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Going through the Quantum mechanics book by Capri, am time and again held with some stupid doubts on this topic of self-adjointness. We have for the momentum operator in finite domain,

$$ p = -i\hbar \frac{\partial }{\partial x} \\ D_p = \big\{f(x),f'(x)\in \mathrm{L_2}(0,L) , f(0) = f(L) = 0 \big\} $$ Now we go on to define, the adjoint of the $p$ operator, with the larger domain

$$ p^{\dagger} = -i\hbar \frac{\partial }{\partial x} \\ D_{p^{\dagger}} = \big\{f(x),f'(x)\in \mathrm{L_2}(0,L) , f(0) = \mathrm e^{i\theta}f(L)\big\} $$

Now, he says, we can see that $p^{\dagger}$ is self-adjoint, which I believe means that $p^\dagger = p ^{{\dagger}^{\dagger}}$ ($D_{p^\dagger} = D_{p ^{{\dagger}^{\dagger}}}$). So we have a self-adjoint extension of $p$. So my question now is that, is it $p ^{{\dagger}^{\dagger}}$ that is the extension of p or what is it that is its extension ?

But he also goes on to say that a symmetric operator $A$ is essentially self-adjoint if $ A^{{\dagger}^{\dagger}} $ is self-adjoint. But is not this true for the above case, given that $p^\dagger = p ^{{\dagger}^{\dagger}}$ and a $p^\dagger $ is self-adjoint ? But however the above case is not essentially self-adjoint since there are infinite ways of choosing $\theta$.

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A closed extension $A_c$ of an operator $A$ is an operator whose action is the same as $A$, the domains satisfy $D(A_c)\supset D(A)$ and $A_c$ is closed.

Given that, the smallest closed extension of a symmetric (densely defined) operator is its double adjoint $A^{**}$. We call it the closure of $A$, and denote it by $\overline{A}$.

An operator $A$ is called essentially self-adjoint if its closure $\overline{A}$ is self-adjoint. This also implies that $\overline{A}$ is the unique self-adjoint extension of $A$ (a self-adjoint extension is a closed extension of $A$ that is self-adjoint). This is a nice feature, since in general a symmetric, densely defined, operator may have zero, one or infinite self-adjoint extensions.

In the case above, it depends on the definition of $D_{p^*}$. If it is dependent on a fixed $\theta$, then something seems not right, but maybe the domain is of functions that satisfy the condition for any $\theta$, and then it could be effectively self-adjoint (and therefore $p$ ess. s-a).

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  • $\begingroup$ What you mean by $A_c$ is closed ? I think from the literature I have gone through $p$ is not essentially self-adjoint in the domain $(0,L)$. When you say same action, is not the action of $p^\dagger$ same as that of $p$ in this case ? $\endgroup$ – user35952 Apr 20 '15 at 6:38
  • $\begingroup$ An operator $A$ on a Hilbert space $\mathscr{H}$ is closed if the graph $\{(\phi,A\phi), \phi\in D(A)\}$ is a closed subspace of $\mathscr{H}\times \mathscr{H}$. Every self-adjoint operator is closed. It is possible that $p$ is not essentially self-adjoint, and that would probably mean that the $p^*$ he defines is a self-adjoint extension of $p$ but not its adjoint (if else it would mean that $p$ is essentially self-adjoint). $\endgroup$ – yuggib Apr 20 '15 at 7:21
  • $\begingroup$ So in this case $p^*$ is the s-a extension by that logic. Thanks ! Am sorry, am from a physics background, what is a graph ? $\endgroup$ – user35952 Apr 20 '15 at 7:34
  • $\begingroup$ No problem, the graph is this set I have defined above. It is called like that because if $\mathscr{H}=\mathbb{R}$ (the reals) it is exactly the cartesian graph of the function $A$ ;-) $\endgroup$ – yuggib Apr 20 '15 at 7:44

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