📂Quantum Mechanics

Hermitian Operator

Definition

An operator $A$ is called a Hermitian operator if it satisfies the following equation.

$$A = A^{\dagger}$$

Here, $A^{\dagger}$ is the conjugate transpose of $A$.

Explanation

$A^{\dagger}$ is read as [A dagger], and a dagger means a small knife.

It is named after the French mathematician Hermite. In English, it is called a Hermitian operator.

All operators in quantum mechanics are Hermitian operators.

The notation for the complex conjugate in mathematics is $\overline{a+ib}$, and the notation for the conjugate transpose is $A^{\ast}$. In physics, the meaning of $\ast$ is often limited to the complex conjugate, and it is also expressed as $A^{\dagger} = (A^{T})^{\ast}$. However, considering the Dirac notation, it can be understood that $\ast$ used in physics includes both ‘conjugate + transpose’ meanings. In other words, in physics, the notation for the complex conjugate and the conjugate transpose both overlap as $\ast$. If the target with $\ast$ is a scalar, it means the complex conjugate, and if it is a matrix or vector, it means the conjugate transpose. If you only consider $\ast$ as a complex conjugate, it can be confusing when multiplying row vectors and column vectors, so remember it as follows.

$\ast =$ Conjugate $+$ Transpose $=\dagger$

Properties

1. The expectation value (eigenvalue) of a Hermitian operator is always real.

2. Two distinct eigenfunctions (eigenvectors) of a Hermitian operator are orthogonal.

3. The following equation holds for a Hermitian operator $A$.

   $$\left\langle A\psi|\phi \right\rangle = \left\langle \psi|A\phi \right\rangle$$

4. The condition for the product $AB$ of two Hermitian operators$A,B$ to be a Hermitian operator is $[A,B]=0$.

5. The following equation is always a Hermitian operator for any operator $A$.

   $$A+A^{\dagger} \\ i(A-A^{\dagger}) \\ AA^{\dagger}$$

Proof

3

For two matrices $A, B$, $(AB)^{\dagger} = B^{\dagger}A^{\dagger}$ holds,

$$\begin{align*} \langle A\psi|\phi \rangle &= \int(A\psi)^{\dagger}\phi dx \\ &= \int \psi^{\dagger} A^{\dagger} \phi dx \\ &= \langle \psi|A^{\dagger}\phi \rangle \\ &= \langle \psi|A\phi \rangle \end{align*}$$

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