Question

If two operators act on a wave function as indicated by \(\hat{A} \hat{B} f(x),\) it is important to carry out the operations in succession with the first operation being that nearest to the function. Mathematically, \(\hat{A} \hat{B} f(x)=\hat{A}(\hat{B} f(x))\) and \(\hat{A}^{2} f(x)=\hat{A}(\hat{A} f(x)) .\) Evaluate the following successive operations \(\hat{A} \hat{B} f(x) .\) The operators \(\hat{A}\) and \(\hat{B}\) are listed in the first two columns and \(f(x)\) is listed in the third column. a. \(\frac{d}{d y} \quad y \quad y e^{-2 y^{3}}\) b. \(y \quad \frac{d}{d y} \quad y e^{-2 y^{3}}\) c. \(y \frac{\partial}{\partial x} \quad x \frac{\partial}{\partial y} \quad e^{-2(x+y)}\) d. \(x \frac{\partial}{\partial y} \quad y \frac{\partial}{\partial x} \quad e^{-2(x+y)}\) Are your answers to parts (a) and (b) identical? Are your answers to parts (c) and (d) identical? As we will learn in Chapter \(17,\) switching the order of the operators can change the outcome of the operation \(\hat{A} \hat{B} f(x)\)

Short answer

The results of successive operations are: (a) \(\hat{A}\hat{B}f(x) = ye^{-2y^3} - 6y^6e^{-2y^3}\) (b) \(\hat{A}\hat{B}f(x) = ye^{-2y^3} - 6y^6e^{-2y^3}\) (c) \(\hat{A}\hat{B}f(x) = -2ye^{-2(x+y)}\) (d) \(\hat{A}\hat{B}f(x) = -2xe^{-2(x+y)}\) The answers to (a) and (b) are identical, and the answers to (c) and (d) are different, showing that switching the order of operations can change the outcome of the operation \(\hat{A}\hat{B}f(x)\).

Step-by-step solution

Apply Operator B

In this case, operator B is the operation of taking the derivative with respect to y. So, apply the derivative operation on the given function \(f(x) = ye^{-2y^3}\): \(\hat{B}f(x) = \frac{d}{dy}(ye^{-2y^3})\) Now simplify the derivative: \(\hat{B}f(x) = e^{-2y^3} - 6y^5e^{-2y^3}\)

Apply Operator A

Now, apply the operation of multiplying by y (Operator A) on \(\hat{B}f(x)\): \(\hat{A}\hat{B}f(x) = y(e^{-2y^3} - 6y^5e^{-2y^3})\) Simplify the expression: \(\hat{A}\hat{B}f(x) = ye^{-2y^3} - 6y^6e^{-2y^3}\) #b. Apply y and then d/dy#

Apply Operator A

In this case, Operator A is the operation of multiplying by 'y.' So, apply Operator A on the given function \(f(x) = ye^{-2y^3}\): \(\hat{A}f(x) = y(ye^{-2y^3})\) Simplify the expression: \(\hat{A}f(x) = y^2e^{-2y^3}\)

Apply Operator B

Now, apply the operation of taking the derivative with respect to y (Operator B) on \(\hat{A}f(x)\): \(\hat{A}\hat{B}f(x) = \frac{d}{dy}(y^2e^{-2y^3})\) Simplify the derivative: \(\hat{A}\hat{B}f(x) = ye^{-2y^3}-6y^6e^{-2y^3}\) #c. Apply y∂/∂x, then x∂/∂y#

Apply Operator B

Apply the operation of taking the partial derivative with respect to x (Operator B) on the given function \(f(x) = e^{-2(x+y)}\): \(\hat{B}f(x) = \frac{\partial}{\partial x}(e^{-2(x+y)})\) Now simplify the partial derivative: \(\hat{B}f(x) = -2e^{-2(x+y)}\)

Apply Operator A

Now, apply the operation of multiplying by 'y' (Operator A) on \(\hat{B}f(x)\): \(\hat{A}\hat{B}f(x) = y(-2e^{-2(x+y)})\) Simplify the expression: \(\hat{A}\hat{B}f(x) = -2ye^{-2(x+y)}\) #d. Apply x∂/∂y, then y∂/∂x#

Apply Operator B

Apply the operation of multiplying by 'x' (Operator B) on the given function \(f(x) = e^{-2(x+y)}\): \(\hat{B}f(x) = -2xe^{-2(x+y)}\)

Apply Operator A

Now, apply the operation of taking the partial derivative with respect to y (Operator A) on \(\hat{B}f(x)\): \(\hat{A}\hat{B}f(x) = \frac{\partial}{\partial y}(-2xe^{-2(x+y)})\) Now simplify the expression: \(\hat{A}\hat{B}f(x) = -2xe^{-2(x+y)}\) In the end, we have the results for problems (a), (b), (c), and (d) as: (a) \(\hat{A}\hat{B}f(x) = ye^{-2y^3} - 6y^6e^{-2y^3}\) (b) \(\hat{A}\hat{B}f(x) = ye^{-2y^3} - 6y^6e^{-2y^3}\) (c) \(\hat{A}\hat{B}f(x) = -2ye^{-2(x+y)}\) (d) \(\hat{A}\hat{B}f(x) = -2xe^{-2(x+y)}\) As we can see, the answers to (a) and (b) are identical, and the answers to (c) and (d) are different. Thus, switching the order of operations can change the outcome of the operation \(\hat{A}\hat{B}f(x)\).

Key concepts

Wave Function Operations

In quantum mechanics, wave functions are crucial as they provide the probabilities of a particle’s position, momentum, and other physical properties. Operating on a wave function typically involves the application of mathematical operators, which can be differentiated as functions, derivatives, or multiplication by variables.

• Operators must be applied in sequence. The order matters because applying one operator could impact how a subsequent operator interacts with the wave function.
• The expression \(\hat{A} \hat{B} f(x)\) should be interpreted as first applying operator \(\hat{B}\) to the function \(f(x)\), and then applying operator \(\hat{A}\) to the result.

When performing wave function operations, ensuring the correct order of operator application is crucial for obtaining accurate results. Switching operators like taking derivatives or multiplying can significantly alter the outcome, which is a core concept in operator algebra.

Partial Derivatives

Partial derivatives allow us to find the rate of change of a function with respect to one variable while keeping others constant. They are crucial in multivariable calculus, especially in the context of quantum mechanics where functions often depend on several variables.

• For a function \(f(x, y) = e^{-2(x+y)}\), taking the partial derivative relative to x involves differentiating while treating y as constant. The operation would be denoted as \(\frac{\partial f}{\partial x}\).
• When evaluating these derivatives, apply the chain rule carefully, especially when dealing with exponential functions.

Partial derivatives help describe the behavior of a quantum system by considering how wave functions change with respect to each variable independently. They form the basis for more complex operations like those seen in differential equations governing quantum mechanics.

Commutative Property of Operators

The commutative property in mathematics refers to the ability to change the order of operations without affecting the result. In quantum mechanics, however, operators can often be non-commutative, meaning that changing the order of operations alters the outcome.

• In expressions such as \(\hat{A} \hat{B} f(x)\) vs. \(\hat{B} \hat{A} f(x)\), it is essential to test whether the operators are commutative. Non-commutative operators produce different results when their order is swapped.
• If two operators do commute, applying them in any order on a wave function will yield the same result.

Understanding whether operators commute is vital, as non-commuting operators can lead to varied interpretations and predictions in the behavior of quantum systems. This principle underpins much of quantum mechanics, given its reliance on operator mathematics to elucidate physical phenomena.