# Momentum Eigenfunctions and Definite Momentum

1. Oct 4, 2012

### jmcelve

Hi everyone,

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 to proper vectors) and come arbitrarily close to having a precise momentum. For example, a wave function that behaves as $e^{i p_0 x / \hbar}$ 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 $p=p_0$ that momentum measurements will only give results practically indistinguishable from $p_0$."

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 $\langle x|p\rangle$ to create the Fourier transform $\psi(p)$ of some $\psi(x)$, 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.

Last edited: Oct 4, 2012
2. Oct 4, 2012

### Staff: Mentor

No, we don't. A properly normalizable wave function for a free particle is a wave packet, a superposition of plane waves covering a continuous range of momentum (and energy). The "width" of the momentum range, Δp, is related to the physical "width" of the packet, Δx, by the Heisenberg uncertainty principle. In fact, the HUP can be derived from theorems about the Fourier transforms that relate the momentum-space wave function $\psi(p)$ to the position-space wave function $\psi(x)$.

3. Oct 4, 2012

### jmcelve

Thanks for the prompt reply. This clarifies it perfectly, actually. So the momentum space equation serves to weight the plane wave and the time dependence term in order to produce the wave packet, correct?

Also -- how exactly is it that we determined that functions of x and p are transforms of one another?

4. Oct 4, 2012

### The_Duck

The momentum operator in the position basis is -i hbar d/dx. Try applying that to a state psi(x), written in terms of its Fourier transform psi(p):

$$-i \hbar \frac{d}{dx}\psi(x) = -i \hbar \frac{d}{dx} \int \frac{dp}{2\pi\hbar}e^{ipx/\hbar}\psi(p) = \int \frac{dp}{2\pi\hbar}e^{ipx/\hbar}p\psi(p)$$

So the Fourier transform of psi(x) just gets multiplied by p when we hit psi(x) with the momentum operator. We call psi(x) the position space wave function because to apply the position operator we just multiply psi(x) by x. Apparently psi(p) is the momentum space wave function in the same sense.