Commutators and Hermiticity - Exam question I'm doing old exam questions, and here is one that on first glance seemed rather simple to me, but I just can't get it:
Given two operators $A$ and $B$, and all we know about them is that
$$[A,B] = B$$
and
$$B^\dagger B = 1 - A^2$$
From this, I must find the "hermiticity properties" of $A$ and $B$. So far, the only progress I have made is to note that $A^2$ must be hermitian, because for any operator, $B^\dagger B$ is hermitian.
However, this does not suffice to determine the hermiticity of $A$ itself, as $A$ could be hermitian, anti-hermitian or even have one hermitian part $A_h^\dagger = A_h$ and one anti-hermitian part $A_a^\dagger = -A_a$ as long as $A_h A_a = -A_a A_h$, i.e. the hermitian and anti-hermitian part of $A$ anti-commute.
I am sure that there must be a really simple trick, but I just don't get it.
 A: The left hand side of the second equation is Hermitian, and therefore so is the right hand side. Therefore $A^2$ is Hermitian. Its eigenvalues are real. It follows that the eigenvalues of $A$ are either real or pure imaginary.
More importantly, the first commutator is known from raising and lowering operators of the $SU(2)$ algebra,
$$ A = J_z, \quad B = k J_+ $$
is a representation of the commutator. By the way, the second equation is also very similar to what $J_\pm, J_z$ satisfy except for the wrong coefficient in front of $A^2$.
Now, let me prove that $A$ actually has no nonzero pure imaginary eigenvalues. Take an eigenstate $\psi$ of $A$ with eigenvalue $a$. The first commutator implies that $B\psi$ is also an eigenstate of $A$ with eigenvalue $a+1$. The commutator means that $B$ carries the $A$-charge equal to one.
However, if $a$ is nonzero and pure imaginary, $a+1$ is neither pure imaginary nor real. It follows that the only way how to satisfy it is to have $B\psi = 0$ which is a trivial eigenstate $A$ with any eigenvalue. However, it is not possible to have $B\psi=0$ because of the second condition: the expectation value of $B^\dagger B$ in the state $\psi$ is the same as the expectation value of $1-A^2$. The former is zero because $B\psi=0$. However, the latter is not zero because the eigenvalue of $A^2$ in $\psi$ is negative, so $1-A^2$ can't vanish.
So we may assume that $A$ only has real eigenvalues. If the eigenvalues of $A$ are nonzero, it's clear that $A$ can't have any non-trivial Jordan blocks with these nonzero eigenvalues because $A^2$ would have them as well so it couldn't be Hermitian. So all the nonzero-eigenvalue portion of $A$ has to be diagonalizable. We've almost proved that $A$ is Hermitian. However, there's a subtle problem.
What about the vanishing eigenvalues of $A$? Well, I clearly can't prove that $A$ isn't allowed to have the nilpotent $2\times 2$ blocks. For example, we may set $B=0$ and $A$ equal to a nilpotent nonzero $2\times 2$ matrix, $((0,1),(0,0))$, and all equations are going to be satisfied. This $A$ is not Hermitian. So the allowed $2\times 2$ nontrivial nilpotent Jordan blocks with a vanishing eigenvalue are the only obstacle in proving that $A$ is Hermitian.
A: You can say a bit more if you for instance assume that the Hilbert space $H$ is a finite dimensional complex vector space.

*

*Spectrum. As Lubos Motl observes, $A^2=1-B^{\dagger}B$ must be Hermitian. In fact, the spectrum $\mathrm{Spec}(A^2)$ of $A^2$ must satisfy
$$ \mathrm{Spec}(A^2) ~\subseteq~ ]-\infty,1], $$
and therefore the spectrum $\mathrm{Spec}(A)$ of $A$ must satisfy
$$ \mathrm{Spec}(A) ~\subseteq~ [-1,1] \cup i\mathbb{R}. $$
As Lubos Motl explains, the equation $[A,B]=B$ implies one of two alternatives:
$$ \lambda\in\mathrm{Spec}(A) \qquad \Leftrightarrow \qquad\exists v\in H\backslash\{0\}: (A-\lambda)v=0  $$
$$\Rightarrow\qquad(A-\lambda-1)Bv=0 \qquad\Rightarrow\qquad \lambda+1\in\mathrm{Spec}(A) \qquad\mathrm{or} \qquad Bv=0.$$
The second alternative $Bv=0$ implies that
$$\lambda^2v=A^2v=(1-B^{\dagger}B)v=v \qquad \Rightarrow \qquad \lambda =\pm 1.$$
This is very restrictive. We conclude that the spectrum $\mathrm{Spec}(A)$ of $A$ must satisfy
$$ \mathrm{Spec}(A) ~\subseteq~ \{-1,0,1\}. $$


