# What are Hermitian conjugates in this context? [closed]

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

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 $$\frac{\mathrm{d}}{\mathrm{d}x}$$.

• 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 Jan 19, 2021 at 23:54
• @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. Jan 19, 2021 at 23:56

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.

• Would you be able to help with the second part? Jan 20, 2021 at 0:45
• 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 Jan 20, 2021 at 0:54