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I am having trouble understanding the definition Hermitian and Hermitian conjugate.

An operator is Hermitian provided that: $\hat{O}^\dagger=\hat{O}$

The Hermitian conjugate of the differentiation operator: $\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^\dagger=-\frac{\mathrm{d}}{\mathrm{d}x}$

We know that the differentiation operator is not a Hermitian then why is it called a Hermitian conjugate and not just a conjugate

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Hermitian is an adjective used to describe an operator which is equal to its Hermitian conjugate.

Hermitian conjugate (sometimes also called Hermitian adjoint) is a noun referring to the generalisation of the conjugate transpose of a matrix.

It doesn't really make sense to say that a particular operator is a Hermitian conjugate without any context. In your example, we would say that $-\frac{\mathrm{d}}{\mathrm{d}x}$ is the Hermitian conjugate of $\frac{\mathrm{d}}{\mathrm{d}x}$.

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  • $\begingroup$ ok thank you i understand. Just because a we found a Hermitian conjugate of an operator it doesn't have to mean that operator is Hermitian unless they are equal $\endgroup$ – Jack Jack Jan 19 at 23:54
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    $\begingroup$ @JackJack Yes. But finding a Hermitean conjugate is a delicate business in general. The modern mathematical terminology speaks of adjoint, instead of Hermitean conjugate. This latter wording is kept in matrix (i.e. finite-dimensional case) algebra. $\endgroup$ – DanielC Jan 19 at 23:56
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The Hermitian Conjugate or Hermitian Transpose of an operator $\hat{O}$ is defined as $\hat{O}^\dagger$.

As you stated in your question an operator $\hat{Q}$ is Hermitian iff $\hat{Q}=\hat{Q}^\dagger$, I know the terminology can be confusing. An operator is Hermitian if it is equal to its Hermitian Conjugate.

Now to the differentiation operator: I assume you know why $\frac{d}{dx}^\dagger=-\frac{d}{dx}$ since its in your question but here is a link that describes it if you are unsure: Explaining why $\mathrm{ d/d}x$ is not Hermitian, but $\mathrm{i~ d/d}x$ is Hermitian

Clearly $\frac{d}{dx}^\dagger \neq \frac{d}{dx}$, so the operator is $\textbf{not}$ Hermitian, but we can still just find its Hermitian Conjugate same as any other operator.

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  • $\begingroup$ Would you be able to help with the second part? $\endgroup$ – Jack Jack Jan 20 at 0:45
  • $\begingroup$ By * do you mean complex conjugate or Hermitian Conjugate? Textbooks sometimes swap notation and its really annoying, im not sure which one you are referring to since I dont think you are interested in only complex conjugating based on your previous question; and you also used the other convention earlier $\endgroup$ – xXx_69_SWAG_69_xXx Jan 20 at 0:54

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