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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?

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    $\begingroup$ 1. $e^{ipx/\hbar}$ is certainly bounded since its absolute value is $1$ for any $p$ and any $x$. 2. By definition and physical assumptions, $p$ is real. 3. $\hat p$ is quantized is the infinite well. Usually the energy is quantized, and this is $p^2/2m+V(x)$. In the case of the infinite well $V(x)=0$ so quantization of energy is the basically the same as quantization of momentum. $\endgroup$ – ZeroTheHero Oct 24 '17 at 12:33
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    $\begingroup$ @ZeroTheHero There is no well defined momentum operator for particle in a box problem. Standard $\hat{p} = -i\hbar \frac{d}{dx}$ does not have any eigen function which satisfy the boundary conditions $\psi(\text{Boundary})=0$. $\endgroup$ – Sunyam Oct 24 '17 at 15:51
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    $\begingroup$ @confused I'm fully aware of this. The question is elementary and I think the OP means that $k$ is quantized, rather than $\hat p$ as a differential operator. $\endgroup$ – ZeroTheHero Oct 24 '17 at 17:13
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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.

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  • $\begingroup$ Thank you very much for your answer! But I still have the following doubt: even if I consider $p$ real, $e^{ipx}$ still diverges to infinity for $+\infty$ and is zero for $-\infty$, by direct substitution of $x$. Do we only consider the magnitude of $e^{ipx}$ ? $\endgroup$ – RicardoP Nov 3 '17 at 15:18
  • $\begingroup$ That depends on what you mean by "diverges", which you haven't specified, so it's impossible to tell. $\endgroup$ – Emilio Pisanty Nov 3 '17 at 18:08
  • $\begingroup$ In the sense that when I directly substitute x by $+\infty$, the wave function goes to infinity. $\endgroup$ – RicardoP Nov 4 '17 at 0:31
  • $\begingroup$ Then no, that is factually incorrect. For real $p$ the eigenfunctions' modulus is constant, which is in direct contradiction to that behaviour. $\endgroup$ – Emilio Pisanty Nov 4 '17 at 9:32
  • $\begingroup$ But then how do I explain, for example, the free particle, whose eigenfunctions can be linear combinations of $Ae^{ikx}$ and $Be^{-ikx}$, are not square integrable? $\int_{-\infty}^{+\infty} dx|Ae^{ikx}+Be^{-ikx}|^2$ diverges for all values of A and B. Btw, thank you very much for the answers. $\endgroup$ – RicardoP Nov 7 '17 at 13:18
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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 $\hat{p}$.
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