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Section 8.6 Linear Operators: Definitions and Examples

Definition 8.6. Linear Operator.

An operator \(\LL\) is linear if it satisfies two conditions when acting on any appropriate vector \(v\) and for any constants \(\alpha\text{:}\)

\begin{align} \LL (\alpha v)\amp =\alpha \LL v\tag{8.6.1}\\ \LL (v_1+v_2)\amp =\LL v_1 + \LL v_2\tag{8.6.2} \end{align}

Derivatives Are Linear Operators.

You have often used the fact that the derivative operator acting on functions is linear:

\begin{equation} \frac{d}{dx} \left(\alpha f\right)= \alpha \left(\frac{d}{dx} f\right)\tag{8.6.3} \end{equation}
\begin{equation} \frac{d}{dx}\left(f+g\right)=\left(\frac{d}{dx}f\right) + \left(\frac{d}{dx}g\right)\tag{8.6.4} \end{equation}

For example,

\begin{equation} \frac{d}{dx} \left(3x^2 + \cos{x}\right) =3 \frac{d}{dx}\, x^2 + \frac{d}{dx}\, \cos{x}\tag{8.6.5} \end{equation}

By a straightforward extension, the differential operator \(\LL\) defined by

\begin{equation} \LL\equiv a_{n}(x)\frac{d^n}{dx^n}+a_{n-1}(x)\frac{d^{n-1}}{dx^{n-1}} +\dots+a_0(x)\tag{8.6.6} \end{equation}

is also linear. The important feature here is that all of the (\(n\)th order) derivatives are to the first power and not inside of any other special functions. (Do not be confused by the notation for \(n\)th order derivatives, which looks like the notation for the \(n\)th power.) Many differential operators in physics ARE linear, so this should look very comfortable and familiar. As counterexamples, the following strange-looking differential operators are NOT linear:

\begin{equation} \left(\frac{d}{dx} f\right)^2\text{,}\tag{8.6.7} \end{equation}
\begin{equation} \sin\left(\frac{d}{dx} f\right)\text{.}\tag{8.6.8} \end{equation}

In particular, you may have seen the equation for the motion of a pendulum:

\begin{equation} \frac{d^2 \theta}{dt^2}+\frac{g}{L}\, \sin\theta=0\tag{8.6.9} \end{equation}

The term \(\sin\theta\) makes this equation non-linear. When you make the small angle approximation (the first term in a power series expansion) \(\sin\theta\approx\theta\text{,}\) the equation becomes linear, and therefore simple to solve. (See Section 9.7 for the method of solving this equation.) But the solution is only approximately true and the approximation is best when the angle \(\theta\) is small.

Other Examples of Linear Operators.

You use the fact that matrix multiplication (acting on vectors that are columns and multiplication by scalars \(\alpha\)) is a linear operator when you do the following common matrix manipulations.

\begin{equation} \begin{pmatrix}2\amp 3\\4\amp 5 \end{pmatrix} \left( \alpha \begin{pmatrix}6\\7 \end{pmatrix} \right) = \alpha\left( \begin{pmatrix}2\amp 3\\4\amp 5 \end{pmatrix} \begin{pmatrix}6\\7 \end{pmatrix} \right)\tag{8.6.10} \end{equation}
\begin{equation} \begin{pmatrix}2\amp 3\\4\amp 5 \end{pmatrix} \left( \begin{pmatrix}6\\7 \end{pmatrix} + \begin{pmatrix}8\\9 \end{pmatrix} \right) = \left( \begin{pmatrix}2\amp 3\\4\amp 5 \end{pmatrix} \begin{pmatrix}6\\7 \end{pmatrix} \right) + \left( \begin{pmatrix}2\amp 3\\4\amp 5 \end{pmatrix} \begin{pmatrix}8\\9 \end{pmatrix} \right)\tag{8.6.11} \end{equation}

You also use the fact that Hermitian operators in quantum mechanics, for example, the Hamiltonian, are linear when you do the following bra/ket manipulations.

\begin{align} H\left(\alpha \vert \psi\rangle\right) \amp =\alpha \left(H \vert \psi\rangle\right)\tag{8.6.12}\\ H\left(\alpha \vert \psi_1\rangle + \alpha \vert \psi_2\rangle\right) \amp =\left(H \vert \psi_1\rangle\right) +\left(H \vert \psi_2\rangle\right)\tag{8.6.13} \end{align}