Translation operator eigenvalues can be real and arbitrary?

Question

Consider the translation in space operator in $1D$: $$D(a)=e^{-ia\hat{p}/\hbar}$$ It is unitary - $D(-a)=D^{\dagger}(a)=D^{-1}(a)$ - which implies that $D(a)$ has eigenvalues on the unit circle like all unitaries do.

$D(a)$ acts on a function $f(x)$ by translating it - $$D(a)f(x)=f(x-a)$$

Now consider the case of $f(x)=e^{\lambda x}$:

$$D(a)f(x)=e^{\lambda(x-a)}=e^{-\lambda a}e^{\lambda x}=e^{-\lambda a}f(x)$$

So $f$ is an eigenfunction of the translation operator with eigenvalue $e^{-\lambda a}$ which may be arbitrary large or small for sufficient $\lambda$.

It appears we arrived at a contradiction. How is it solved? Is it enough to consider only eigenfunctions that are normalized wavefunctions? Is it possible that $D(a)$ has eigenstates that are not orthogonal to each other?

Answer 1

Are you working on the whole real line? If so, then $e^{\lambda x}$ is not normalizable even in the rigged Hilbert space (i.e. delta function normalizations) sense, so $e^{\lambda x}$ is not in the domain of the translation operator. If you are working with periodic boundary condition $\psi(x)=\psi(x+L)$ then $e^{\lambda x}$ is not in the Hilbert space unless $\lambda= 2\pi i n/L$ for integer $n$. If you are working on a finite interval with boundary consitions such as $\psi(0)=\psi(L)=0$, then $\hat p$ has no eigenvalues because it not self-adjoint and $e^{i\hat p a}$ is not unitary.