# Plane waves not physical states?

1. Apr 12, 2008

### jdstokes

Hi all,

In quantum mechanics, it is postulated that the physical states of a system correspond to the points on the unit sphere in the associated Hilbert space.

And yet, people freely talk about e.g. scattering of plane waves by a potential? What does $e^{\mathrm{i} \vec{k}\cdot\vec{x}}$ correspond to physically?

2. Apr 12, 2008

### G01

Plane waves are the eigensolutions of the free particle Hamiltonian, and therefore correspond to free particles.

As you point out though, a plan wave is not square integrable and is thus not normalizable. This does pose a problem, but there are several ways around it. For instance, we can work with traveling wave packets or apply periodic boundary conditions to obtain a normalizable form of the solution.

3. Apr 12, 2008

### jdstokes

Yep, we can form wave packets out of superpositions of planes waves of differing frequency, but at the end of the day the plane wave itself is non-physical right? Ie we don't see plane waves in the real world.

4. Apr 12, 2008

### lbrits

The more precisely the momentum of the particle is defined, the more exactly it's wavefunction will look like a plane wave. For reasons above, the wavefunction will never be a true plane wave, but a very dispersed wavepacket will look like a plane wave near the center.

If you put your particle in a box, then plane waves are normalizable, and depending on the state, may indeed be a good approximation. When blackbody radiation is discussed, the modes of the cavity are plane waves, although of course these aren't strictly wavefunctions.

Finally, when we do scattering calculations we often assume an approximately monoenergetic beam of particles scattering off of some target (perhaps another beam). Although the plane wave description isn't good everywhere, the calculation receives its dominant contribution from the center of the scattering potential, and, as long as the potential is weak, we can say that the incident wave is well approximated by a plane wave. This is because the wavefunction

$$:\psi(x,t) = \frac{1}{\sqrt{2\pi}} \int^{\,\infty}_{-\infty} A(k) ~ e^{i(kx-\omega(k)t)} \,dk$$

is appproximately a plane wave as long as $$A(k)$$ has support mainly around a single value of $$k$$.

5. Apr 13, 2008

### comote

Seeing in the real world? But you are commenting as if we actually see a regular wave function in the real world, but this is of course another discussion.

My guess is that the fact that the plane waves are not in the Hilbert space but can be approximated arbitrarily close is related to the idea that measurements in the continuous spectrum can not be measured to perfect accuracy.