How do operators combine in quantum mechanics?

[omiros]

Hello everyone I've got these things buzzing in my head and not exactly knowing how to solve them.

Homework Statement

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

 

[CAF123]

Hello everyone I've got these things buzzing in my head and not exactly knowing how to solve them.

Homework Statement

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

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

 

[omiros]

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

 

[berkeman]

Hello everyone I've got these things buzzing in my head and not exactly knowing how to solve them.

Homework Statement

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

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.

 

[CAF123]

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

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

 

[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

 

[omiros]

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

 

[CAF123]

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

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.

 

[omiros]

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

 

[CAF123]

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

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.