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On page 138 of my Shankar text, Shankar states:

"...since the plane waves are eigenfunctions of P, does it mean that states of well-defined momentum do not exist? Yes, in the strict sense. However, there do exist states that are both normalizable to unity (i.e. correspond topropervectors) and come arbitrarily close to having a precise momentum. For example, a wave function that behaves as [itex] e^{i p_0 x / \hbar} [/itex] over a large region of space and tapers off to zero beyond, will be normalizable to unity and will have a Fourier transform so sharply peaked at [itex]p=p_0[/itex] that momentum measurements will only give results practically indistinguishable from [itex]p_0[/itex]."

My question is: What about momentum in the case of a free particle? Don't we have states of well defined energy here? And if we do, doesn't that imply (by E = p^2/2m) that we have states of well defined momentum as well? Perhaps I'm simply misunderstanding the distinction between "well defined" momentum and "arbitrarily close" to well defined momentum. I thought the whole point of being "arbitrarily close" was that you could have momentum as well defined as you wanted.

My whole point in asking this question is -- since we *can't* associate plane waves with a physically sensible probability distribution, how do we obtain a physically sensible probability distribution for momentum? I know we use the complex exponential [itex]\langle x|p\rangle[/itex] to create the Fourier transform [itex]\psi(p)[/itex] of some [itex]\psi(x)[/itex], but it seems so odd to me that the plane wave is simply a mathematical entity that enters the *definition* of a Fourier transform for *all* transforms but just so happens to be the eigenfunction of the momentum operator in x-space. Can someone draw the connection for me? I'm failing to see it.

Many thanks in advance.

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# Momentum Eigenfunctions and Definite Momentum

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