Issue with squaring operators

  • #1

Main Question or Discussion Point

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

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.
 

Answers and Replies

  • #2
1,006
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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
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
dextercioby
Science Advisor
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The mathematical operation of <squaring> a linear operator means <applying it twice>:

[tex] \hat{p}^2 := \hat{p}\circ\hat{p} [/tex]

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

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