# Domain of symmetric momentum operator vs self-adjoint momentum operator

Is there an example of a function that is not in the domain of the 'naive' symmetric (but not self-adjoint) momentum operator $$p:=-i\frac{d}{dx}$$ but is in the 'true' self-adjoint momentum operator $$p:= \left(-i\frac{d}{dx}\right)^\dagger$$.

I am trying to understand the mathematical differences b/w symmetric and self-adjointness and thought that this would be an enlightening example. Could you also show why this example in the domain of the self-adjoint momentum operator using the definition of the adjoint domain: $$D(A^\dagger) := \{ \phi \in {\cal H}\:|\: \exists \phi_1 \in {\cal H} \mbox{ with} \: \langle \phi_1 |\psi \rangle = \langle \phi | A \psi\rangle \:\: \forall \psi \in D(A)\}$$

This question was motivated by reading this fantastic answer by Valter Moretti.

• I like the example in Branimir Ćaćić's Answer to a similar question "the momentum operator $i\tfrac{d}{dx}$ on the interval $[0,1]$ (or your finite interval of choice) is symmetric but not essentially self-adjoint; you have a full circle's worth of different self-adjoint extensions, each coming precisely from imposing a quasi-periodic boundary condition of the form $f(0) = e^{2\pi i \beta}f(1)$". – Keith McClary Dec 8 '18 at 23:30
• You have a misunderstanding: the adjoint of a symmetric operator does not mean the same as its selfadjoint extension. They may be the same operator, but not necessarily. – Keith McClary Dec 8 '18 at 23:39
• @KeithMcClary Isn't the adjoint of a symmetric operator always an extension? My question does not rely on the adjoint being self-adjoint (even though I believe it is for the momentum operator). My question is just about a vector in the domain of the adjoint but not in the domain of the original operator. – Jacob Schneider Dec 9 '18 at 0:04
• – Keith McClary Dec 9 '18 at 2:28

Define $$P \equiv -i\frac{d}{dx} \tag{1}$$ and $$\psi(x) = |x|\,\exp(-x^2). \tag{2}$$ Let $$D(P)$$ denote the domain of $$P$$. Clearly, $$\psi$$ is not in $$D(P)$$, because it is not differentiable at $$x=0$$. However, $$\psi$$ is in the domain of $$P^\dagger$$, because $$\langle \psi|P\phi\rangle \equiv -i\int^\infty_{-\infty}dx\ \psi^*(x)\frac{d}{dx}\phi(x) \tag{3}$$ is well-defined for all $$\phi\in D(P)$$, and $$P^\dagger$$ is defined by the condition $$\langle P^\dagger\psi|\phi\rangle = \langle \psi|P\phi\rangle \tag{4}$$ for all $$\phi\in D(P)$$. The domain $$D(P)$$ is dense, so $$\psi$$ can be arbitrarily well-approximated by a function in $$D(P)$$, just by smoothing out the "kink" in an arbitrarily small neighborhood of $$x=0$$, but $$\psi$$ itself is not in $$D(P)$$, not even after accounting for the fact that vectors in this Hilbert space are represented by functions modulo zero-norm functions. We can't smooth out the "kink" at $$x=0$$ in $$\psi$$ by adding any zero-norm function.
The the value of $$P^\dagger$$ acting on the example (2) is $$P^\dagger\psi = -i\big(s(x)-2x|x|\big)\exp(-x^2),$$ where $$s(x)=\pm 1$$ is the sign of $$x$$. This can be checked by checking that it satisfies (4) for arbitrary differentiable functions $$\phi$$, which is possible because the point $$x=0$$ can be omitted from the integrand without changing the value of the integral.
• @JacobSchneider $P^\dagger$ is not a differential operator in the ordinary sense. It coincides with the differential operator $P$ when acting on differentiable functions, but $P^\dagger$ is also defined as an abstract operator on other functions. The Hilbert space isn't a set of functions. It's a set of abstract vectors. An abstract vector can be (non-uniquely) represented by a function, but it isn't a function. Similarly, $P$ and $P^\dagger$ are linear op's on the Hilbert (vector) space. $P$ can be represented by a diff op, and so can $P^\dagger$ on some func's, but not in general. – Chiral Anomaly Dec 9 '18 at 0:14
• @JacobSchneider That comment addresses why the defined function is in the domain of the adjoint. To address what $P^\dagger\psi$ would be, I added an appendix to the answer. (Good question!) – Chiral Anomaly Dec 9 '18 at 0:25
• Any hints on how you determined $P^\dagger\psi$? I've tested it for a few values of $\phi(x)$ and it works! But it seems like magic to me. – Jacob Schneider Dec 9 '18 at 0:49