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

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In summary, the conversation is discussing whether the product of two linear self-adjoint operators is also self-adjoint. The first few steps of the solution show that this is true if the operators commute. The conversation then shifts to finding conditions for two linear operators to commute, and it is pointed out that being linear differential operators is not a sufficient condition for them to commute. Finally, the conversation touches on using a functional defined as J[f] = (A[f], f) to find stationary points, and it is questioned whether the last equality is true (A is linear).
  • #1
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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##.
 
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  • #2
The last equality requires A and B to commute. That two operators are linear is certainly no guarantee for them to commute.
 
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  • #3
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?
 
  • #4
Given the problem formulation, are you sure that what you want to do is to prove that A and B commute?
 
  • #5
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).
 
  • #6
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.
 
  • #7
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?
 
  • #8
joshmccraney said:
Well both operators are linear differential operators, so they should commute then right?
No.
 
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  • #9
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)?
 

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

1. What is a self-adjoint operator?

A self-adjoint operator is a linear operator that is equal to its own adjoint. In other words, the operator and its adjoint have the same action on any vector in the vector space.

2. How do you solve for the product of two self-adjoint operators?

To solve for the product of two self-adjoint operators, you can use the properties of self-adjoint operators. First, you can rewrite the product as a sum of two operators, each of which is self-adjoint. Then, you can use the property that the product of two self-adjoint operators is also self-adjoint. Finally, you can apply the definition of a self-adjoint operator to solve for the product.

3. What is the significance of self-adjoint operators in quantum mechanics?

In quantum mechanics, self-adjoint operators represent physical observables such as position, momentum, and energy. This means that the eigenvalues of a self-adjoint operator correspond to measurable quantities in a quantum system, making them essential in the study of quantum mechanics.

4. Can all operators be written as a product of self-adjoint operators?

No, not all operators can be written as a product of self-adjoint operators. The product of two self-adjoint operators is also self-adjoint, but the converse is not always true. Some operators may not have a self-adjoint representation, which means they cannot be written as a product of self-adjoint operators.

5. How can self-adjoint operators be used to solve differential equations?

Self-adjoint operators can be used to solve differential equations by transforming the equation into an eigenvalue problem. This involves writing the differential equation as a linear operator and then solving for the eigenvalues and eigenvectors of the operator. The solutions to the differential equation can then be expressed as a linear combination of the eigenvectors.

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