Timeline for Translation operator eigenvalues can be real and arbitrary?

7 events

when toggle format	what		by	license	comment
May 12, 2020 at 17:56	vote	accept	proton
May 12, 2020 at 17:53	comment	added	mike stone		You can define translation in $C^{\infty}[{\mathbb R}]$ (smooth functions) for example, and then your exponential function is certainly an eigenfunction. But there is no inner product, and no physical applications that I can think of.
May 12, 2020 at 17:13	comment	added	proton		Is it a mathematical issue or a physical one? Doesn't $e^{\lambda x}$ live in some function space where the translation operator is not unitary?
May 12, 2020 at 17:08	comment	added	mike stone		Yes. Normalized so that $\langle p\|p'\rangle = 2\pi \delta(p-p')$.
May 12, 2020 at 16:45	comment	added	proton		I had the real line in mind but your answer is good. On the real line, it is it correct that free particle states $\psi(x)=e^{ikx}$ are the only eigenfunctions, so $e^{ika}$ (the entire unit circle) are the only eigenvalues?
May 12, 2020 at 16:42	history	edited	mike stone	CC BY-SA 4.0	added 1 character in body
May 12, 2020 at 16:37	history	answered	mike stone	CC BY-SA 4.0