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Section 4.3 Linear Operators

Definition 4.3.1.

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{4.3.1}\\ \LL (v_1+v_2)\amp =\LL v_1 + \LL v_2\tag{4.3.2} \end{align}

The word linear comes from linear equations, i.e. equations for straight lines. The equation for a line through the origin \(y=mx\) comes from the operator \(f(x)=mx\) acting on vectors which are real numbers \(x\) and constants that are real numbers \(\alpha\text{.}\) The first property:

\begin{equation} f(\alpha x)=m(\alpha x)=\alpha (mx)=\alpha f(x)\tag{4.3.3} \end{equation}

is just commutativity of the real numbers. The second property:

\begin{equation} f(x_1+x_2)=m(x_1+x_2)=m(x_1)+m(x_2)=f(x_1)+f(x_2)\tag{4.3.4} \end{equation}

is just distributivity of multiplication over addition. These are properties of real numbers that you've used since grade school, even if you didn't know to call the properties by these fancy names! Note that the equation for a line NOT through the origin \(y=mx+b\text{,}\) leading to the operator \(g(x)=mx+b\text{,}\) is NOT linear.

\begin{equation} g(x_1+x_2)=m(x_1+x_2)+b \ne (mx_1+b)+(mx_2+b)=g(x_1)+g(x_2)\text{.}\tag{4.3.5} \end{equation}

It will be helpful to remember the example of linear equations with zero and non-zero \(y\)-intercept when you are learning the difference between homogeneous and inhomogeneous linear differential equations in In Section 9.1.

I will not prove it here, but 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{4.3.6} \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{4.3.7} \end{equation}

I will not prove it here, but you use the fact that Hermitian operators in quantum mechanics, e.g. 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{4.3.8}\\ 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{4.3.9} \end{align}