# Normalisation of wavefunction given by the form $Ae^{i(kx-wt)}$

Question 1.

Let's say that the wavefunction is given in the form

$$\Psi(x, t) = Ae^{i(kx-wt)}$$

Then because of the normalisation condition, the following should hold.

$$\int \Psi^*\Psi dx = A^2 \int_{-\infty}^{\infty} e^{-i(kx-wt)}\times e^{i(kx-wt)} \ dx = 1$$

Because $$e^{-i(kx-wt)}\times e^{i(kx-wt)} = e^{-i(kx-wt) + i(kx-wt)} = 1$$, the condition demands that

$$A^2 \int_{-\infty}^{\infty} dx = 1$$

As the integral value diverges to $$+\infty$$, we reach the conclusion that $$A$$ should converge to zero.

What's wrong here?

Question 2.

This is another question that should be classified and asked separately but as it is a short one I will just put this one into here. When expressing the wavefunction as a linear combination of basis functions, especially in discrete cases, is it that the index varies from $$-\infty$$ to $$\infty$$? That means, is it that

$$\Psi(x) = \sum_{-\infty}^{\infty} c_i \psi_i \ ?$$

Apologies in advance if the questions are trivial. I am a newcomer to quantum mechanics.

## 1 Answer

(1) Nothing wrong there. Plane waves are states of infinitely precise momentum and cannot be properly normalized in position space due to having infinite spread from Heisenberg uncertainty. In practice they still help e.g. in the scattering matrix formalism to get an amplitude for reflection and an amplitude for transmission, and to settle e.g. the basic physics of an Aharonov-Bohm ring where the actual lengths one cares about are finite.

(2) You always can and you never have to. There is a bijection between $$\mathbb Z$$ and $$\mathbb N$$ so however you number things is up to you. There is a slight reason to prefer $$\mathbb N$$ which is that a large class of these basis states are eigenfunctions of a Hamiltonian which is bounded from below, and thus these eigenvalues go on infinitely in one direction but not the other.

• Thanks for the answer! I have two questions following up from your respective answers. For (1), does that become reasoning for writing a wavefunction in terms of linear combination in an infinite-dimensional space, i.e. $\int \ dp \ c(p) \ \Psi_p(x)$? For (2), so is it that the choice of indices from 1 to $\infty$ or from $-\infty$ to $\infty$ depends on my preference but that anyway the number of indices should be infinite? – curious Feb 7 at 5:20
• (1) The Schrödinger equation is linear, is why you might consider superpositions of wavefunctions. But yeah there's nothing wrong in physics with using an integral for that superposition; we like to assume physics is “nice” and lacks pathological counterexamples because the world is noisy and those counterexamples are noise-adjacent to normal examples. (2) Doesn't have to be, a qubit is a 2-dimensional space for example rather than $\mathbb N$-dimensional. Arguably any quantum system “is really” finite dimensional, try to put too much energy in a particle in a box and it will blackhole itself. – CR Drost Feb 7 at 17:26