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

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The discussion focuses on the interpretation of the complex conjugate of integrals in the context of Hermitian operators in quantum mechanics. The textbook defines Hermiticity using integrals of wave functions, leading to confusion about the nature of these integrals as complex numbers rather than functions of x. Participants clarify that evaluating the integral is not strictly necessary to understand the complex conjugate, as it relates to the properties of the bilinear, self-adjoint operator. The conversation also touches on the importance of the inner product in defining Hermiticity. Overall, the thread emphasizes the need for careful interpretation of complex conjugates in quantum mechanics.
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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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