Hermittian Operators

Definition

The conjugate transpose of a matrix M $$M = \sum_n\sum_k m_{nk} \left| e_n \right> \left< e_k \right|$$ is defined as $$M^{\dagger} = \sum_n\sum_k m^*_{kn} \left| e_n \right> \left< e_k \right| .$$ If the conjugate transpose of a matrix is equal to itself, $$M^{\dagger}=M,$$ then the matrix is said to be Hermittian.

Properties of Hermittian Operators

• For any kets \(\left| \psi_1 \right> \) and \(\left| \psi_2 \right> \) $$\left< \psi_1 \right| M \left| \psi_2 \right> = \left< \psi_1 \right| M^{\dagger} \left| \psi_1 \right> = \left< \psi_2 \right| M \left| \psi_1 \right>^*,$$ by the definition of Hermittian operator.

• All eigenvalues of a Hermittian operator are real $$M \left| n \right> &= \lambda_n \left| n \right>$$ Then, $$\left< n \right| M \left| n \right> = \lambda_n$$ \begin{align} M \left| n \right> &= \lambda_n \left| n \right> \\ \left< n \right| M^{\dagger} &= \lambda^*_n \left< n \right| \\ \left< n \right| M &= \lambda^*_n \left< n \right| .\\ \end{align} Then, $$\left< n \right| M \left| n \right> = \lambda^*)_n.$$ So, $$\lambda_n = \lambda^*_n,$$ i.e. it is real.

• Eigenvectors of Hermittian operators with distinct eigenvalues are orthogonal to each other. Let $$M \left| n \right> = \lambda_n \left| n \right>$$ and $$M \left| k \right> = \lambda_k \left| k \right>,$$ where \(\lambda_n \neq \lambda_k \). Then, $$ \left< k \right| M \left| n \right> = \lambda_n \left< k \middle| n \right>$$ and \begin{align} \\ \left< n \right| M \left| k \right> &= \lambda_k \left< n \middle| k \right>\\ \left< k \right| M \left| n \right> &= \lambda^*_k \left< k \middle| n \right> .\\ \end{align} So, $$0 = (\lambda_n - \lambda_k)\left< k \middle| n \right> .$$ Since \(\lambda_n \neq \lambda_k\), we have $$\left< k \middle| n \right> = 0$$

• If the eigenvectors of the Hermittian operator have same eigenvalue, one can always construct a new set of eigenvectors in terms of linear combination of the original eigenvector set to achieve orthogonality