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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 ?

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  • $\begingroup$ What about $Asinkx$? $\endgroup$
    – Urgje
    Apr 15, 2015 at 9:12
  • $\begingroup$ I don't its an eigenfunction of $p_x$ operator !! $\endgroup$
    – user35952
    Apr 15, 2015 at 16:51

1 Answer 1

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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.

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  • $\begingroup$ 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, 2015 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, 2015 at 2:15

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