# I How exactly does squaring operators (e.g. <p^2> work?

1. Feb 27, 2016

### Zacarias Nason

I'm having to essentially piece together a framework of background knowledge to understand parts of a QM class in which I'm lacking prerequisites; one of the things that I noticed that really confused me was the square of the momentum operator <p>, and how that translated into the integrand of an expectation value; Squaring an operator as far as I'm as aware isn't as simple as just squaring its components, because, for the momentum operator defined as $$\langle p^2 \rangle = \int_{-\infty}^{\infty}\psi^*(x)p^2\psi(x) = \int_{-\infty}^{\infty}\psi^*(x)\bigg(-\hbar^2 \frac{\partial^2}{\partial x^2}\psi(x)\bigg)$$
The differential part of the operator changed, too! This isn't just squaring, what is it? $$\frac{d}{dx}\frac{d}{dx} \neq \frac{d^2}{dx^2}$$ If I'm given some arbitrary operator A and told to calculate $$\langle A^2\rangle$$ How do I know what the operator "squared" takes the form of?

2. Feb 27, 2016

### drvrm

The differential part of the operator changed, too! This isn't just squaring, what is it?

operators are supposed to operate in a space - so a A^2 must be equivalent to A(op). A(op) , a product say operating on the right....

3. Feb 27, 2016

### Zacarias Nason

I don't understand what that meant; I don't have any working knowledge of spaces, regardless of what they are.
I also have no idea of what A(op) is supposed to indicate. A is the operator, and I'm only guessing "op" is shorthand for "operator", so unless A(op) is a really unnecessarily roundabout way of saying (A)(A) or something, I don't get it.

4. Feb 27, 2016

### Zacarias Nason

Alright then, thanks for the help(?).

5. Feb 28, 2016

### Staff: Mentor

$p^2\psi=pp\psi=p(p\psi)$, that is, you apply $p$ to $\psi$ to get a new function $\psi_1$, then you apply $p$ to $\psi_1$.

Likewise for any other operator.

6. Mar 1, 2016

### Zacarias Nason

Thank you very much!