What does complex conjugate of a derivate mean?

[jstrunk]

An exercise asks me to determine whether the following operator is Hermitian:
${\left( {\frac{d}{{dx}}} \right)^ * }.$

I don't even know what that expression means.
a) Differentiate with respect to x, then take the complex conjugate of the result?
b) Take the complex conjugate, then differentiate with respect to x?
c) ${\left( {\frac{d}{{dx}}} \right)^ * } = \frac{d}{{d{x^*}}} = \frac{d}{{dx}}$?

Can someone clarify?

 

[andrewkirk]

If $x$ is real then (a) and (b) are the same, so it probably means those.

 

[Mentz114]

An exercise asks me to determine whether the following operator is Hermitian:
${\left( {\frac{d}{{dx}}} \right)^ * }.$

I don't even know what that expression means.
a) Differentiate with respect to x, then take the complex conjugate of the result?
b) Take the complex conjugate, then differentiate with respect to x?
c) ${\left( {\frac{d}{{dx}}} \right)^ * } = \frac{d}{{d{x^*}}} = \frac{d}{{dx}}$?

Can someone clarify?

The derivative wrt x of $i\,g\left( x\right) +f\left( x\right)$ is $i\,\left( \frac{d}{d\,x}\,g\left( x\right) \right) +\frac{d}{d\,x}\,f\left( x\right)$.
Nuff said ?

 

[PeterDonis]

An exercise

In what textbook?

 

[strangerep]

If $x$ is real then (a) and (b) are the same, so it probably means those.

If the original question is the context of QM, then I'll bet it doesn't mean either of those. Rather, the exercise probably intends to determine whether $d/dx$ is self-adjoint on the space of square-integrable functions.

In that case, @jstrunk: you should probably take a look at the Wikipedia page for "hermitian operators".

 

[DrClaude]

See https://www.physicsforums.com/threads/hermitian-operators-referencing-griffiths.929259/#post-5867817

 

[A. Neumaier]

The star applied to operators generally means the adjoint operator. For differential operators you can find it using integration by parts.