Question

Let \(D\) stand for \(d / d x,\) that is, \(D y=d y / d x ;\) then $$D^{2} y=D(D y)=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d^{2} y}{d x^{2}}, \quad D^{3} y=\frac{d^{3} y}{d x^{3}}, \text { etc. }$$ \(D\) (or an expression involving \(D\) ) is called a differential operator. Two operators are equal if they give the same results when they operate on \(y\). For example, $$D(D+x) y=\frac{d}{d x}\left(\frac{d y}{d x}+x y\right)=\frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=\left(D^{2}+x D+1\right) y$$ so we say that $$D(D+x)=D^{2}+x D+1$$ In a similar way show that: (a) \(\quad(D-a)(D-b)=(D-b)(D-a)=D^{2}-(b+a) D+a b\) for constant \(a\) and \(b\). (b) \(\quad D^{3}+1=(D+1)\left(D^{2}-D+1\right)\) (c) \(\quad D x=x D+1 .\) (Note that \(D\) and \(x\) do not commute, that is, \(D x \neq x D .\) ) (d) \(\quad(D-x)(D+x)=D^{2}-x^{2}+1,\) but \((D+x)(D-x)=D^{2}-x^{2}-1\) Comment: The operator equations in (c) and (d) are useful in quantum mechanics; see Chapter 12, Section 22.

Short answer

(a) (D - a)(D - b) = D^2 - (a + b)D + ab. (b) D^3 + 1 = (D + 1)(D^2 - D + 1). (c) D x = x D + 1. (d) (D - x)(D + x) = D^2 - x^2 + 1, but (D + x)(D - x) = D^2 - x^2 - 1.

Step-by-step solution

Part (a) Step 1: Expand the Differentials

First, we need to expand (D - a)(D - b). We do this by applying the operators in sequence to a function y.

Part (a) Step 2: Apply the Operators

When applying (D - a)(D - b) to y, we get: (D - a)(D - b)y = (D - a) (Dy - by) = (D(Dy - by) - a(Dy - by)).

Part (a) Step 3: Simplify the Expression

Simplify the expression inside the parentheses: D(Dy - by) - a(Dy - by) = D^2y - D(by) - aDy + aby.Next, we calculate each term separately: D(by) = bDy since b is a constant.Thus, the expression becomes: D^2y - bDy - aDy + aby.

Part (a) Step 4: Combine Like Terms

Combine the terms involving Dy:D^2y - (b + a)Dy + aby.The resulting operator equation is: (D - a)(D - b) = D^2 - (b + a)D + ab.

Part (b) Step 1: Expand the Expression

We start with the expression on the right-hand side: (D + 1)(D^2 - D + 1).

Part (b) Step 2: Apply the Operators

Applying (D + 1) to (D^2 - D + 1)y:(D + 1)((D^2 - D + 1)y) = D(D^2y - Dy + y) + (D^2y - Dy + y).

Part (b) Step 3: Distribute and Simplify

Distribute D and 1:D(D^2y) - D(Dy) + Dy + D^2y - Dy + y = D^3y - D^2y + Dy + D^2y - Dy + y.

Part (b) Step 4: Combine Like Terms

Combine the identical D^2y and Dy terms:D^3y + y.Thus, the left-hand side simplifies to: D^3 + 1.

Part (c) Step 1: Apply the Operators

We need to show D x = x D + 1. Begin by applying D to xy:Dx(y) = xDy + y, because of the product rule: \( \frac{d}{dx}(xy) = x \frac{dy}{dx} + y.\)

Part (c) Step 2: Rearrange and Simplify

Rearrange the equation to isolate Dx:Dx(y) = xDy + y.Thus, we get the operator equation: D x = x D + 1.

Part (d) Step 1: Expand the First Expression

Expand (D - x)(D + x):(D - x)(D + x)y = D(Dy + xy) - x(Dy + xy).

Part (d) Step 2: Simplify

Simplify inside the parentheses applying D and -x:DDy + D(xy) - xDy - x(xy).Applying the product rule for D(xy) = xDy + y gives:D^2y + xDy + y - xDy - x^2y.

Part (d) Step 3: Combine Like Terms

Combine and simplify the terms to yield: D^2y - x^2y + y.Thus, we get D^2 - x^2 + 1 as the operator equation.

Part (d) Step 4: Expand the Second Expression

Expand (D + x)(D - x):(D + x)(D - x)y = D(Dy - xy) + x(Dy - xy).

Part (d) Step 5: Simplify

Simplify similarly using the product rule and combining terms:D^2y - xDy - Dyx + x^2y = D^2y - x^2y - y.Thus, we get: D^2 - x^2 - 1.

Key concepts

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative measures how a function changes as its input changes. In this context, the differential operator is represented by the symbol \( D \). This operator replaces the notion of taking derivatives. For instance, when \( D \) is applied to a function \( y \), it results in the derivative of \( y \) with respect to \( x \), denoted as \( D y = \frac{d y}{dx} \).
Operators can be applied multiple times, such as \( D^2 y = \frac{d^2 y}{dx^2} \), meaning the second derivative of \( y \). This concept is foundational because it provides a streamlined way to discuss and manipulate derivatives in mathematical problems.
Let's consider an example: If we apply \( D \) to the function \( x^2 \), it results in \( D(x^2) = 2x \). Applying \( D \) again, we get \( D(2x) = 2 \). Thus, \( D^2(x^2) = 2 \).

Operator Algebra

Operator algebra deals with the manipulation of differential operators. Operators can be combined, much like algebraic expressions, to create new operators with specific properties. For example, adding operators or multiplying them can reveal interesting relationships.
In the provided exercise, we see how operators can expand and simplify to yield new forms. For instance, consider the expression \( (D - a)(D - b) \). By applying these operators in sequence to a function \( y \), we expand and simplify to find an equivalent operator, \( D^2 - (b + a)D + ab \).
This sequence of steps demonstrates that, despite their non-commutative nature, differential operators follow predictable algebraic rules which can be used to simplify complex differential equations.
Another key example is \( D x = x D + 1 \). Here, operator algebra shows how applying a differential operator to a product of variables leads to a new relationship, highlighting the rules of differentiation and product manipulation.

Product Rule

The product rule is a fundamental rule in calculus used to differentiate products of two or more functions. It states that: \[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \].
This rule is essential when dealing with differential operators. For instance, in the exercise, when proving \( D x = x D + 1 \), the product rule is applied: \( D (xy) = x D y + y \).
This is particularly important in operator algebra where we need to handle expressions like \( D(by) \). Since \( b \) is a constant,\( D(by) = b Dy \).
The product rule helps break down complex expressions into simpler parts, making it easier to manipulate and simplify operator expressions.

Quantum Mechanics

Quantum mechanics often uses differential operators, especially in relation to the Schrödinger equation and commutation rules. Operators in quantum mechanics, like the position operator \( x \) and the momentum operator \( p = -i \frac{d}{dx} \), do not commute.
This non-commutativity is illustrated in the provided exercise: \( D x eq x D \). Instead, \( D x = x D + 1 \). This result is crucial in quantum mechanics, where the order of operations affects outcomes significantly.
Another example from the exercise, \( (D - x)(D + x) \) versus \( (D + x)(D - x) \), shows how different ordering of operators can lead to different results. In quantum mechanics, such differences can impact the physical interpretation of operators and their effects on quantum states.
Therefore, understanding differential operators and their algebra is pivotal for students studying quantum mechanics, as these concepts frequently appear in problem-solving and theory development.