Positive unbounded operator with zero not as an eigenalue

Question

I am currently doing Quantum Mechanics and I am supposed to show that zero is an eigenvalue of a positive operator. I have no knowledge of Functional Analysis at that kind of level, so I was wondering whether there is somebody who has some knowledge about these kind of operators and could for example list a few properties that these kind of operators have, so that I could find a contradiction? This operator is given by $N=A^*A$, where $A$ is another unbounded operator.

Answer 1

The question seems to be, "Show that zero can be a nontrivial eigenvalue of a positive unbounded operator." To do this, modify Martin Argerami's answer: Choose an orthonormal basis $\{e_n\}_{n=1}^\infty$ of your Hilbert space and define $Ne_n = (n-1)e_n$. This in particular has $Ne_1 = 0 = 0e_1$, so $0$ is a nontrivial eigenvalue.

(This operator $N$ is positive its eigenvalues are positive. Also, $N = A^*A$ for the operator $A$ defined by $Ae_n = \sqrt{n-1}e_n$.)

This is an important observation because positive bounded operators cannot have zero as a nontrivial eigenvalue. Being positive for a bounded symmetric operator $T$ is the same as saying that for any nonzero $x$, $\langle Tx,x\rangle > 0$. In particular, if $x$ is a $\lambda$-eigenvector of $T$, then $$0 < \langle Tx,x\rangle = \lambda\|x\|^2,$$ so $\lambda > 0$.

Answer 2

Having zero as an eigenvalue is a synonym of having non-trivial kernel. But you can easily have a positive operator that is injective, i.e. such that zero is not an eigenvalue. For example, fix an orthonormal basis $\{e_n\}$ and let $Ne_n=ne_n$.

Answer 3

For $A$ as you state, if $A^{\star}Af=0$ for some non-zero $f$ in the space, then $$ 0 = (A^{\star}Af,f) = (Af,Af)=\|Af\|^{2}, $$ implies that $Af=0$. So the null space of $A^{\star}A$ is the same as the null space of $A$. This has a definite effect on the ground state solutions. For example, consider $H=-\frac{d^{2}}{dx^{2}}$ on the domain consisting of twice continuously differentiable functions $f$ on $[0,2\pi]$ with periodic function and first derivative. This operator is $A^{\star}A$ where $A=-i\frac{d}{dx}$ is defined on the domain of continuously differentiable periodic functions on $[0,2\pi]$. The eigenvalues of $H$ are $0,1^{2},2^{2},3^{2},\ldots$. For $\lambda = n^{2}$ for $n > 0$, there are two linearly-independent eigenfunctions. However, for $n=0$, the ground state solution is reduced to the null space of $A$, which is a lower-order differential operator. The constant function is, up to a multiplicative constant, the only eigenfunction for eigenvalue $0$.