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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

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

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