Self-adjoint Operators and their Products: Solving for ##AB##

  • Thread starter Thread starter member 428835
  • Start date Start date
  • Tags Tags
    Operator
Click For Summary
The discussion centers on whether the product of two linear self-adjoint operators, A and B, is also self-adjoint. It is established that for the product AB to be self-adjoint, A and B must commute, which is not guaranteed solely by their linearity. Participants explore the conditions under which two linear operators commute and emphasize that finding a counterexample would disprove the assertion. Additionally, a related question arises about the stationary points of a functional defined using a self-adjoint operator. The conversation concludes with clarification on the validity of an equality involving the linear operator A and its action on variations of the function.
member 428835

Homework Statement


Given two linear self-adjoint operators ##A,B##, is it true ##AB## is also self-adjoint.

Homework Equations


Self adjoint implies ##(A[f],g) = (f,A[g])##

The Attempt at a Solution


I'm not really sure. I'm stuck almost right away: ##(AB[f],g) = (A[B[f]],g) = (B[f],Ag) = (f,BAg) = (f,ABg)##. Last equality is from linearity of ##B,A##. The first few steps I think follow from self-adjoint of ##A,B##.
 
Last edited by a moderator:
Physics news on Phys.org
The last equality requires A and B to commute. That two operators are linear is certainly no guarantee for them to commute.
 
  • Like
Likes member 428835
Orodruin said:
The last equality requires A and B to commute. That two operators are linear is certainly no guarantee for them to commute.
Riiiiight, good call! So what should I do to prove two operators commute? Are there any sufficient conditions?
 
Given the problem formulation, are you sure that what you want to do is to prove that A and B commute?
 
Orodruin said:
Given the problem formulation, are you sure that what you want to do is to prove that A and B commute?
I guess I'm mostly interested to know when two linear operators are able to commute. Unless there is a better way to know if ##AB## is self adjoint (I realize I had the question stem all bold, so I corrected it).
 
The point is that you have essentially shown that AB is self adjoint if A and B commute. So in order to answer the question ”is AB self adjoint?” you must ask yourself ”is it true that A and B always commute”. If you can find a single counter example, then you will have shown it to not be the case.
 
Orodruin said:
The point is that you have essentially shown that AB is self adjoint if A and B commute. So in order to answer the question ”is AB self adjoint?” you must ask yourself ”is it true that A and B always commute”. If you can find a single counter example, then you will have shown it to not be the case.
Well both operators are linear differential operators, so they should commute then right?
 
joshmccraney said:
Well both operators are linear differential operators, so they should commute then right?
No.
 
  • Like
Likes member 428835
Orodruin said:
No.
Thanks, I see what I'm looking for now! I really appreciate the help!
 
  • #10
Orodruin said:
No.
I do have a related question. Suppose we have a functional defined as $$J[f] = (A[f],f)$$ and then I want to find the stationary points of the functional, so that ##\delta J = (\delta A[f],f) = (A[\delta f],f)##. Is the last equality true (##A## is linear as mentioned above)?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
9
Views
2K
  • · Replies 19 ·
Replies
19
Views
6K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K