Position space wave function of an electron

Question

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?

Answer 1

How come a free particle (take a free electron) has a representation as a plane wave and not as something with spherical symmetry?

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).

Now what? How can the wavepacket be the same as a plane wave?

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 $\psi(x)$.

Answer 2

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.

How come a free particle (take a free electron) has a representation as a plane wave and not as something with spherical symmetry.

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.