Math Problem Statement

Which of the following operators are linear?

Solution

To determine which of the given operators are linear, let's first recall the definition of a linear operator. An operator OO is linear if it satisfies the following properties for any functions ψ(x)\psi(x) and ϕ(x)\phi(x), and any constants aa and bb:

  1. Additivity: O(ψ(x)+ϕ(x))=O(ψ(x))+O(ϕ(x))O(\psi(x) + \phi(x)) = O(\psi(x)) + O(\phi(x)).
  2. Homogeneity: O(aψ(x))=aO(ψ(x))O(a \psi(x)) = a O(\psi(x)).

Now, let's analyze each operator to see if it meets these conditions:

  1. (a) O1ψ(x)=x3ψ(x)O_1 \psi(x) = x^3 \psi(x):
    This operator is not linear because it involves multiplication by x3x^3, which depends on xx. It does not satisfy the linearity properties since O1(aψ(x)+bϕ(x))=x3(aψ(x)+bϕ(x))=ax3ψ(x)+bx3ϕ(x)O_1(a \psi(x) + b \phi(x)) = x^3 (a \psi(x) + b \phi(x)) = a x^3 \psi(x) + b x^3 \phi(x), but it fails to separate as a result of a differential or integral linear transformation.

  2. (b) O2ψ(x)=xddxψ(x)O_2 \psi(x) = x \frac{d}{dx} \psi(x):
    This operator is linear because differentiation is a linear operation, and multiplying by xx still maintains linearity. For example, O2(aψ(x)+bϕ(x))=xddx(aψ(x)+bϕ(x))=axddxψ(x)+bxddxϕ(x)O_2(a \psi(x) + b \phi(x)) = x \frac{d}{dx} (a \psi(x) + b \phi(x)) = a x \frac{d}{dx} \psi(x) + b x \frac{d}{dx} \phi(x).

  3. (c) O3ψ(x)=λψ(x)O_3 \psi(x) = \lambda \psi^*(x):
    This operator is not linear because it involves the complex conjugate ψ(x)\psi^*(x). The conjugate operation does not preserve linearity, as (ψ(x)+ϕ(x))ψ(x)+ϕ(x)(\psi(x) + \phi(x))^* \neq \psi^*(x) + \phi^*(x) in general.

  4. (d) O4ψ(x)=eψ(x)O_4 \psi(x) = e^{\psi(x)}:
    This operator is not linear because the exponential function of ψ(x)\psi(x) does not satisfy the additivity and homogeneity properties of linear operators.

  5. (e) O5ψ(x)=ddxψ(x)+aO_5 \psi(x) = \frac{d}{dx} \psi(x) + a:
    This operator is not linear because of the additional constant term aa. Linear operators cannot have constant terms added to them as it breaks the homogeneity condition.

  6. (f) O6ψ(x)=xdxψ(x)xO_6 \psi(x) = \int_{-\infty}^x dx' \, \psi(x') x':
    This operator is linear because integration is a linear operation. The integral of a linear combination of functions is the linear combination of their integrals, so this satisfies the conditions for linearity.

Conclusion:

The linear operators among the given options are:

  • (b) O2ψ(x)=xddxψ(x)O_2 \psi(x) = x \frac{d}{dx} \psi(x)
  • (f) O6ψ(x)=xdxψ(x)xO_6 \psi(x) = \int_{-\infty}^x dx' \, \psi(x') x'

Would you like further explanation on any of these operators or their linearity?


Here are five related questions that might expand your understanding:

  1. How does the property of homogeneity influence the classification of an operator as linear?
  2. Why does taking the complex conjugate of a function generally break linearity?
  3. Can a linear operator include constants, and under what conditions?
  4. How does integration affect the linearity of an operator?
  5. Are all differentiation operations inherently linear, or are there exceptions?

Tip: Remember that linearity often simplifies solving equations because it allows superposition of solutions. This is particularly useful in physics and engineering.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Differential Equations
Complex Analysis

Formulas

Additivity: O(ψ(x) + φ(x)) = O(ψ(x)) + O(φ(x))
Homogeneity: O(aψ(x)) = aO(ψ(x))

Theorems

Linearity Conditions
Properties of Differential and Integral Operators

Suitable Grade Level

Undergraduate Mathematics