# 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.

(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