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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.

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  • $\begingroup$ It's simply easier for a general understanding to fix the uncertainty for one variable (momentum) and have the wave become infinite in space than to calculate with a mixture of both being uncertain. Think about explaining the vibrations of a guitar string: are you picturing standing waves with certain frequencies or do you think of a dispersing wave packet? Also: simply because we are taught plane waves does not mean we are obsessed with them. There will always be more accurate (more complex) solutions to specific physical problems but you have to start abstracting somewhere... $\endgroup$
    – Asmus
    Nov 20 '20 at 8:05
  • $\begingroup$ In my case the premise is wrong. The solution to the free particle was found to be/given as an infinite sum (integral) of plane waves, a wave packet that one could normalize. We saw that it "decayed" with time and that the expectation value of the position moved as the CoM of a classical particle, etc. So yeah, this question is opinion based. Not a question suited for this website. $\endgroup$ Nov 20 '20 at 9:07
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    $\begingroup$ As a side point, you can have normalizable momentum eigenstates in compact spaces. As others have pointed out, the cash value of momentum eigenstates is not necessarily as physically realizable states (i.e., normalizable states) but rather as a useful basis for the Hilbert space. Useful because they are generators of translations in space and consequently diagonalize the free Hamiltonian, etc. $\endgroup$
    – Dvij D.C.
    Nov 21 '20 at 1:12
  • $\begingroup$ Have you ever tried doing a practical calculation using a wave-packet basis? $\endgroup$
    – tparker
    Nov 22 '20 at 4:39
  • $\begingroup$ @tparker I keep hearing this word basis and I'm somewhat ashamed to admit that I have no clue what people mean when they say it. Why would you need to worry about basis vectors for a complex vector space that is at most four dimensions? $\endgroup$ Nov 22 '20 at 15:34
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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 variance 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.

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    $\begingroup$ Also their use as the underlying fields on which creation and annihilation operators work in quantum field theory, wave packets would be an unnecessary complication in the very successful calculations of Feynman diagrams. $\endgroup$
    – anna v
    Nov 19 '20 at 13:25
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    $\begingroup$ +1: To spell out the connection between translations and momentum more explicitly, momentum eigenstates are generators of translation in space. So even if the theory is not translational invariant, you would end up needing them one way or the other. $\endgroup$
    – Dvij D.C.
    Nov 21 '20 at 1:08
  • $\begingroup$ @annav: I misread that as "very stressful calculations" ... $\endgroup$
    – user21299
    Nov 21 '20 at 2:42
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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.

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    $\begingroup$ Slight technicality: being orthogonal is not a requirement for being a basis. $\endgroup$
    – Javier
    Nov 21 '20 at 1:45
  • $\begingroup$ @Javier Generally in inner product spaces we are interested not in bases but orthonormal bases, because that's the notion that respects the structure of the space. $\endgroup$ Nov 21 '20 at 5:05
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    $\begingroup$ @MarioCarneiro I know, but that's an orthogonal basis, not just a basis. I did say it was a slight technicality! $\endgroup$
    – Javier
    Nov 21 '20 at 13:50
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    $\begingroup$ 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. But that is not a peculiar property at all; bases are a dime a dozen and not particularly meaningful. It's their other properties that make them useful. $\endgroup$ Nov 21 '20 at 18:31
  • $\begingroup$ @all If you think you can provide a clearer presentation then you're obviously welcome to write your own. $\endgroup$ Nov 21 '20 at 20:41
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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.

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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.

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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.

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