# Find Adjoints of certain Operators

## 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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TSny
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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##?

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?

TSny
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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?

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TSny
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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.

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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)?

TSny
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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?

TSny
Homework Helper
Gold Member
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!

I really appreciate the help. Thank you :]