of operator algebra in quantum mechanics
  • Familiarity with the Baker-Campbell-Hausdorff theorem
  • Knowledge of commutators and their properties
  • Basic calculus and series expansions
NEXT STEPS
  • Study the Baker-Campbell-Hausdorff theorem in detail
  • Learn about the properties of commutators in quantum mechanics
  • Explore examples of operator exponentials in quantum mechanics
  • Investigate Lie algebras and their applications in physics
USEFUL FOR

Students and researchers in quantum mechanics, particularly those focusing on operator theory and mathematical physics. This discussion is beneficial for anyone looking to deepen their understanding of operator exponentials and commutation relations.

chill_factor
Messages
898
Reaction score
5

Homework Statement



Prove the following for operators A and B.

e^A B e^-A = B + [A,B] + (1/2!) * [A,[A,B]] + (1/3!) * [A,[A,[A,B]]] + ...

Homework Equations



e^A = 1 + A + (1/2!)A^2 + (1/3!)A^3 + ...

The Attempt at a Solution



I have no clue how to start.

For the highly special case of [A,B] = constant and [A,B] commutes with A and B, we can prove that e^A B e^-A = B + [A,B]

through the following:

Take the given identity [A,F(B)] = [A,B] dF/dB and set F = e^B

[A,e^B] = Ae^B - e^B*A = e^B * [A,B]

multiply both sides by e^-B to left.

e^-B A e^B - A = [A,B]

e^-B A e^B = A + [A,B]

This is the first 2 terms of the series and if we take [A,B] = c, then the other terms are zero. How do I prove it for general [A,B] then?
 
Physics news on Phys.org
any help with this? i am seriously stuck and everything I've looked up involves Lie algebras or some abstract math stuff I've never learned in physics before =(
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
1K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K