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After solving the eigenvalue equation for the momentum operator, I get $u(x)=Ce^{ipx/\hbar}$, just like in Gasiorowicz's chapter 3. And then it says there:

"...and the eigenvalue $p$ real, so that the eigenfunction does not blow up at either $+\infty$ or $-\infty$....".

Doesn't the obtained solution always blow up at $+\infty$ and is 0 for $-\infty$? What happens if $p$ is imaginary?

Also, he says further on "this is the only constraint on $p$: we say that $\hat{p}$ has a continuous spectrum". This only happens for the free particle right? For a particle in a box, for example, the momentum is quantized, correct?

Qmechanic
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RicardoP
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2 Answers2

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Doesn't the obtained solution always blow up at $+\infty$ and is 0 for $− \infty$? What happens if $p$ is imaginary?

That's precisely the point. The momentum $p$ needs to be real because if it has any imaginary part then the obtained solution will diverge at either sign of infinity. If $p$ is real, however, $|u(x)|=|e^{ipx}|=1$ is constant, and there is no blow-up at either infinity.

On your second question,

"this is the only constraint on $p$: we say that $\hat{p}$ has a continuous spectrum". This only happens for the free particle right? For a particle in a box, for example, the momentum is quantized, correct?

that's half right and half wrong. If you impose a finite box, then the momentum becomes rather problematic, and you can't quite say that it's quantized (but if you impose periodic boundary conditions then you can).

However, having the real line as your configuration space (instead of a bounded interval) does not equate to the particle being free, because you can well have a potential that makes the hamiltonian eigenfunctions bounded. In those situations $\hat p$ still has a continuous spectrum (because $\hat p$ knows nothing of the potential) but you still don't call the situation a "free particle" either.

Emilio Pisanty
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  1. For the standard momentum operator the eigenvalues are real simply because it is a hermitian operator.
  2. Momentum is quantized, and its possible values are those of the eigenvalues of $\hat{p}$.
Daniel Duque
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