# Find Adjoints of certain Operators

1. May 26, 2013

### rebork

1. The problem statement, all variables and given/known data
Find the adjoints for x2d/dx and d2/dx2

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

3. 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.

2. May 27, 2013

### TSny

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 $B^\dagger$?

3. May 27, 2013

### rebork

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?

4. May 27, 2013

### TSny

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

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).

5. May 27, 2013

### rebork

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: May 27, 2013
6. May 27, 2013

### TSny

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: May 27, 2013
7. May 27, 2013

### rebork

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)?

8. May 27, 2013

### TSny

Yes, that's right.

9. May 27, 2013

### rebork

Thank you very much!

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

10. May 27, 2013

### TSny

Second part is correct. Good work!

11. May 27, 2013

### rebork

I really appreciate the help. Thank you :]