*We can write $A$ on Jordan canonical block form by a choice of basis. Note that the basis is not necessarily orthonormal. The diagonal elements satisfy $A_{ii}\in \{-1,0,1\}$. Now square the Jordan canonical block form to get $A^2$ on triangular form. The diagonal elements satisfy $(A^2)_{ii}\in \{0,1\}$. From the characteristic polynomial for $A^2$,
$$p_{A^2}(\lambda)=\det(A^2-\lambda)= \prod_i ((A^2)_{ii}-\lambda),$$
we deduce that
$$ \mathrm{Spec}(A^2) ~\subseteq~ \{0,1\}. $$


*Define eigenspaces
$$H_0 :=\mathrm{ker}(A^2)=\mathrm{ker}(B^{\dagger}B-1),$$
and
$$H_1 :=\mathrm{ker}(A^2-1)=\mathrm{ker}(B^{\dagger}B)=\mathrm{ker}(B),$$
each of which may be zero. (The equality $\mathrm{ker}(B^{\dagger}B)=\mathrm{ker}(B)$ can be deduced from, e.g., polar decomposition of $B$.) Since $A^2$ is Hermitian, we have that $H=H_0\oplus H_1$ is an orthogonal decomposition of $H$. Moreover, the operator $A^2$ is the orthogonal projection on the subspace $H_1$, and $B^{\dagger}B$ is the orthogonal projection on the subspace $H_0$.


*Next, the equation $[A,B]=B$ implies that $A\mathrm{ker}(B) \subseteq\mathrm{ker}(B)$, i.e., that $AH_1 \subseteq H_1$. Furthermore, $A\mathrm{ker}(A^2)\subseteq\mathrm{ker}(A)\subseteq\mathrm{ker}(A^2)$, i.e., that $AH_0 \subseteq H_0$. So $A$ is stable under the decomposition $H=H_0\oplus H_1$. Let us introduce operators
$$A_1 := A |_{H_{1}} \qquad \mathrm{and} \qquad A_0 := A |_{H_{0}}, $$
with spectra
$$ \mathrm{Spec}(A_1) ~\subseteq~ \{\pm 1\} \qquad \mathrm{and} \qquad \mathrm{Spec}(A_0) ~\subseteq~ \{0\},$$
respectively. It follows from the fact that
$$ (A_1\pm 1)^2 = \underbrace{(A_1^2-1)}_{=0} \pm 2(A_1\pm 1), $$
that the series of generalized eigenspaces are just ordinary eigenspaces
$$\mathrm{ker}(A\pm 1)=\mathrm{ker}(A_1\pm 1)\subseteq\mathrm{ker}(A_1\pm 1)^2\subseteq\mathrm{ker}(A_1\pm 1)^3\subseteq\ldots =\mathrm{ker}(A_1\pm 1). $$
In other words, the operator $A_1:H_1\to H_1$ is diagonalizable (but not necessarily in an orthonormal basis),
$$ H_1 = \mathrm{ker}(A+1)+\mathrm{ker}(A-1). $$
Similarly, it follows from the fact that $A_0^2 = 0$, that the series
$$\mathrm{ker}(A)=\mathrm{ker}(A_0)\subseteq\mathrm{ker}(A_0^2)\subseteq\mathrm{ker}(A_0^3)\subseteq\ldots =\mathrm{ker}(A_0^2)=H_0 $$
stabilizes after at most two steps.


*Next, the equation $[A,B]=B$ implies after a straightforward calculation that $B(H_0) \subseteq H_1$, and hence $B^2=0$ nilpotent. In fact, a refined argument yields that $B(H_0) \subseteq \mathrm{ker}(A-1)$. Therefore $BA_0=0$, and since $B|_{H_{0}}$ is injective on $H_0$, we finally conclude that
$$A_0 = 0\qquad \Leftrightarrow \qquad H_0=\mathrm{ker}(A).$$
In other words, the operator $A:H\to H$ is diagonalizable (but not necessarily in an orthonormal basis),
$$ H = (\mathrm{ker}(A-1)+\mathrm{ker}(A+1))\oplus\mathrm{ker}(A). $$


*If we choose ortonormal bases for $\mathrm{ker}(A-1)$ and $\mathrm{ker}(A)$, and then extend with an ortonormal basis for
$$H'_{-1}:= \left( \mathrm{ker}(A-1)\oplus\mathrm{ker}(A) \right)^{\perp}, $$
it is not hard to see that the operators $A$ and $B$ are of the block form
$$A= \left[\begin{array}{ccc}1&T&0 \\ 0&-1&0 \\ 0&0&0  \end{array}\right], \qquad B= \left[\begin{array}{ccc}0&0&U \\ 0&0&0 \\ 0&0&0  \end{array}\right], \qquad U^{\dagger}U=1,$$
where the row and columns are ordered as $\mathrm{ker}(A-1)\oplus H'_{-1}\oplus\mathrm{ker}(A)$.


*Two special cases:
$$A= \left[\begin{array}{cc}1&T \\ 0&-1 \end{array}\right], \qquad B= \left[\begin{array}{cc}0&0 \\ 0&0 \end{array}\right].$$
$$A= \left[\begin{array}{cc}1&0 \\ 0&0 \end{array}\right], \qquad B= \left[\begin{array}{cc}0&U \\ 0&0 \end{array}\right], \qquad U^{\dagger}U=1.$$
