I know this is essentially a mathematic question, but I received no answer on math SE. Moreover it has a direct application in physics, so I thought to ask this here too.

The momentum operator in one dimension in quantum mechanics is $P=-i\frac{d}{dx}$ (with $\hbar=1$). Consider it as an operator on $L_2(0,2\pi)$, the space of square-integrable functions on $(0,2\pi)$. It isn't continuous in fact if I consider the sequence $$ g_n(x)=\frac{e^{inx}}{\sqrt{2\pi n}} $$ it's a Cauchy sequence but $\{Pg_n\}_n$ does not converge.

I am searching a domain where $P$ is a continuous functional. My professor gave me the example $$D_P=\{\varphi \in L_2(0,2\pi):\varphi(0)=\varphi(2\pi) \}$$ but I'm not convinced because if I consider the function $\psi$

\begin{equation} \begin{cases} -\frac{x}{\sqrt{\pi}}+\sqrt{\pi},\hspace{0.5cm} 0\leq x<\pi\\\sqrt{x-\pi},\hspace{0.5cm} \pi\leq x \leq 2\pi \end{cases} \end{equation} it belongs to $D_P$, but if I apply $P$ on it I obtain

\begin{equation} \begin{cases} \frac{i}{\sqrt{\pi}},\hspace{0.5cm} 0\leq x<\pi\\ -\frac{i}{2\sqrt{x-\pi}},\hspace{0.5cm} \pi\leq x \leq 2\pi \end{cases} \end{equation} that is not square-integrable on $(0,2\pi)$.

Am I wrong? If not, which is a correct continuity domain for $P$?


Sorry, the answer to this technical question needs some mathematical technology.

The space you are looking for is $H^1(\mathbb R)$, the first Sobolev Hilbert space. It is made of the functions in $L^2(\mathbb R)$ admitting the weak first derivative represented by a $L^2$ function in turn.

$H^1(\mathbb R)$ is a complex Hilbert space if equipped with the scalar product: $$\langle \psi| \phi \rangle := \int_{\mathbb R} \overline{\psi}(x) \phi(x) dx + \int_{\mathbb R} \overline{\frac{d\psi}{dx}} \frac{d\phi}{dx} dx\qquad (1)$$ where $d/dx$ denotes the weak derivative (see below).

Equivalently, $H^1(\mathbb R)$ can be defined as the space of $L^2$ functions $\psi(x)$ whose Fourier (-Plancherel) transform $\hat{\psi}(k)$ admit finite $L^2$ norm with respect to the measure $(1+ k^2) ~ dk$ instead of the simpler $dk$.

Indeed it holds, where the scalar product is the same as in (1): $$\langle \psi| \phi \rangle := \int_{\mathbb R} \overline{\hat{\psi}}(k) \hat{\phi}(k)(1+ k^2) dk \qquad (2)\:.$$

Obviously it also holds: $H^1(\mathbb R)\subset L^2(\mathbb R, dx)$

Sticking to work in the physically sensible Hilbert space $L^2(\mathbb R, dx)$ for QM, it turns out that $H^1(\mathbb R)$ is the natural domain where the momentum operator is self-adjoint (not only Hermitian or symmetric). However, in that Hilbert space, the momentum operator, whose correct definition is:

$$P = -i \frac{d}{dx}\quad \mbox{in weak sense, and with domain $D(P)= H^1(\mathbb R)$}$$

is always unbounded, i.e. discontinuous.

So, to see $P$ as a bounded (i.e. continuous) operator it is not enough to restrict it to an appropriate domain, but you also have to change the topology (norm) of the domain, passing from that of the simple $L^2$ to that of $H^1(\mathbb R)$. The topology in the co-domain remains that of $L^2$.

NOTE 1. A (measurable) function $f: \mathbb R \to \mathbb C$ is said to admit weak derivative $\frac{df}{dx}=g$, where $g$ is another (measurable) function, if, for every $h: \mathbb R \to \mathbb C$ of class $C^\infty$ and compactly supported, one has: $$\int_{\mathbb R} f(x) \frac{dh}{dx} dx = - \int_{\mathbb R} g(x) h(x) dx \:.$$ For instance $f(x)= |x|$ does not admit derivative at $x=0$ however admits weak derivative given by $\mbox{sgn}(x)$. The Dirichlet function $d(x) = 1$ if $x$ is rational $d(x)= 0$ if $x$ is not rational does not admit a derivative anywhere, but it admits weak derivative given by the zero function.

NOTE 2. If dealing with the particle in $[0,2\pi]$ the situation is analogous. The self-adjointness domain of $P$ is $H^1((0,2\pi))$ and $P$ is defined as before. The only change is that passing in Fourier rep. one has to use Fourier series instead of Fourier transform. In this case the $H^1((0,2\pi))$ scalar product becomes: $$\langle \psi | \phi \rangle = \sum_{n \in \mathbb Z} \overline{\psi_n} \phi_n (1+ n^2)\:.$$ In the Hilbert space $H^1((0,2\pi))$ as domain and with values in $L^2$, $P$ is continuous, otherwise, it is unbounded as usual.

What it is true in the statement of your professor, is that the set of $C^1$ functions on $[0,2\pi]$ with $f(0)= f(2\pi)$ is included in $H^1((0,2\pi))$.

  • $\begingroup$ I did not think these so technical issues were so popular... $\endgroup$ – Valter Moretti Feb 3 '14 at 17:19
  • $\begingroup$ Thank you very much! I miss some math here but I think I understood. This is really useful and interesting to me! $\endgroup$ – LG06 Feb 3 '14 at 17:38
  • $\begingroup$ I am puzzled. Is the last H^1 supposedly the same as the first mentioned one? Then I think this cant be true. A bounded operator has to have a bounded spectrum. (With periodic boundary condition I assume point spectrum, aka the Eigenvalues have to be bounded). But Eigenvalues dont change if you change the scalar product! So changing a scalar product alone can never make an operator bounded (while keeping the self-adjoint property). Could you elaborate? $\endgroup$ – lalala Aug 8 '17 at 12:25
  • 1
    $\begingroup$ What I wrote is that $P: H^1 \to L^2$ (not $H^1$) is bounded. This is not an operator defined over a unique Hilbert space so you cannot apply the standard results of spectral theory. $\endgroup$ – Valter Moretti Aug 8 '17 at 12:40
  • $\begingroup$ Indeed it was not clear in my answer, I will improve it. $\endgroup$ – Valter Moretti Aug 8 '17 at 12:45

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