# Eigenvectors of $p_x$ in a particular domain

Defining the $p_x$ operator for the problem of particle in a infinite well. In the book by Capri on Quantum mechanics, the domain of the operator is given by,

$$p = -i\hbar \frac{\partial }{\partial x} \\ D_p = \big\{f(x),f'(x)\in \mathrm{L_2}(0,L) , f(0) = f(L) = 0 \big\}$$ Then later on he goes to define, $p^{\dagger}$ which has a bigger domain (Why ?) or with rather more general conditions on the functions given by, $$f(0) = \text e^{i\theta}f(L)$$ for the domain $D_{p^{\dagger}}$.

My question is concerned with the fact if I chose the domain $D_p$ (for the moment considering that $p$ is not self-adjoint i.e. $D_p \ne D_{p^{\dagger}}$ but rather $D_p \subset D_{p^{\dagger}}$), then there won't be any eigenfunctions for $p$ operator as such, since if there was it has to be trivially zero. Since for an eigenfunction $A \text e^{ikx}$ to be zero at $x=0$, $A$ has to be zero.

So how to address this fact that there is no eigenfunction for $p$ operator in the case when its not self-adjoint ?

Also is there a theorem on existence of eigenvectors for an operator ?

• What about $Asinkx$? – Urgje Apr 15 '15 at 9:12
• I don't its an eigenfunction of $p_x$ operator !! – user35952 Apr 15 '15 at 16:51

## 1 Answer

You're right, there is no eigenfunction. The eigenfunctions of a self-adjoint operator form a complete basis for the Hilbert space, but this is simply not true for symmetric operators. Therefore if an operator is not self-adjoint, it may not have any eigenfunctions.

• The operator's eigenfunction forming a complete basis is a different business, but I am still not able digest the fact that the operator doesn't have eigenvectors at all !! – user35952 Apr 15 '15 at 16:52
• I think even if the $p_x$ is self-adjoint, it will not have eigenfunctions, since its non-compact. Thanks for the answer !! – user35952 Apr 19 '15 at 2:15