# Eigenvalues of Product of 2 hermitian operators [closed]

Let $$A$$ and $$B$$ be two Hermitian operators. Let $$C$$ be another operator such that $$C = AB$$. What can we say about Eigenvalues of $$C$$? Will they be real/imaginary/complex? What I did was to search for examples. The following were examples (in matrix representation) I looked for: $$A = \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$ and $$B = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$ to get a hermitian matrix and so real eigenvalues.
Next I tried:
$$A = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}$$ and $$B = \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$ to get Anti-Hermitian matrix and so imaginary eigenvalues.
Is there a more concrete way of solving this? Can we have a general complex number as eigenvalues for the product of the Hermitian Matrices?

In general, we can say that $$C=AB$$ will have real, imaginary and complex eigenvalues (complex of the form $$z=a+ib$$ where and $$\{a,b\in \mathbb{R}\mid a,b \ne 0\}$$ as shown in the comments by Mark and Qmechanic's answer). For example, if $$A=\begin{bmatrix} 0 &1 \\ 1& 0 \end{bmatrix}\ \ \text{and}\ \ B=\begin{bmatrix} 1 & 0\\ 0& -1 \end{bmatrix}$$ where $$AB=\begin{bmatrix} 0 &-1 \\ 1& 0 \end{bmatrix}$$ will not have real, but imaginary eigenvalues.

However, one thing we can say is that if $$A$$ and $$B$$ commute then $$C=AB$$ will always have real eigenvalues, since the eigenvalues of all Hermitian operators are real.

So if $$C=AB$$ then $$C^\dagger =(AB)^\dagger =B^\dagger A^\dagger =BA$$ since $$A$$ and $$B$$ are Hermitian, and clearly $$C^\dagger =C$$ if $$[A,B]=AB-BA=0\rightarrow AB=BA$$ This means that $$C^\dagger =C$$ only if $$A$$ and $$B$$ commute in which case $$C$$ will have real eigenvalues.

• Can $C$ ever have eigenvalues of the form $a + ib$, where $a, b \in \mathbb{R}$ and $a, b \neq 0$? Sep 15 at 8:15
• That is the same as asking if $C$ has a complex eigenvalue $z$ with $Re(z)=0$ so yes. Sep 15 at 8:18
• sir, I meant that can we have eigenvalues of the form $a + ib$ where $a,b \in \mathbb{R}$ and $a, b \neq 0$ Sep 15 at 8:23
• I'd like to add in @josephh ' s answer that if anti-commutator of $A$ and $B$ is 0, then C will be anti-hermitian and thus will have purely imaginary eigenvalues Sep 15 at 10:28
• @josephh It's not true that the eigenvalues are either real or imaginary. The matrix $AB$ generally has both Hermitian and anti-Hermitian components, so the eigenvalues can be any complex number. Consider for example $A = \sigma_z$ and $B=\sigma_z + \sigma_x$, then $AB = \mathbb{1} + i \sigma_y$, where $\sigma_{x,y,z}$ are Pauli matrices. The eigenvalues of $AB$ are $1\pm i$. Sep 15 at 14:31

TL;DR: Assuming that $$A,B$$ are self-adjoint, the product $$AB$$ does not need to be diagonalizable. And if $$AB$$ is diagonalizable, the eigenvalues need not be real or imaginary.

Example 1: $$AB$$ is not diagonalizable: $$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix} \quad\wedge\quad B~=~\begin{pmatrix} 1 & 0 \cr 0 & 0 \end{pmatrix}\quad\Rightarrow\quad AB~=~\begin{pmatrix} 0 & 0 \cr 1 & 0 \end{pmatrix}.$$

Example 2: $$AB$$ has complex eigenvalues: $$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix}\quad\wedge\quad B~=~\begin{pmatrix} 0 & b \cr b^{\ast} & 0 \end{pmatrix}\quad\Rightarrow\quad AB~=~\begin{pmatrix} b^{\ast} & 0 \cr 0 & b \end{pmatrix}.$$