In Wikipedia I find the wave function of a free particle to be $$ \Psi(\vec{r},t) = A\,e^{i(\vec{k}\vec{r}-\omega t)}$$ This is is a plane wave moving in the direction of $\vec{k}$ with speed (phase velocity) $v=\omega/|\vec{k}|$.

In particular $\Psi(\vec{r}, t)$ for a fixed pair $\vec{r}$ and $t$ is constant and $\neq 0$ on an infinite plane.

For $|\Psi|^2$ to be a probability of position, some trickery is obviously involved to get out something finite, but be that as it may.

What I really would like to have a hint about: How come a free particle (take a free electron) has a representation as a plane wave and not as something with spherical symmetry. The wikipedia article just says that the plane wave is not normalizable but the wavepacket is and keeps describing the wavepacket. Now what? How can the wavepacket be the same as a plane wave?

  • $\begingroup$ The electron is not described using a plane wave as this has infinite amplitude, the wave packet uses constructive interference to model the electron and produce a finite amplitude. A wave packet is a collection of plane waves, with a gaussian like function included. $\endgroup$ – user74893 Mar 29 '15 at 14:36
  • $\begingroup$ To expand a bit, whatever function you use to model the electron has to be square integrable, and you need to normalise it to keep the probability of finding it less than or equal to 1. You may have to use Wikipedia to follow these terms, sorry....regards $\endgroup$ – user74893 Mar 29 '15 at 14:41
  • $\begingroup$ @irishphysics Thanks, that makes perfect sense. Just wonder what the plane wave as a solution of the Schrödinger equation for a free particle than may mean. $\endgroup$ – Harald Mar 29 '15 at 14:47


Because we specified a single outward momentum vector $\vec{k}$, with a direction to it. This breaks spherical symmetry, but it's not required. We can do spherical waves too, and we often do in scattering theory: $$ \psi(\vec{r}) = \frac{e^{ikr}}{4 \pi r} $$

This has the nice advantage that if we calculate the probability at $r$: $$ \int \mathrm{d} \Omega | \psi(r) |^2 = 4 \pi r^2 \frac{1}{4 \pi r^2} = 1 $$ So the probability stays constant as the wave expands outwards (still not normalizable over all space, though).


Forget the complex exponentials for a second. Think of a sine wave. One sine wave is non-renormalizable--if I integrate over infinity, it never comes out finite. Two sine waves, still non renormalizable. But if you add together a bunch of them, they start to look less and less like sine waves, and start to look like something localized. That link shouldn't be taken to have any physical significance, by the way, it's just something I cooked up playing around in Wolfram.

My point here is that, by a technique known as Fourier transform, we can take nearly any function and turn it into a sum over sines and cosines, or rather, an integral. This means that given an arbitrarily shaped position wavefunction, we can decompose it into a sum over these momentum basis states. And this Fourier transform, even though it's an integral over functions which are not normalizable, can itself be finite and have a well-defined, normalized amplitude. This is a great trick, because we know a lot about how to handle momentum basis states--they're very simple! All we have to do to know how $\psi(x)$ behaves is to turn it into $\int \mathrm{d} k \ A(k) e^{i k r}$, and then apply what we know to those individual plane waves. Adding them back up gives us knowledge about $\psi(x)$.

  • $\begingroup$ I take it that even a spherical symmetric $\Psi$ with sufficient smoothness properties can be decomposed into an (in)finite sum of plane waves? And does this "fuzzy ball" then travel without dispersion? $\endgroup$ – Harald Mar 29 '15 at 15:30
  • $\begingroup$ I believe so, yes, you should just get a big "shell" of probability fading out as it gets spread over a larger and larger area. But, yes, it's the same thing--in practice you might use a different basis (spherically symmetric waves instead of plane waves). You could do the plane waves as well, but your integral would run over $\vec{k}$ instead of $k$ because the problem is no longer 1D. $\endgroup$ – zeldredge Mar 29 '15 at 15:33
  • $\begingroup$ Harald: I can't answer your question, but I would suggest that you look at the next section of the Wikipedia article, which is about a particle in a box, and think of pair production and electron diffraction and the Einstein-de Haas effect, then think of the electron as a 511keV photon in a spin ½ spindle-sphere "box" of its own making. $\endgroup$ – John Duffield Mar 29 '15 at 16:24

To be sure, the TISE solutions (in Cartesian coordinates) for a free particle are not normalizable and thus are not physical states. Quoting Griffiths

  • "there is no such thing as a free particle with a definite energy"

However, these free particle solutions are a basis for normalizable states including the wave packet. That is to say, by Fourier decomposition, we can write the general wavefunction as a weighted 'sum' (integral) of free particle solutions.

This is important because these free particle solutions are the eigenfunctions of the (linear) momentum operator.


Solving the TISE for the free particle in spherical coordinates gives solutions that have definite angular momentum rather than definite linear momentum. The angular dependent part of these solutions are the spherical harmonics.

The spherically symmetric wavefunctions are then states with zero angular momentum.


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