Operator algebra with commutators

[chill_factor]

Homework Statement

Consider the operator A and its Hermitian adjoint A*.

If [A,A*] = 1, evaluate: [A*A,A]

Homework Equations

standard rules of linear algebra, operator algebra and quantum mechanics

The Attempt at a Solution

[A,A*] = AA* - A*A = 1

A*A = (1+AA*)

[A*A,A] = AA*A - A*AA = ??

How do I even start?

 

[jmcelve]

Hi chill_factor,

If we expand [A*A,A], we obtain A*AA-AA*A. Do you see a way to factor this that might allow you to use the [A,A*] commutator?

 

[chill_factor]

Hi chill_factor,

If we expand [A*A,A], we obtain A*AA-AA*A. Do you see a way to factor this that might allow you to use the [A,A*] commutator?

Thank you for answering. I think I see it. There is an A on the right side of both the A*AA and AA*A operators so A*AA - AA*A = (A*A-AA*)A = -[A,A*]A = -A. Is that right?

I guess my mistake was writing the commutator out incorrectly.

 

[jmcelve]

Thank you for answering. I think I see it. There is an A on the right side of both the A*AA and AA*A operators so A*AA - AA*A = (A*A-AA*)A = -[A,A*]A = -A. Is that right?

I guess my mistake was writing the commutator out incorrectly.

Yep. You've got it. Nice work.

 

[paranormal]

The commutator [A,A*] represents the non-commutativity of the operators A and A*. This means that their order matters in the product, and so we cannot simply rearrange them as we would with ordinary numbers.

To evaluate [A*A,A], we can use the standard rules of operator algebra and the definition of the adjoint operator to simplify the expression. First, we can rewrite A*A as (A*)^2, and then use the definition of the adjoint to rewrite [A*A,A] as [A*^2,A]. From there, we can use the definition of the commutator [X,Y] = XY-YX to expand the expression and simplify it. Ultimately, we should end up with [A*A,A] = A*^2A-A*A* = A.

In summary, [A*A,A] = A represents the commutator of the operators A and its adjoint A*, which is not surprising given that they are Hermitian conjugates of each other.