Question

Form the operator \(\hat{A}^{2}\) if \(\hat{A}=x-d / d x .\) Be sure to include an arbitrary function on which the operator acts.

Short answer

The operator \(\hat{A}^{2}\) acting on an arbitrary function \(\psi(x)\) is given by: \(\hat{A}^{2} \psi(x) = x^{2}\psi(x) - 3x\frac{d\psi(x)}{dx} + \psi(x) + \frac{d^{2}\psi(x)}{dx^{2}}\)

Step-by-step solution

Determine the Operator \(\hat{A}^{2}\)

To find the square of the operator \(\hat{A}\), we need to apply the operator \(\hat{A}\) on itself. That is, we need to compute \(\hat{A}(\hat{A})\): \(\hat{A}^{2} = \hat{A}(\hat{A}) = \hat{A}(x - \frac{d}{dx})\) Now, we will apply the operator \(\hat{A}\) to an arbitrary function, say \(\psi(x)\): \(\hat{A} \psi(x) = \left( x - \frac{d}{dx} \right) \psi(x)\) This expression gives us the result of the operator \(\hat{A}\) acting on the function \(\psi(x)\).

Apply the Operator \(\hat{A}\) to \(\hat{A} \psi(x)\)

Now, we need to apply the operator \(\hat{A}\) to the expression obtained in Step 1. This will give us the operator \(\hat{A}^{2}\) acting on the function \(\psi(x)\): \(\hat{A}^{2} \psi(x) = \hat{A}(\hat{A} \psi(x)) = \hat{A} \left( x\psi(x) - \frac{d\psi(x)}{dx} \right)\) Now, we apply the operator \(\hat{A} = x - \frac{d}{dx}\) to the expression inside the parenthesis: \(\hat{A}^{2} \psi(x) = \left( x - \frac{d}{dx} \right) \left( x\psi(x) - \frac{d\psi(x)}{dx} \right)\)

Compute the Resulting Expression

Now, we distribute the operator \(\hat{A}\) to both terms inside the parenthesis: \(\hat{A}^{2} \psi(x) = \left( x - \frac{d}{dx} \right)(x\psi(x)) - \left( x - \frac{d}{dx} \right)\left(\frac{d\psi(x)}{dx}\right)\) Now, we need to apply the operator to each term separately: \[ \begin{aligned} \hat{A}^{2} \psi(x) &= x(x\psi(x)) - \frac{d}{dx}(x\psi(x)) - x\frac{d\psi(x)}{dx} + \frac{d^{2}\psi(x)}{dx^{2}} \\ &= x^{2}\psi(x) - x\frac{d\psi(x)}{dx} - \frac{d(x\psi(x))}{dx} - x\frac{d\psi(x)}{dx} + \frac{d^{2}\psi(x)}{dx^{2}} \end{aligned} \] Now, we can simplify the expression further: \[ \begin{aligned} \hat{A}^{2} \psi(x) &= x^{2}\psi(x) - 2x\frac{d\psi(x)}{dx} - \left(\psi(x) + x\frac{d\psi(x)}{dx}\right) + \frac{d^{2}\psi(x)}{dx^{2}} \\ &= x^{2}\psi(x) - 3x\frac{d\psi(x)}{dx} + \psi(x) + \frac{d^{2}\psi(x)}{dx^{2}} \end{aligned} \] Thus, the operator \(\hat{A}^{2}\) acting on an arbitrary function \(\psi(x)\) is given by: \(\hat{A}^{2} \psi(x) = x^{2}\psi(x) - 3x\frac{d\psi(x)}{dx} + \psi(x) + \frac{d^{2}\psi(x)}{dx^{2}}\)

Key concepts

Differential Operators

Differential operators are mathematical objects used to specify how functions change. They are central to the field of calculus, especially in the context of differential equations. In the given problem, the operator \( \hat{A} = x - \frac{d}{dx} \) is a simple example of a differential operator. Understanding how these operators work is fundamental to solving complex mathematical problems, especially those involving functions of one or more variables.

Here are a few key points about differential operators:

• A differential operator takes a function as input and provides another function as output.
• The simplest form of a differential operator is the derivative, represented as \( \frac{d}{dx} \), indicating how the function changes with respect to \( x \).
• In the problem, the operator \( \hat{A} \) is actually a combination of multiplication by \( x \) and the derivative operator.

Understanding the behavior of differential operators involves applying them to a function and observing the transformation. This helps in predicting how functions behave under continuous changes.

Quantum Mechanics

In quantum mechanics, operators are used to represent physical quantities such as position, momentum, and energy. This field heavily relies on operator theory, as systems are described by wave functions and operators act on these functions to extract meaningful physical information.

The exercise with \( \hat{A} \) is a simplified example akin to quantum operators. Here's how these concepts link:

• Operators like \( \hat{A} \) are similar to position and momentum operators in quantum mechanics, where they manipulate wave functions.
• Physically, acting with an operator on a wave function can correspond to observable properties, hence why understanding how to combine and square operators is crucial.
• This exercise models the action of operators in filtering functions into measurable quantities, a fundamental principle in quantum measurement.

Overall, getting acquainted with how operators work mathematically offers insight into their physical interpretations in quantum systems.

Mathematical Physics

Mathematical physics is the field that applies mathematical methods to solve physical problems. It uses advanced mathematical structures to formulate theories and derive predictions.

Operators like \( \hat{A} \) are essential tools here, bridging abstract mathematical concepts with physical phenomena:

• They help in expressing complex physical laws compactly, such as those found in quantum mechanics and field theories.
• Mathematical physics often deals with differential equations, where operators like \( \frac{d}{dx} \) describe dynamic systems.
• This exercise shows how operators can be squared and combined to solve and simplify physical equations, illustrating their role in analysis and problem-solving.

Through these mathematical manipulations, theoretical predictions become possible, serving as the foundation for further experimental validation in physical sciences. Recognizing the power of these operators in modeling real-world phenomena is a cornerstone of mathematical physics.