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## Main Question or Discussion Point

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

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

Then he observes that the operator on the right (which I construe to mean [itex]p^2_{op}/2m[/itex]) 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 [itex]p^2_{op}[/itex] isn't saying use the momentum squared, but apply the momentum operator twice.

What am I missing here?

Sylvia.

*i*[itex]\hbar[/itex][itex]\frac{\partial ψ(x, t)}{\partial t}[/itex] = -[itex]\frac{\hbar^{2}}{2m}[/itex][itex]\frac{\partial ^{2}ψ(x, t)}{\partial x^2}[/itex]He then makes the identification [itex](h/i)(\partial/\partial x) = p_{op}[/itex], which he has previously identified as the momentum operator, and rewrites the equation as

*i*[itex]\hbar[/itex][itex]\frac{\partial ψ(x, t)}{\partial t}[/itex] = [itex]\frac{p^2_{op}}{2m}[/itex][itex]ψ(x, t)[/itex]Then he observes that the operator on the right (which I construe to mean [itex]p^2_{op}/2m[/itex]) 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 [itex]p^2_{op}[/itex] isn't saying use the momentum squared, but apply the momentum operator twice.

What am I missing here?

Sylvia.