Find Adjoints of certain Operators

[rebork]

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.

 

[TSny]

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 $B^\dagger$?

 

[rebork]

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]

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).

 

[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?

 

[TSny]

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.

 

[rebork]

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]

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.

 

[rebork]

Thank you very much!

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

 

[TSny]

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!

 

[rebork]

I really appreciate the help. Thank you :]