# Normal operators

References: edX online course MIT 8.05 Week 6.

Hermitian and unitary operators are actually special cases of a more general type of operators known as normal operators. An operator is normal if it commutes with its adjoint:

$\displaystyle \left[M^{\dagger},M\right]=0 \ \ \ \ \ (1)$

A hermitian operator is normal because ${M^{\dagger}=M}$. A unitary operator ${U}$ is normal because ${U^{\dagger}=U^{-1}}$ and every operator commutes with its inverse.

Theorem 1 A normal operator ${M}$ remains normal after a similarity transformation with a unitary operator ${U}$.

Proof: Suppose we transform ${M}$ according to

$\displaystyle M_{1}=U^{-1}MU=U^{\dagger}MU \ \ \ \ \ (2)$

Then

 $\displaystyle \left[M_{1}^{\dagger},M_{1}\right]$ $\displaystyle =$ $\displaystyle \left(U^{\dagger}M^{\dagger}U\right)\left(U^{\dagger}MU\right)-\left(U^{\dagger}MU\right)\left(U^{\dagger}M^{\dagger}U\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{\dagger}M^{\dagger}MU-U^{\dagger}MM^{\dagger}U\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{\dagger}\left[M^{\dagger},M\right]U\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (6)$

Thus ${M_{1}}$ is normal.$\Box$

Theorem 2 If ${w}$ is an eigenvector of the normal operator ${M}$ with eigenvalue ${\lambda}$, it is also an eigenvector of the adjoint ${M^{\dagger}}$ with eigenvalue ${\lambda^*}$.

Proof: Suppose

 $\displaystyle Mw$ $\displaystyle =$ $\displaystyle \lambda w\ \ \ \ \ (7)$ $\displaystyle \left(M-\lambda I\right)w$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (8)$

Consider the vector

$\displaystyle u\equiv\left(M^{\dagger}-\lambda^*I\right)w \ \ \ \ \ (9)$

The norm of this vector satisfies

 $\displaystyle \left\langle u\left|u\right.\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \left(M^{\dagger}-\lambda^*I\right)w\left|\left(M^{\dagger}-\lambda^*I\right)w\right.\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle w\left|\left(M-\lambda I\right)\left(M^{\dagger}-\lambda^*I\right)w\right.\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle w\left|\left(M^{\dagger}-\lambda^*I\right)\left(M-\lambda I\right)w\right.\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (13)$

where in the third line we used 1 and in the last line we used 8. Since the norm of ${u}$ is 0, ${u=\left|0\right\rangle }$ and since ${w\ne0}$, from 9 we have

 $\displaystyle \left(M^{\dagger}-\lambda^*I\right)w$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (14)$ $\displaystyle M^{\dagger}w$ $\displaystyle =$ $\displaystyle \lambda^*w \ \ \ \ \ (15)$

as required.$\Box$