Why do physicists use plane waves so much?

Question

When looking at solutions of the Dirac equation people tend to give solutions as $$\psi^{(1)} = e^{\frac{-imc^2t}{\hbar}}\begin{pmatrix}1\\0\\0\\0\\\end{pmatrix},\psi^{(2)} = e^{\frac{-imc^2t}{\hbar}}\begin{pmatrix}0\\1\\0\\0\\\end{pmatrix},\psi^{(3)} = e^{\frac{imc^2t}{\hbar}}\begin{pmatrix}0\\0\\1\\0\\\end{pmatrix},\psi^{(4)} = e^{\frac{imc^2t}{\hbar}}\begin{pmatrix}0\\0\\0\\1\\\end{pmatrix}$$

To me this seems useless because you cannot normalize it. Doesn't it represent a situation where there is infinite uncertainty in position and zero uncertainty in momentum? How is that useful? Surely it would be more useful to give a wave packet solution to the Dirac equation?

It's the same problem when looking at solutions of the Schrodinger equation for a free particle. There, the given solution is the plane wave $e^{i(kx-\omega t)}$, which you cannot normalize. I understand that the equation is linear and that you can represent the solution as a sum of these stationary states, but wouldn't it be more logical to give the general solution, the Gaussian wave packet? $$\psi=\frac{1}{\sqrt{\pi+\frac{i\hbar t}{m}}}e^{\frac{-x^2}{2(\pi+\frac{i\hbar t}{m})}}$$

You can also construct solutions with sums of this and it makes much more sense because you can actually normalize it. You can add them, see how the particles interfere with each other, understand the role of complex numbers, etc. I feel like there is some concept that I am missing because otherwise, I wouldn't see this plane wave solution so much.

Answer 1

Plane waves in quantum mechanics are usually the eigenstates of the momentum operator, which is what makes them very useful. Momentum conservation is the manifestation of the translational invariance in space, which is arguably what makes plane waves also very useful in classical contexts, whenever one deals with a homogeneous media.

Mathematically, plane waves correspond to the Fourier expansion, which is also a very convenient mathematical tool.

On a more general level: expanding in terms of the appropriate orthogonal basis is often a good idea.

Answer 2

I understand that the equation is linear and that you can represent the solution as a sum of these stationary states

That's the whole story right there. Plane-wave solutions are useful because every other solution can be built up as a decomposition of plane-wave contributions.

However,

wouldn't it be more logical to give the general solution, the Gaussian wave packet?

there is no meaningful or useful sense in which the Gaussian wavepacket is a "general" solution.

You can also construct solutions with sums of this and it makes much more sense because you can actually normalize it. You can add them, see how the particles interfere with each other, understand the role of complex numbers, etc.

This is indeed true, and Gaussian-wavepacket solutions are very useful in understanding the dynamics, but they are of very limited usefulness in studying the behaviour of an arbitrary initial condition. Gaussian wavepackets are not a basis, because they are not mutually orthogonal. Moreover, while they do span the space in the sense that $\frac1\pi\int |\alpha⟩⟨\alpha|\mathrm d^2\alpha = \mathbb I$ using coherent-state notation, they are overcomplete, and this completeness relationship is not particularly useful $-$ basically because the Segal-Bargmann transform is not a particularly convenient tool, especially when compared with the Fourier transform.

Answer 3

Obsession:

Plane waves diagonalise the free hamiltonian and are useful as a basis for perturbation expansions of scattering problems or periodic systems. For atomic phsysics they are not useful.

Fiasco:

Since the plane waves are periodic, you can think of these as solutions in a box normalised by $1/\sqrt{V}$. Since the normalisation factor does not add anything it is often dropped.

Answer 4

In classical fields, the solution of the wave differential equation describes the real stuff, after adding the specific boundary conditions. So, a sinusoidal plane wave (SPW) can be a real solution.

But other plane waves, also solution for the differential equation, are not necessarily sinusoidal. In this case the SPW's change its status from a real solution to a basis for the real solution.

I understand that in QM that change of status is complete. SPW are no more real solutions, but they only define basis for them.

It seems more precise to call them eingenfunctions instead of solutions, to avoid taking them as physical entities. But as they are solutions of the differential equations, the wording ambiguity is here to stay.

Answer 5

There, the given solution is the plane wave ei(kx−ωt), which you cannot normalize.

You can't normalize it in isolation, but you don't need to normalize the basis vectors, you just need to normalize the actual vectors you're working with. Do you have the same qualms with waves given in position space? Position space and momentum space are dual. If we give a wave as a function that assigns a complex amplitude to each position in physical space, then we're using the states with zero uncertainty in position and infinite uncertainty in momentum as basis vectors, and those states can't be normalized either. When you write $\psi=\frac{1}{\sqrt{\pi+\frac{i\hbar t}{m}}}e^{\frac{-x^2}{2(\pi+\frac{i\hbar t}{m})}}$, strictly speaking, that's a function. To make it a vector, we have to treat that function as giving the coefficients of an uncountable number of vectors: $\psi = \sum_{x \in X}f(x) \delta_x $ where $\delta_x$ is a state with definite position of $x$.

Also, wouldn't we need more parameters to have a general solution, such as

$\psi=\frac{1}{\sqrt{\pi+\frac{i\hbar t}{m}}}e^{\frac{-(x-x_0)^2}{2(\pi+\frac{i\hbar t}{m})}}$?

As other answers have said, plane waves are eigenstates of the momentum operator, which means that any operator based on the momentum operator will be diagonal in terms of this basis. If you have $\hat {\mathcal H}\psi = \lambda \psi$, then $e^{-\frac i {\hbar}\hat {\mathcal H}t}\psi$ is just $e^{-\frac i {\hbar}\lambda t}\psi$.

That means that each state evolves independently. If you had a Gaussian basis, the time evolution of one time-independent basis state will have to involve other states. And if you have a medium with frequency-dependent propagation speeds, any state that doesn't have a fixed frequency is going to exhibit dispersion. There can also be frequency-dependent damping.