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

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