# Homework Help: Operator acting on an operator

1. Nov 8, 2013

### omiros

Hello everyone I've got these things buzzing in my head and not exactly knowing how to solve them.
1. The problem statement, all variables and given/known data
Operator Ahat = (d/dx + x) and Bhat = (d/dx - x)

a. Chat = AhatAha
b. Chat = AhatBhat

What do the position and momentum operator Xhat = x and Phat = -i*hbar*d/dx, give when they combine as Phat^2Xhat^2?

Thanks a lot

2. Nov 8, 2013

### CAF123

Given $\hat{P}$, do you know how to find $\hat{P}^2$?

3. Nov 8, 2013

### omiros

well I can assume it is -H^2*d^2/dx^2 (without being sure)

4. Nov 8, 2013

### Staff: Mentor

Please remember that you must show your attempt at solving the problem before we can be of tutorial help. That's in the PF Rules link at the top of the page under Site Info.

5. Nov 8, 2013

### CAF123

It's correct, assuming H = $\hbar$. What makes you unsure?

6. Nov 8, 2013

### omiros

Well what makes me unsure is the way that this acts on x^2. as it goes like d/dx d/dx x x

7. Nov 8, 2013

### omiros

And on the Ahat*Ahat I get a constant in an operator, which I don't know if it is right.

8. Nov 8, 2013

### CAF123

The important thing to remember about operators is that they act on functions. I think what you have to do is to express the operator $\hat{P^2} \hat{X^2}$ in terms of $\hat{A}$ and $\hat{B}$.
As you said, $$\hat{P^2}\hat{X^2} \equiv -\hbar^2 \frac{d^2}{dx^2} x^2$$ Now act on a function f(x). You will need product rule.

9. Nov 9, 2013

### omiros

So is it left like that or is it written as -2*hbar^2 (differentiate twice x^2)?

10. Nov 10, 2013

### CAF123

What did you get when you computed $$-\hbar^2 \frac{d^2}{dx^2} (x^2 \psi)?$$
I don't know if $\hat{A}$ or $\hat{B}$ have any relevance to the problem or not.

It is okay if you have a constant as part of the operator, as long as you are adding it to a dimensionless quantity.