Let $\mathcal{L}$ be a second-order self-adjoint differential operator. Then $\mathcal{L}u(x)$ may be written as

$$\mathcal{L}u(x)=\frac{d}{dx}\left[p(x)\frac{du(x)}{dx}\right]+q(x)u(x)\ \ \ \ \ (1)$$ as we discussed here. Multiply (1) by $v^\ast$ ($v^\ast$ is the complex conjugate of $v$) and integrate

\begin{align*}

\int_a^bv^\ast\mathcal{L}udx&=\int_a^bv^\ast\frac{d}{dx}\left[p(x)\frac{du(x)}{dx}\right]dx+\int_a^bv^\ast qudx\\

&=\int_a^bv^\ast d\left[p(x)\frac{du(x)}{dx}\right]+\int_a^bv^\ast qudx\\

&=v^\ast p\frac{du}{dx}|_a^b-\int_a^b {v^\ast}^\prime pu’dx+\int_a^bv^\ast qudx

\end{align*}

We may impose

$$v^\ast p\frac{du}{dx}|_a^b=0\ \ \ \ \ (2)$$

as a boundary condition.

\begin{align*}

-\int_a^b {v^\ast}^\prime pu’dx&=-\int_a^b {v^\ast}^\prime pdu\\

&=-{v^\ast}^\prime pu|_a^b+\int_a^b u(p{v^\ast}^\prime)’dx

\end{align*}

We may also impose

$$-{v^\ast}^\prime pu|_a^b=0\ \ \ \ \ (3)$$

as a boundary condition. Then

\begin{align*}

\int_a^bv^\ast\mathcal{L}udx&=\int_a^b u(p{v^\ast}^\prime)’dx+\int_a^bv^\ast qudx\\

&=\int_a^b u\mathcal{L}v^\ast dx

\end{align*}

*Definition*. A self-adjoint operator $\mathcal{L}$ is called a *Hermitian operator* with respect to the functions $u(x)$ and $v(x)$ if

$$\int_a^bv^\ast\mathcal{L}udx=\int_a^b u\mathcal{L}v^\ast dx\ \ \ \ \ (4)$$

That is, a self-adjoint operator $\mathcal{L}$ which satisfies the boundary conditions (2) and (3) is a Hermitian operator.

**Hermitian Operators in Quantum Mechanics**

In quantum mechanics, the differential operators need to be neither second-order nor real. For example, the momentum operator is given by $\hat p=-i\hbar\frac{d}{dx}$. Therefore we need an extended notion of Hermitian operators in quantum mechanics.

*Definition*. The operator $\mathcal{L}$ is *Hermitian* if

$$\int \psi_1^\ast\mathcal{L}\psi_2 d\tau=\int(\mathcal{L}\psi_1)^\ast\psi_2 d\tau\ \ \ \ \ (5)$$

Note that (5) coincides with (4) if $\mathcal{L}$ is real. In terms of Dirac’s braket notation (5) can be written as

$$\langle\psi_1|\mathcal{L}\psi_2\rangle=\langle\mathcal{L}\psi_1|\psi_2\rangle$$

The *adjoint operator* $A^\dagger$ of an operator $A$ is defined by

$$\int \psi_1^\ast A^\dagger \psi_2 d\tau=\int(A\psi_1)^\ast\psi_2 d\tau\ \ \ \ \ (6)$$ Again in terms of Dirac’s braket notation (6) can be written as

$$\langle\psi_1|A^\dagger\psi_2\rangle=\langle A\psi_1|\psi_2\rangle$$

If $A=A^\dagger$ then $A$ is said to be *self-adjoint*. Clearly, self-adjoint operators are Hermitian operators. However the converse need not be true. Although we will not delve into this any deeper here, the difference is that Hermitian operators are always assumed to be bounded while self-adjoint operators are not necessarily restricted to be bounded. That is, bounded self-adjoint operators are Hermitian operators. Physicists don’t usually distinguish self-adjoint operators and Hermitian operators, and often they mean self-adjoint operators by Hermitian operators. In quantum mechanics, observables such as position, momentum, energy, angular momentum are represented by (Hermitian) linear operators and the measurements of observables are given by the eigenvalues of linear operators. Physical observables are regarded to be bounded and continuous, because the measurements are made in a laboratory (so bounded) and points of discontinuity are mathematical points and nothing smaller than the Planck length can be observed. As well-known any bounded linear operator defined on a Hilbert space is continuous.

**For those who are interested:** This may cause a notational confusion, but in mathematics the complex conjugate $a^\ast$ is replaced by $\bar a$ and the adjoint $a^\dagger$ is replaced by $a^\ast$. Let $\mathcal{H}$ be a Hilbert space. By the Riesz Representation Theorem, it can be shown that for any bounded linear operator $a:\mathcal{H}\longrightarrow\mathcal{H}’$, there exists uniquely a bounded linear operator $a^\ast: \mathcal{H}’\longrightarrow\mathcal{H}$ such that

$$\langle a^\ast\eta|\xi\rangle=\langle\eta|a\xi\rangle$$ for all $\xi\in\mathcal{H}$, $\eta\in\mathcal{H}’$. This $a^\ast$ is defined to be the *adjoint* of the bounded operator $a$. ${}^\ast$ defines an involution on $\mathcal{B}(\mathcal{H})$, the set of all bounded lineart operators of $\mathcal{H}$ and $\mathcal{B}(\mathcal{H})$ with ${}^\ast$ becomes a C${}^\ast$-algebra. In mathematical formulation of quantum mechanics, observables are represented by self-adjoint operators of the form $a^\ast a$, where $a\in\mathcal{B}(\mathcal{H})$. Note that $a^\ast a$ is positive i.e. its eigenvalues are non-negative.

*Definition*. The *expectation value* of an operator $\mathcal{L}$ is

$$\langle\mathcal{L}\rangle=\int \psi^\ast\mathcal{L}\psi d\tau$$

$\langle\mathcal{L}\rangle$ corresponds to the result of a measurement of the physical quantity represented by $\mathcal{L}$ when the physical system is in a state described by $\psi$. The expectation value of an operator should be real and this is guaranteed if the operator is Hermitian. To see this suppose that $\mathcal{L}$ is Hermitian. Then

\begin{align*}

\langle\mathcal{L}\rangle^\ast&=\left[\int \psi^\ast\mathcal{L}\psi d\tau\right]^\ast\\

&=\int\psi\mathcal{L}^\ast\psi^\ast d\tau\\

&=\int(\mathcal{L}\psi)^\ast\psi d\tau\\

&=\int\psi^\ast\mathcal{L}\psi d\tau\ (\mbox{since $\mathcal{L}$ is Hermitian})\\

&=\langle\mathcal{L}\rangle

\end{align*}

That is, $\langle\mathcal{L}\rangle$ is real.

*There are three important properties of Hermitian (self-adjoint) operators:*

- The eigenvalues of a Hermitian operator are real.
- The eigenfunctions of a Hermitian operator are orthogonal.
- The eigenfunctions of a Hermitian operator form a complete set.

**References:**

- G. Arfken,
*Mathematical Methods for Physicists*, 3rd Edition, Academic Press 1985 - W. Greiner,
*Quantum Mechanics, An Introduction*, 4th Edition, Springer-Verlag 2001 - P. Szekeres,
*A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry*, Cambridge University Press 2004

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