How Does Hermitian Property Lead to \langle A^2 \rangle in Quantum Mechanics?

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SUMMARY

The discussion clarifies the relationship between the Hermitian property of an operator A and the expression \langle A^2 \rangle in quantum mechanics. It establishes that for a self-adjoint (Hermitian) operator A, the equality \langle Aψ|Aψ\rangle = \langle ψ|A^{\dagger}A|ψ\rangle holds true, leading to \langle ψ|A^{2}|ψ\rangle. This derivation is crucial for understanding quantum expectation values and the implications of Hermitian operators in quantum mechanics.

PREREQUISITES
  • Understanding of Hermitian operators in quantum mechanics
  • Familiarity with quantum state notation, specifically Dirac notation
  • Knowledge of expectation values in quantum mechanics
  • Basic principles of linear algebra as applied to quantum states
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  • Study the properties of Hermitian operators in quantum mechanics
  • Learn about the significance of expectation values in quantum systems
  • Explore the mathematical foundations of Dirac notation
  • Investigate the implications of self-adjoint operators in quantum theory
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Quantum physicists, students of quantum mechanics, and anyone interested in the mathematical foundations of quantum theory will benefit from this discussion.

spaghetti3451
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I know that <Aψ|Aψ> = <A2>. Can anyone tell please tell me how you get one line to the another?
 
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Ok so, <Aψ|Aψ> = <AAψ|ψ> because A is self adjoint (hermitian). What next?
 
failexam said:
Ok so, <Aψ|Aψ> = <AAψ|ψ> because A is self adjoint (hermitian). What next?

[tex]\langle A\psi \mid A\psi \rangle = \langle \psi \mid A^{\dagger}A\mid \psi \rangle[/tex]

If A is Hermintian:

[tex]\langle \psi \mid A^{\dagger}A\mid \psi \rangle = \langle \psi \mid A^{2}\mid \psi \rangle[/tex]

And that's it :)
 

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