What is the Complex Conjugate of a Hermitian Integral in QM?

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The discussion centers on the interpretation of the complex conjugate of a Hermitian integral in quantum mechanics, specifically the expression $$\int f^*\hat{\Omega}g dx = \left \{ \int g^*\hat{\Omega}f dx \right\}^*$$. Participants clarify that these integrals are definite integrals representing complex numbers rather than functions of x. The necessity of evaluating the integral to extract an explicit form of the complex conjugate is debated, with consensus that evaluation is not strictly required. The conversation also touches on the definition of Hermiticity and its implications in quantum mechanics.

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Mayhem
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My QM textbook defines Hermiticity as $$\int f^*\hat{\Omega}g dx = \left \{ \int g^*\hat{\Omega}f dx \right\}^*$$ where f and g are any two wave functions, and * denotes the complex conjugate.

I am having a little trouble interpreting the complex conjugate of the RHS integral. Usually the complex conjugate of a function is defined as ## \psi^* = (f+gi)^* = f-gi ## (here f and g are not necessarily related to the above definition). Can I make a similar decomposition of the integral and is this even useful?
 
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Mayhem said:
My QM textbook defines Hermiticity as $$\int f^*\hat{\Omega}g dx = \left \{ \int g^*\hat{\Omega}f dx \right\}^*$$ where f and g are any two wave functions, and * denotes the complex conjugate.

I am having a little trouble interpreting the complex conjugate of the RHS integral. Usually the complex conjugate of a function is defined as ## \psi^* = (f+gi)^* = f-gi ## (here f and g are not necessarily related to the above definition). Can I make a similar decomposition of the integral and is this even useful?
Your textbook omits to show that those integrals are definite integrals and hence complex numbers, not functions of ##x##.
 
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PeroK said:
Your textbook omits to show that those integrals are definite integrals and hence complex numbers, not functions of ##x##.
Ah, that makes sense. Actually they do show an example where there are limits. So this means that it is necessary to evaluate the integral in order to extract an explicit form of the complex conjugate?
 
Mayhem said:
Ah, that makes sense. Actually they do show an example where there are limits. So this means that it is necessary to evaluate the integral in order to extract an explicit form of the complex conjugate?
You don't necessarily have to evaluate the integral.
 
Mayhem said:
My QM textbook defines Hermiticity as $$\int f^*\hat{\Omega}g dx = \left \{ \int g^*\hat{\Omega}f dx \right\}^*$$ where f and g are any two wave functions, and * denotes the complex conjugate.

I am having a little trouble interpreting the complex conjugate of the RHS integral. Usually the complex conjugate of a function is defined as ## \psi^* = (f+gi)^* = f-gi ## (here f and g are not necessarily related to the above definition). Can I make a similar decomposition of the integral and is this even useful?
You have an inner product defined by a bilinear, self-adjoint operator ##\hat{\Omega}=\hat{\Omega}^*=\overline{\hat{\Omega}}##
$$
\langle \overline{f}\, , \,g \rangle_\hat{\Omega}= \int f^*\hat{\Omega}g dx = \left \{ \int g^*\hat{\Omega}f dx \right\}^*=\overline{\langle \overline{g},f \rangle}_\hat{\Omega}
$$
 
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fresh_42 said:
You have an inner product defined by a bilinear, self-adjoint operator ##\hat{\Omega}=\hat{\Omega}^*=\overline{\hat{\Omega}}##
$$
\langle \overline{f}\, , \,g \rangle_\hat{\Omega}= \int f^*\hat{\Omega}g dx = \left \{ \int g^*\hat{\Omega}f dx \right\}^*=\overline{\langle \overline{g},f \rangle}_\hat{\Omega}
$$
That doesn't look right. That's the definition of Hermicity of ##\Omega##, using the standard inner product on the space of square-integrable functions.

That's what the OP gets for posting QM in a maths forum!
 
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Let me guess: The textbook is Griffiths...?
 
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vanhees71 said:
Let me guess: The textbook is Griffiths...?
No, Atkin's Physical Chemistry. QM is interesting though. Might take a graduate level elective if I pass this course with a decent grade.
 
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