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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?

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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.

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  • $\begingroup$ Can $C$ ever have eigenvalues of the form $a + ib$, where $a, b \in \mathbb{R}$ and $a, b \neq 0$? $\endgroup$ Sep 15 at 8:15
  • $\begingroup$ That is the same as asking if $C$ has a complex eigenvalue $z$ with $Re(z)=0$ so yes. $\endgroup$
    – joseph h
    Sep 15 at 8:18
  • $\begingroup$ sir, I meant that can we have eigenvalues of the form $a + ib $ where $a,b \in \mathbb{R}$ and $a, b \neq 0$ $\endgroup$ Sep 15 at 8:23
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    $\begingroup$ 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 $\endgroup$ Sep 15 at 10:28
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    $\begingroup$ @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 $1\pm i$. $\endgroup$ Sep 15 at 14:31
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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: $$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix}\quad\wedge\quad B~=~\begin{pmatrix} 0 & b \cr b^{\ast} & 0 \end{pmatrix}\quad\Rightarrow\quad AB~=~\begin{pmatrix} b^{\ast} & 0 \cr 0 & b \end{pmatrix}. $$

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