3
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ What about $Asinkx$? $\endgroup$ – Urgje Apr 15 '15 at 9:12
  • $\begingroup$ I don't its an eigenfunction of $p_x$ operator !! $\endgroup$ – user35952 Apr 15 '15 at 16:51
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 !! $\endgroup$ – user35952 Apr 15 '15 at 16:52
  • $\begingroup$ I think even if the $p_x$ is self-adjoint, it will not have eigenfunctions, since its non-compact. Thanks for the answer !! $\endgroup$ – user35952 Apr 19 '15 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.