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# 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

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