# Explain step in Galindo and Pascual (Quantum Mechanics I) proof of self-adjointness of the momentum operator in QM

In the book Quantum Mechanics (Volume I) by Galindo & Pascual, they define the domain of the QM momentum operator on the Hilbert space $$\mathcal{H}=L^2(\mathbb{R})$$ as $$\begin{equation*} D(P)=\Biggl\{\psi\in\mathcal{H}: \psi\text{ absolutely continuous,} \int_{-\infty}^\infty\!dx\,\Biggl\lvert\frac{d\psi(x)}{dx}\Biggr\rvert^2<\infty\Biggr\} \end{equation*}$$ and the momentum operator $$P$$ by $$(P\psi)(x)=-i\frac{d\psi(x)}{dx}.$$ They go on to prove that $$P$$ is densely defined and symmetric. To prove that $$P$$ is self-adjoint, they attempt to show that $$D(P^\dagger)\subseteq D(P)$$. Here are the next couple of lines of the proof:

... consider a function $$\psi\in D(P^\dagger)$$ and define $$\psi_1=P^\dagger\psi$$; then $$\langle\psi|P|\varphi\rangle=\langle\psi_1|\varphi\rangle,\quad\forall\varphi\in D(P)$$ can be rewritten as $$\begin{equation*} \begin{split} \langle\psi|P|\varphi\rangle &=\int_{-\infty}^\infty\!dx\,\psi_1^*(x)\varphi(x)\\ &=i\int_{-\infty}^\infty\!dx\Biggl[\frac{d}{dx} \Biggl(i\int_0^x\!dt\,\psi_1(t)+c\Biggr)^*\Biggr]\varphi(x), \end{split} \end{equation*}$$ where $$c$$ is an arbitrary constant. Choosing $$\varphi\in C^\infty_0$$, integrating by parts, and taking into account that $$\varphi$$ is zero outside a finite interval, we obtain $$$$\tag{2.16} \int_{-\infty}^\infty\!dx\Biggl(\psi(x)-i\int_0^x\!dt\,\psi_1(t)-c\Biggr)^* \Biggl(-i\frac{d\varphi(x)}{dx}\Biggr)=0,\quad\forall\varphi\in C^\infty_0.$$$$

[So far, this seems OK to me. It is the next statement that I don't follow:]

Since $$C^\infty_0$$ is dense in $$L^2(\mathbb{R})$$, the first factor of the integrand in (2.16) must be a constant and hence, with a convenient choice $$c_0$$ for $$c$$, we can write almost everywhere $$$$\tag{2.17} \psi(x)=c_0+i\int_0^x\!dt\,\psi_1(t),$$$$ [and it goes on from there]

I want to concentrate on the validity of going from (2.16) to (2.17). I understand that, with total lack of rigor, if we have $$\int h\varphi'=0\quad\forall\varphi\in C^\infty_0$$ we'd like to do an integration by parts and write $$\int h'\varphi=\int h\varphi'=0\quad\forall\varphi\in C^\infty_0$$ from which we would get that $$h'=0$$ almost everywhere hence $$h=c$$ almost everywhere. But I don't see how to apply that here since I don't know that $$\psi$$ is differentiable a.e. or even a.e. on a compact interval.

It even looks like a version of the DuBois-Raymond theorem from variational calculus, but I only know that for continuous functions on a compact interval, so it would seem to not apply here.

So, my questions are:

1. how do you get from (2.16) to (2.17)?

2. what element of $$L^2(\mathbb{R})$$ would they be talking about when they say that $$C^\infty_0$$ is dense in $$L^2(\mathbb{R})$$?

The fact that $$C_0^\infty$$ is dense in $$L^2$$ allows to have $$(2.16)$$ but with any function from $$L^2$$ in place of the second bracket (that is, given any $$g\in L^2$$ there is a $$\varphi\in C_0^\infty$$ such that $$-i\varphi'$$ is as close to $$g$$ as you want).

In particular you can use the function $$g= \psi(x)-i\int_0^x\!dt\,\psi_1(t)-c$$, and you get $$\int_{-\infty}^{\infty}|g|^2=0,$$ so $$g=0$$. That is, $$\psi(x)-i\int_0^x\!dt\,\psi_1(t)-c=0,$$ which is $$(2.17)$$.

• $\psi$ is certainly in $L^2$, but why is $\psi(x)-i\int_0^x dt\,\psi_1(t)-c$ in $L^2$? And how do you get $\lvert\lvert-i\varphi'-g\rvert\rvert_2$ to go to zero? May 15, 2020 at 23:15

I found an answer in this post: https://math.stackexchange.com/a/428744/527829 I'll convert it to the current situation.$$\newcommand{\loc}{\mathop{\rm loc}}$$

Let $$\varphi\in C^\infty_0(\mathbb{R})$$. Choose any $$\varphi_1\in C^\infty_0(\mathbb{R})$$ such that $$\int_{-\infty}^\infty\!\varphi_1=1$$. Let $$[a,b]$$ contain the supports of both $$\varphi$$ and $$\varphi_1$$. Put $$\alpha=\int_{-\infty}^\infty\!\varphi$$ and $$\varphi_0(x)=\int_a^x\!(\varphi(t)-\alpha\varphi_1(t))\,dt$$. Then $$\varphi_0'(x)=\varphi(x)-\alpha\varphi_1(x)$$ for all $$x\in\mathbb{R}$$, so $$\varphi_0\in C^\infty(\mathbb{R})$$. For $$t\leq a$$, $$\varphi(t)-\alpha\varphi_1(t)=0$$, so for $$x\leq a$$, $$\varphi_0(x)=0$$. For $$x\geq b$$, $$\varphi_0(x)=\varphi_0(b)=\int_a^b\!\varphi-\Biggl(\int_{-\infty}^\infty\!\varphi\Biggr)\Biggl(\int_a^b\!\varphi_1\Biggr)=0,$$ so $$\varphi_0\in C^\infty_0(\mathbb{R})$$.

Let $$u\in L^1_{\loc}(\mathbb{R})$$ such that $$\int_{-\infty}^\infty\!u\varphi'=0$$ for all $$\varphi\in C^\infty_0(\mathbb{R})$$. Put $$c=\int_{-\infty}^\infty\!u\varphi_1$$ and $$\tilde{u}=u-c\in L^1_{\loc}(\mathbb{R})$$. In the following, all the limits of integration are from $$-\infty$$ to $$\infty$$: $$\begin{equation*} \begin{split} \int\!\tilde{u}\varphi &=\int\!(u-c)(\varphi_0'+\alpha\varphi_1) =\int\!u\varphi_0'-c\int\!\varphi_0'+\alpha\int\!u\varphi_1-\alpha c\int\varphi_1\\ &=0-c\cdot(0-0)+\alpha\int\!u\varphi_1-\alpha\biggl(\int\!u\varphi_1\biggr)\cdot 1 =0.\qquad\qquad\qquad\qquad\qquad\qquad(1) \end{split} \end{equation*}$$ Since $$\tilde{u}\in L^1_{\loc}(\mathbb{R})$$ and (1) holds for all $$\varphi\in C^\infty_0(\mathbb{R})$$, we have (for example by Lang Real and Functional Analysis Chapter VI Corollary 9.5) that $$u-c=\tilde{u}=0$$ almost everywhere.

Apply this in (2.16) with $$u(x)=\Biggl(\psi(x)-i\int_0^x\!dt\,\psi_1(t)-c\Biggr)^*.$$