- #1

Haynes Kwon

Gold Member

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Trying to prove Hermiticity of the operator AB is not guaranteed with Hermitian operators A and B and this is what I got:

$$<\Psi|AB|\Phi> = <\Psi|AB\Phi> = ab<\Psi|\Phi>=<B^+A^+\Psi|\Phi>=<BA\Psi|\Phi>=b^*a^*<\Psi|\Phi>$$

but since A and B are Hermitian eigenvalues a and b are real,

Therefore we have

$$ ab<\Psi|\Phi>=b^*a^*<\Psi|\Phi>$$ since multiplication of the numbers commutes, this is same as

$$b^*a^*<\Psi|\Phi>=a^*b^*<\Psi|\Phi> = <AB\Psi|\Phi>$$

So Hermiticity is guaranteed?

$$<\Psi|AB|\Phi> = <\Psi|AB\Phi> = ab<\Psi|\Phi>=<B^+A^+\Psi|\Phi>=<BA\Psi|\Phi>=b^*a^*<\Psi|\Phi>$$

but since A and B are Hermitian eigenvalues a and b are real,

Therefore we have

$$ ab<\Psi|\Phi>=b^*a^*<\Psi|\Phi>$$ since multiplication of the numbers commutes, this is same as

$$b^*a^*<\Psi|\Phi>=a^*b^*<\Psi|\Phi> = <AB\Psi|\Phi>$$

So Hermiticity is guaranteed?

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