Hermitian conjugate of differential operator

[Eismc[]]

Hi everyone!

How can I find the Hermitian conjugate of the differential operator D, with

D psi = 1/i dpart/dpart(x) psi?

I know you can do this with partial integration starting from

<phi|D|psi>* = <phi|D+|psi>

but how exactly does it work?

I'm sorry for using such an ugly notation but I am new to this forum and not yet familiar with other more beautiful notations.

Thanks a lot in advance, everyone!

Best regards
Torsten

 

[Hans de Vries]

Hi everyone!
How can I find the Hermitian conjugate of the differential operator D, with
D psi = 1/i dpart/dpart(x) psi?
I know you can do this with partial integration starting from
<phi|D|psi>* = <phi|D+|psi>
but how exactly does it work?

This should be:

$$\left( \frac{1}{i}\frac{\partial}{\partial x} \right)^\dag\ = \ \left( -\frac{1}{i}\frac{\partial}{\partial x} \right)^*\ = \ \left( \frac{1}{i}\frac{\partial}{\partial x} \right)$$

thus

$$D^\dag \psi = \ D \psi$$If you consider that the differential has two diagonals (+1 and -1) where the
unity operator has a single diagonal (+1). The two diagonals exchange place
when the operator is transposed. This inverses the sign once. The conjungate
then reverses the sign for the second time.

The background is that the operator is "Hermitian" or "SelfAdjoint" Regards, Hans

P.S. Thus: $$\frac{\partial}{\partial x}\ =\ \left( \begin{array}{cccccc} & 1 & 0 & 0 & 0 & . \\ -1 & & 1 & 0 & 0 & . \\ 0 & -1 & & 1 & 0 & . \\ 0 & 0 & -1 & & 1 & . \\ 0 & 0 & 0 & -1 & & . \\ . & . & . & . & . & . \end{array} \right)$$

 

[Hurkyl]

You can click on any of the images to see what you have to type to get them to appear.

I know you can do this with partial integration starting from

<phi|D|psi>* = <phi|D+|psi>

but how exactly does it work?

First off, I think you meant <phi|D|psi>* = <psi|D+|phi>,
or maybe <phi|D psi> = <D+ phi| psi>

Well, starting with the definitions is almost always a good first step:$\langle \phi | D \psi \rangle = \langle D^\dag \phi | \psi \rangle$ is the same thing as

$$\int \phi(x)^* (D \psi)(x) \, dx = \int (D^\dag \phi)(x)^* \psi(x) \, dx$$

and thus

$$\int \phi(x)^* \frac{1}{i} \frac{\partial \psi}{\partial x}(x) \, dx = \int (D^\dag \phi)(x)^* \psi(x) \, dx$$

isn't it? You had a hint to use partial integration here, and as far as I can tell, there's only one way to try it that makes sense. Have you at least gotten this far? (You should generally show what you have tried when asking a question, even if you think you haven't gotten anywhere) Have you tried to apply partial integration at all?

 

[Eismc[]]

Hi everyone,

thank you very much for your answers!

Indeed I should have been more precise, I'm sorry. I did get to the point at which the partial integration can be done and I've actually done it. Two questions arise here:

1. What happens to 1/i phi* psi with x between plus and minus infinity? I know these functions are elements of a Hilbert space but does this mean they disappear? At a first look, only the integral of the both disappears.

2. If I have to take the complex conjugate of an integral, can I put the star into it, placing it behind the functions in the scalar product? What if there is an operator aiming at one of the two - can I pull the star behind the operator?

Sorry again for my vague question. I hope you can help me from here! Many thanks in advance...

Best regards
Torsten

P.S. And thank you also, Hurkyl, for the tip for LaTeX! I will try to get into that, doesn't seem to be too complicated...

 

[Hurkyl]

It might help if you were able to post what you got.


For 1, do you mean

$$\left[ \frac{1}{i} \phi(x)^* \psi(x) \right]_{x = -\infty}^{x = +\infty}$$

? That's just a limit, isn't it? Use the limit properties! I'm confused about why you mentioned integral later in the paragraph, though.


For 2, it is a fact about integrals of complex functions that:

$$\left( \int_a^b f(x) \, dx \right)^* = \int_{a}^{b} f(x)^* \, dx$$

but I don't understand the rest of your question.

 

[Eismc[]]

Hi Hurkyl,

thanks again for the quick answer!

For 2: Coming back to your example (the complex conjugate of an integral of a single function), what if there was an operator like dpart/dpartx before f(x)? Could I still pull the star behind the f(x)?

Thanks in advance
Torsten