How Do You Prove the Adjoint of a Product of Operators?

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Homework Help Overview

The discussion revolves around proving the property of the Hermitian adjoint for the product of two operators, A and B, specifically that (AB)^t = (B)^t(A)^t. The original poster expresses confusion about applying concepts from linear algebra to operators in Dirac notation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the proof to matrix transposes but struggles with the transition to Dirac notation. They present a line of reasoning involving inner products and express uncertainty about the validity of their steps. Another participant suggests a continuation of the proof, while a third participant questions the validity of manipulating the operators in the context of the proof.

Discussion Status

The discussion is active, with participants exploring different aspects of the proof. Some guidance is offered regarding the manipulation of operators, and there is a sense of progress as participants engage with the concepts. However, there is no explicit consensus on the proof's completion.

Contextual Notes

Participants are navigating the complexities of operator notation and the rules governing the manipulation of operators in the context of Hermitian adjoints. There is an acknowledgment of potential confusion stemming from the transition between different mathematical frameworks.

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Homework Statement



For operators A, B, prove that (AB)^t = (B)^t(A)^t where ^t is representing the Hermitian adjoint.

I know that this should be similar to proofs I did about matrix transposes in linear algebra, but I'm not sure how to do it without seeing the operators as matrices with indices. I've been trying to do it with dirac notation but that's been confusing...

Homework Equations





The Attempt at a Solution



Well... it's a long shot. I don't think this works:

<(AB)^t psi1| psi2> = ((AB)^t)* < psi1|psi2> = B^t A^t < psi1|psi2> = <psi1| B^tA^t psi2> ==> (AB)^t = B^t A^t.

I think I kind of made it up at the part with the complex conjugate, so yeah, basically I'm confused, even though this is supposed to be the easiest proof ever...
 
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<(AB)^T(psi1)|psi2>=<psi1|(AB)(psi2)>=<psi1|A(B(psi2))>=<A^T(psi1)|B(psi2)>. Can you continue?
 
aha... I wasn't sure if I was allowed to "split them up", so to speak, at the point where the A moves back to the rhs of the statement. If I'm allowed to do that I'm done! :biggrin:
 
oooh wait, it's just composition of functions! hehehe
 

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