If $\psi$ is an eigenvector of $\Delta$, is $|\psi|$ an eigenvector as well? I tried to make the title as succinct as possible, now let me explain the motivation for this.
In a beautiful paper The Stability Of Matter: From Atoms To Start by Elliott Lieb it is proved (in Part IV) that the energy eigenvector with the lowest possible eigenvalue is always symmetric with respect to any permutation of electrons. The crucial observation is that $\|\nabla\psi\|^2$ doesn't change when we swap $\psi$ with $|\psi|$. Here, $\psi \in L^2(\mathbb R^n)$ which we may assume to be real, and the energy operator is (omitting the physical constants) $H = -\frac{1}{2} \Delta + V(x_1,\dots,x_n)$. However, I feel that some domain difficulties may arise.
I suppose the sketch of this idea goes like this: for $\psi$ to be in $L^2$, it should tend towards zero sufficiently fast when the argument goes to infinity, so by integration by parts
$$\langle \psi |\Delta| \psi \rangle = \int\limits_{\mathbb{R}^n}(\psi\cdot\Delta\psi) dx = \lim\limits_{r \rightarrow \infty} \int\limits_{B(r)}(\psi\cdot\Delta\psi) dx = \\
= \lim\limits_{r \rightarrow \infty} \left[ \int\limits_{S(r)}\psi(\nabla\psi\cdot n) d\omega - 
 \int\limits_{B(r)}\|\nabla\psi\|^2dx \right] = -\int\limits_{\mathbb{R}^n}\|\nabla\psi\|^2dx$$
Here, $B(r)$ is a ball of radius $r$ centered in the origin, $S(r)$ is the corresponding sphere.
I am not completely sure about the term that integrates over $S(r)$, but the idea is that it should tend to zero.
Now, since $\|\nabla(|\psi|)\|^2$ is the same as $\|\nabla\psi\|^2$ almost everywhere (assuming that $\nabla \psi = 0$ happens on a set of measure 0), this completes the proof.
I have several concerns about this argument:


*

*While it is obvious that $\psi \in L^2$ implies $|\psi| \in L^2$, why is it true that $\psi \in D(\Delta)$ (domain of $\Delta$) implies $|\psi| \in D(\Delta)$?

*How to prove the equations above rigorously, probably with some more assumptions on $\psi$?

*How to prove that $\int\|\nabla \psi\|^2 dx$ doesn't change upon taking the absolute value, considering that $\nabla |\psi|$ becomes undefined in points where $\psi=0$?


The last question can be traced to a simple one-dimensional example. In $L^2(0,\pi)$ the function $\sin(2x)$ is an eigenvector of $\frac{d^2}{dx^2}$ (with eigenvalue $-4$). Is $|\sin(2x)|$ an eigenvector as well?
 A: The general answer to the question title is no and you should be able to verify that yourself on your own example.
However, for the laplacian the eigenvector with the smallest eigenvalue, and only that one, is everywhere positive (or negative) and therefore equal to its absolute value up to sign. I'm not sure whether the remaining questions and calculations in your question relate to this fact. 
Edit: see fact 7 in this review for my claim about the first eigenfunction arxiv.org/pdf/1206.1278.pdf
A: *

*As already noted by Bananach, in general the absolute value of an eigenfunctions is not an eigenfunction. In fact, since eigenfunctions of a self-adjoint operator to different eigenvalues are orthogonal, this can happen for at most one eigenfunction.

*The Laplacian on $L^2(\mathbb{R}^n)$ does not have any eigenfunctions, as can be easily seen under Fourier transform. The Schrödinger operator $-\Delta+V$ may or may not have eigenfunctions, depending on the potential $V$, but this is not important for the argument. Instead, Lieb uses the (Courant-Fischer) min-max principle, that asserts
$$
\inf \sigma(H)=\inf_{\psi\in D(h)}\frac{h(\psi,\psi)}{\|\psi\|^2},
$$
where $h$ is the sesquilinear form associated with the lower bounded self-adjoint operator $H$.

