# Issue with squaring operators

1. May 17, 2013

### Sylvia Else

I'm, slowly, working through "Quantum Physics" by Stephen Gasiorowicz, second edition. On page 49, he gives the equation

i$\hbar$$\frac{\partial ψ(x, t)}{\partial t}$ = -$\frac{\hbar^{2}}{2m}$$\frac{\partial ^{2}ψ(x, t)}{\partial x^2}$

He then makes the identification $(h/i)(\partial/\partial x) = p_{op}$, which he has previously identified as the momentum operator, and rewrites the equation as

i$\hbar$$\frac{\partial ψ(x, t)}{\partial t}$ = $\frac{p^2_{op}}{2m}$$ψ(x, t)$

Then he observes that the operator on the right (which I construe to mean $p^2_{op}/2m$) is just the energy for a free particle. And that's where I get confused. As I understand it, the square of the momentum divided by twice the mass gives the energy, but $p^2_{op}$ isn't saying use the momentum squared, but apply the momentum operator twice.

What am I missing here?

Sylvia.

2. May 17, 2013

### The_Duck

Consider a state with a definite value of momentum. In QM, this means a state that is an eigenstate of the momentum operator. For example, a state $\psi(x)$ satisfying

$\hat{p} \psi(x) = k \psi(x)$

(where $\hat{p}$ is the momentum operator) has a momentum of $k$. Suppose we apply the momentum operator twice:

$\hat{p}^2 \psi(x) = \hat{p} \hat{p} \psi(x) = \hat{p} (k \psi(x)) = k \hat{p} \psi(x) = k^2 \psi(x)$.

So an eigenstate of the operator $\hat{p}$ with eigenvalue $k$ is an eigenstate of the operator $\hat{p}^2$ with eigenvalue $k^2$. The point is that applying the operator twice gives us exactly what we want: it squares the eigenvalue. We can legitimately think of the operator $\hat{p}^2$ as the "momentum-squared" operator.

3. May 17, 2013

### Sylvia Else

OK, I can follow that, thanks. He starts discussing eigenfunctions and eigentvalues a bit later. Perhaps readers are meant to take it on trust until then.

Sylvia.

4. May 18, 2013

### dextercioby

The mathematical operation of <squaring> a linear operator means <applying it twice>:

$$\hat{p}^2 := \hat{p}\circ\hat{p}$$

(For simplicity, let's neglect hat the momentum operator is unbounded).