I'm reading a book and there is an argument that the ground state of a Schrödinger operator is unique. The problem is I think the argument is complete non-sense! These are lecture notes by Witten, I will mark the places I have comments with numbers in sup-scripts.
Here our Schrödinger operator is an operator $H$ on $L^2(\mathbb R)$ that is of the form $-\Delta + U(x)$, where $U(x) \to \infty$ as $|x| \to \infty$. Consider for any $f \in L^2(\mathbb R)$ $^{[1]}$: $$(f,Hf) = \int_{-\infty}^\infty |f'(x)|^2 +U(x) |f(x)|^2$$ The ground state is the global minimum of this energy functional on the sphere $\|f\|=1$.$^{[2]}$ Let $f$ be a ground state with energy $E_0$, then $f$ has constant sign:
Suppose that $f$ changes sign at some point $x_0$, since $f$ satisfies the Euler-Lagrange equation $H f= E_0$ we have $f'(x_0)\neq0$ $^{[3]}$. Consider the function $|f|$. It is clear that $(|f|,H|f|)=E_0$ but $|f|$ is not smooth, so it doesn't satisfy the Euler-Lagrange equation $H\psi = E_0 \psi$ and thus cannot be a minimum, $^{[4]}$ so the smallest value of $(\psi,H\psi)$ on the sphere is smaller than $E_0$ which is a contradiction.
Uniqueness follows easily from that consideration, two groundstates must have constant sign so $(f_1,f_2)\neq0$ $^{[5]}$ whenever $f_1, f_2$ are grounds states. But in any subspace that is not one dimensional we can find two orthogonal vectors, thus arriving at a contradiction.
My problems:
Strictly speaking $H$ is defined on a subspace of $L^2(\mathbb R)$, not on the entire space. It doesn't seem like as if it would be a problem to restrict $H$ onto just the Schwartz-space (if $U$ has polynomial growth), but is it really ok?
Why is it clear that the ground state exists from this consideration?
If $f$ satisfies the Euler Lagrange equation (which is $f''(x)=U(x)f(x)$ here), then $f'(x)$ cannot be zero unless $f$ itself is zero everywhere, that is true. But we are looking for minima of $(f,Hf)$ in $L^2(\mathbb R)$, the functions that satisfy EL definitely do not need to be square normalisable. Why can we assume that $f$ satisfies EL?
If you for example look at the harmonic oscillator, you have $U(x)=x^2$ and the ground state looks like $e^{-c x^2}$ for some $c$. If you check you find that this does not solve the EL equations.
Again, why do the EL equations have to be satisfied for the minimum? How is this not non-sense?
What if $f_1, f_2$ are localised at different locations? Ie if $f_1=0$ in the region where $f_2 \neq 0$. The EL equations forbid $f$ being zero in an open region, ok. But we have seen that the ground state does not actually need to satisfy the EL equations!
Is the proof wrong or am I missing essential details?
I would be happy if somebody could point me in the direction of how to fix my problems.
I'm not exactly sure what level of rigour you're aiming for; the problem may at least in part be that you're expecting more rigour than physicists usually care for; but here are two (possibly related) points at which I don't follow your objections: You write that the Euler-Lagrange equation is $f''(x)=U(x)f(x)$ – that's missing the term $Ef(x)$, which arises if you take into account the normalisation constraint using a Lagrange multiplier (or equivalently if you divide the functional by the square of the norm). And the harmonic oscillator ground state does satisfy the Euler-Lagrange equation.