*In general, $\psi\in D(\Delta)$ does not imply $|\psi|\in D(\Delta)$. As an example, take $\psi\in H^2(\mathbb{R})$ that looks like $x$ around the origin. Then $|\psi(x)|=|x|$ around the origin, which is not in $H^2(\mathbb{R})$. However, $\psi\in H^1(\mathbb{R}^n)$ does imply $|\psi|\in H^1(\mathbb{R}^n)$ since $\nabla |\psi|=(\operatorname{sgn} \psi)\cdot \nabla \psi$ (in the sense of weak derivatives).

*When talking about unbounded operators, you always have to be careful with the domain. In many cases, the most elegant way to define $H=-\Delta+V$ is to declare it as the generator of the sesquilinear form $h$ with 
$$D(h)=H^1(\mathbb{R}^n)\cap\{\psi\in L^2\mid V|\psi|^2\in L^1\}, 
h(\phi,\psi)=\int \nabla \phi\cdot \nabla\psi+\int V \bar\phi \psi.$$
This makes the identity
$$
\langle \phi,H \psi\rangle=\int \nabla \phi\cdot \nabla\psi+\int V \bar\phi \psi
$$
true for all $\phi,\psi\in D(H)$. Moreover, this makes it clear why we only need $|\psi|\in H^1$ for the min-max principle, which we have by 3.

*If you want to check $-\langle \psi,\Delta \psi\rangle=\int |\nabla \psi|^2$ for $\psi\in D(\Delta)=H^2(\mathbb{R}^n)$, it is probably the best to Fourier transform it. Once you have shown that $\mathcal{F}(\Delta \psi)(\xi)=-|\xi|^2\hat \psi(\xi)$ for $\psi\in H^2$ and $\mathcal{F}(\partial_j \psi)(\xi)=-i\xi_j \hat\psi(\xi)$ for $\psi\in H^1$, the identity is trivial.

A: 1.Let's say the boundary conditions are $\lim_{||x||\to \infty} \psi(x) =0$, as you have already suggested. Rigorously 
$$ D(H) = L^2(\mathbb R^n) \cap C^2(\mathbb R^n) .$$
Suppose $\psi$ is an eigenfunction of H, i.e. there exists a $\lambda$ scalar s.t. $H(x) \psi(x) = \lambda \psi(x)\ \ \forall x$. So $\psi$ has to be twice differentiable. Since $|\psi|$ fails to be differentiable at some points, rigorously it cannot be an eigenvector. 
But: let's look at $\psi^+:= \max (0, \psi)$ and $\psi^-:=\max (0, -\psi)$. Then $\psi = \psi^+ - \psi^-$ and $|\psi| = \psi^++\psi^-$. H is linear, hence the 0 function is also in $D(H)$ and $D(H)$ is a linear space. So technically $H(x)\psi^{\pm}(x) = \lambda \psi^{\pm}(x)$ whenever x is s.t. $\psi(x)\ne 0$. So due to linearity $H|\psi(x)| = \lambda |\psi(x)|$ whenever x is s.t. $\psi(x)$ is not 0. Rigorously $|\psi|$ is not in $D(H)$, but still it can satisfy the eigenvalue eqn on a big set. 


*The surface are of $S(r)$ scales with $r^{n-1} $ so you will need that $r^{n-1} ||\psi (x)\nabla \psi(x)||\to 0 $ as $r\to \infty$ for $||x|| =r$. 

*Let's say $x$ is such that $\nabla| \psi(x)|$ is well defined. Then $ || \nabla |\psi(x)| || = || \pm \nabla \psi(x) || = || \nabla \psi(x) ||$. Assuming that the measure of the set of x's where $\nabla |\psi|$ is not defined you get that $\int || \nabla \psi||^2 = \int ||\nabla |\psi| ||^2$. 
Hope this helps.
