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 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 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 at 0:04
up vote 4 down vote accepted

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.


Edit:

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.

  • I am having a little bit of trouble understanding why the defined function would still be in the domain of the adjoint. What would be the value of the adjoint acting on the defined example? – Jacob Schneider Dec 8 at 23:29
  • @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. – Dan Yand Dec 9 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!) – Dan Yand Dec 9 at 0:25
  • This makes sense thanks. The appendix really helped me b/c in order to be in the domain of the adjoint that value has to be well defined, so seeing that value convinced me. Thanks for the answer! – Jacob Schneider Dec 9 at 0:33
  • 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 at 0:49

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