I have taken linear algebra and have some experience with basic application to quantum mechanics. I am reading a paper in quantum mechanics and am having trouble following some of the math. Here is a link to the paper. Here I refer to equations (5) and (6) on page 412.
I am given a vector $\Psi$ and operators $A, A', B,$ and $B'$. A fifth operator is defined:
$ F = A \otimes B + A \otimes B' + A' \otimes B – A' \otimes B' $.
It is shown that $\Psi$ is an eigenvector of $F$ with eigenvalue $2 \sqrt{2}$.
Then $ F^{2} \Psi = 8 \Psi \implies i(AA' – A'A) i(BB' – B'B) \Psi = 4 \Psi $.
The first equality is trivial, but I do not understand why it implies the second. I attempted to expand $F^{2}$ but don't see a relationship with $i(AA' – A'A) i(BB' – B'B)$. Can someone help me to understand this?
I have tried asking others and searching online, but I have failed to find adequate search terms.
I should note that $i(AA' – A'A)$ and $i(BB' – B'B)$ are referred to in the paper as Hermitian combinations.
Best Answer
The trick is in an additional assumption between (4) and (5):
Thus as $A$,$A'$,$B$ and $B'$ are Hermitian operators we can deduce that: $$A^2=A'^2=B'^2=B^2=I$$ So: \begin{equation} (AB+AB'+A'B-A'B')^2=((A+A') B +(A-A')B')^2=(A+A')^2B^2+(A-A')^2 B'^2+(A+A')(A-A')BB'+(A-A')(A+A')B'B=(AA'+A'A+2 I)+(2I -AA'-A'A)+(A'A-AA')BB'+(AA'-A'A)B'B \end{equation} thus: $$F^2=4 I -(AA'-A'A)(BB'-B'B)$$ And: $$F^2 \psi= 8 \psi=4\psi+i(AA'-A'A)i(BB'-B'B) \psi$$