How to Find Adjoints of Differential Operators?

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Homework Help Overview

The discussion revolves around finding the adjoints of differential operators, specifically for the operators \(x^2 \frac{d}{dx}\) and \(\frac{d^2}{dx^2}\). Participants explore the properties of adjoint operators in the context of differential calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the adjoint of the operator \(x^2 \frac{d}{dx}\) and question the validity of separating the second derivative operator into its components. There is also exploration of the general property of adjoints concerning the product of operators.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's reasoning. Some guidance has been offered regarding the correct application of adjoint properties, and there is acknowledgment of the need to consider the order of operations when dealing with products of operators.

Contextual Notes

Participants are navigating the complexities of operator adjoints without complete information on the underlying functions they are applying these operators to, which influences their reasoning and conclusions.

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Homework Statement


Find the adjoints for x2d/dx and d2/dx2


Homework Equations


I know that (x2)_dagger=x2 and that (d/dx)_dagger=-(d/dx).


The Attempt at a Solution


I solved x2d/dx by doing the following:

(x2)_dagger * (d/dx)_dagger= (x2) * (-d/dx)

thus the answer should be -x2d/dx


For the second adjoint can I do the same step as I did previously?

Is it right for me to separate (d2/dx2)_dagger into (d/dx)_dagger * (d/dx)_dagger?

I got d2/dx2 as the final answer.
 
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rebork said:
I solved x2d/dx by doing the following:

(x2)_dagger * (d/dx)_dagger= (x2) * (-d/dx)

thus the answer should be -x2d/dx

In general, it is not true that ##(AB)^\dagger = A^\dagger B^\dagger##. Have you learned the correct way to express ##(AB)^\dagger## in terms of ##A^\dagger## and ##B^\dagger##?
 
TSny said:
In general, it is not true that ##(AB)^\dagger = A^\dagger B^\dagger##. Have you learned the correct way to express ##(AB)^\dagger## in terms of ##A^\dagger## and ##B^\dagger##?

Yes I believe I figured it out yesterday!

##(AB)^\dagger##=(##B^\dagger##)(##A^\dagger##)

But for this d2/dx2, it will be the same result right or should I not be able to use that rule?
 
rebork said:
Yes I believe I figured it out yesterday!

##(AB)^\dagger##=(##B^\dagger##)(##A^\dagger##)

Yes, that's right. The order of the operators is switched.

But for this d2/dx2, it will be the same result right or should I not be able to use that rule?

No. The operator O1 = (x2 d/dx) is not the same as the operator O2 = (d/dx x2), as you can see if you compare the results of letting O1 and O2 operate on an arbitrary function f(x).
 
So I messed up for the first part:

(##(x2d/dx)^\dagger##) = (##(d/dx)^\dagger##)(##(x2)^\dagger##)

Then the result would be:

(-d/dx)*(x2)= -2x?
 
Last edited:
rebork said:
Then the result would be:

(-d/dx)*(x2)= -2x?

The left side is correct, but the right side isn't. See what you get if you let the operator (-d/dx*x2) act on a function f(x). First you will multiply f(x) by x2 and then take the derivative -d/dx.
 
Last edited:
TSny said:
The left side is correct, but the right side isn't. See what you get if you let the operator (-d/dx*x2) act on a function f(x). First you will multiply f(x) by x2 and then take the derivative -d/dx.

I see so I should've included the function as well.

So it would be -d(x2f(x))/dx= -2xf(x)-x2df(x)/dx. So what should my operator look like? Is it everything except for the f(x)?
 
rebork said:
So it would be -d(x2f(x))/dx= -2xf(x)-x2df(x)/dx. So what should my operator look like? Is it everything except for the f(x)?
Yes, that's right.
 
Thank you very much!

Is my second part correct where I applied the same concept but this time separate the second derivative?
 
  • #10
rebork said:
Thank you very much!

Is my second part correct where I applied the same concept but this time separate the second derivative?

Second part is correct. Good work!
 
  • #11
I really appreciate the help. Thank you :]
 

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