- #1
Mentz114
- 5,432
- 292
This matrix, I had hoped was a good candidate for (a representation of ) a unitary, self-adjoint operator
\begin{align*}
\hat{A}&= \frac{1}{D} \left[ \begin{array}{cc}
a^2 & iy^2 \\\
-iy^2 & a^2\end{array} \right]
\end{align*}
##a## and ##y## are real with ##D^2=a^2-y^2\ >\ 0## . ##\hat{A}## has determinant ##1##, eigenvalues ##(a-y)/D,\ (a+y)/D## and eigenvectors ##\vec{e}_0=(1,i),\ \vec{e}_1=(1,-i)##. Also ##\hat{A}^{\dagger}=\hat{A}^{T}=adjoint(\hat{A})=\hat{A}^{-1}##.
When the unit vector ##(\cos(\theta),\sin(\theta))## is acted on by ##\hat{A}## its length changes to ##(a^2+y^2)/(a^2-y^2)##. This is unexpected because the determinant of ##\hat{A}=1##.
Is my expectation wrong, or is there some flaw in ##\hat{A}## that I have missed ?
\begin{align*}
\hat{A}&= \frac{1}{D} \left[ \begin{array}{cc}
a^2 & iy^2 \\\
-iy^2 & a^2\end{array} \right]
\end{align*}
##a## and ##y## are real with ##D^2=a^2-y^2\ >\ 0## . ##\hat{A}## has determinant ##1##, eigenvalues ##(a-y)/D,\ (a+y)/D## and eigenvectors ##\vec{e}_0=(1,i),\ \vec{e}_1=(1,-i)##. Also ##\hat{A}^{\dagger}=\hat{A}^{T}=adjoint(\hat{A})=\hat{A}^{-1}##.
When the unit vector ##(\cos(\theta),\sin(\theta))## is acted on by ##\hat{A}## its length changes to ##(a^2+y^2)/(a^2-y^2)##. This is unexpected because the determinant of ##\hat{A}=1##.
Is my expectation wrong, or is there some flaw in ##\hat{A}## that I have missed ?