# 3.2: Linear Operators in Quantum Mechanics

Skills to Develop

- Classical-Mechanical Quantities Are Represented by Linear Operators in Quantum Mechanics

The bracketed object in the time-independent Schrödinger equation (Equation 3.1.19) is called an *operator*. An operator is a generalization of the concept of a function. Whereas a function is a rule for turning one number into another, an operator is a rule for turning one function into another. The Laplacian is an example of an operator. We often (but not always) indicate that an object is an operator by placing a `hat' over it, eg, \(\hat{A}\). The action of an operator that turns the function \(f(x)\) into the function \(g(x)\)is represented by

\[\hat{A}f(x)=g(x)\label{3.2.1}\]

The most common kind of operator encountered are *linear operators* which satisfies the following two conditions:

\[ \underset{\text{Condition A}}{\hat{O}(f(x)+g(x)) = \hat{O}f(x)+\hat{O}g(x)} \label{3.2.2a}\]

and

\[\underset{\text{Condition B}}{\hat{O}cf(x) = c \hat{O}f(x)} \label{3.2.2b}\]

where

- \(\hat{O}\) is a linear operator,
- \(c\) is a constant that can be a complex number (\(c = a + ib\)), and
- \(f(x)\) and \(g(x)\) are functions of \(x\)

Is an operator fails to satisfy the conditions in either Equations \(\ref{3.2.2a}\) or \(\ref{3.2.2b}\), then it is not a linear operator.

Example \(\PageIndex{1}\)

Is the differentiation operator \( \left(\dfrac{d}{dx} \right) \) linear?

**Solution**

To confirm is an operator is linear, both conditions in Equation \(\ref{3.2.2b}\) much be demonstrated.

Condition A (Equation \(\ref{3.2.2a}\)):

\[\hat{O}(f(x)+g(x)) = \dfrac{d}{dx} \left( f(x)+g(x)\right)\]

From basic calculus, we know that we can use the **sum rule for differentiation**

\[= \dfrac{d}{dx} f(x) + \dfrac{d}{dx} g(x) = \hat{O}f(x)+\hat{O}g(x) \;\;\; \checkmark \]

Condition A is confirmed. Does Condition B (Equation \(\ref{3.2.2b}\)) hold?

\[\hat{O}cf(x) = \dfrac{d}{dx} c f(x) \]

Also from basic calculus, this can be factored out of the derivative

\[ = c \dfrac{d}{dx} f(x) = c \hat{O}f(x) \;\;\; \checkmark\]

Yes. The differentiation operator is a linear operator

Exercise \(\PageIndex{1}\)

Confirm if the square root operator \(\sqrt{f(x)}\) linear or not?

The most common kind of operator encountered in quantum mechanics are *linear operators*.

### The Momentum Operator

Equation 3.1.11 from Section 3.1 implies that the operator for the \(x\)-component of momentum can be written

\[\hat{p}_{x}=-i\hbar\dfrac{\partial}{\partial x}\label{3.2.3a}\]

and by analogy, we must have the other two operators representing the other dimentions.

\[\hat{p}_{y}=-i\hbar\dfrac{\partial}{\partial y} \label{3.2.3b}\]

and

\[\hat{p}_{z}=-i\hbar\dfrac{\partial}{\partial z}\label{3.2.3c}\]

### Contributors

Seymour Blinder (Professor Emeritus of Chemistry and Physics at the University of Michigan, Ann Arbor)