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Chapter 4: Problem 16

Evaluate the commutator \([\hat{A}, \hat{B}],\) where \(\hat{A}\) and \(\hat{B}\) are given below. $$\begin{array}{llll} & \hat{A} & \hat{B} \\\\\hline(\mathrm{a}) & \frac{d^{2}}{d x^{2}} & x \\\\(\mathrm{b}) & \frac{d}{d x}-x & \frac{d}{d x}+x \\\\(\mathrm{c}) & \int_{0}^{x} d x & \frac{d}{d x} \\\\(\mathrm{d}) & \frac{d^{2}}{d x^{2}}-x & \frac{d}{d x}+x^{2}\end{array}$$

Short Answer

Expert verified
(a) 2\frac{d}{dx}, (b) 4x \frac{d}{dx} + 2, (c) -f(0), (d) 4x^2 \frac{d}{dx} - (x^3 - 1).

Step by step solution

01

Understanding the Commutator

The commutator of operators \( \hat{A} \) and \( \hat{B} \) is defined as \( [\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A} \). We will evaluate this for each given pair of operators.
02

Evaluating Case (a)

For case (a), the operators are \( \hat{A} = \frac{d^2}{dx^2} \) and \( \hat{B} = x \). We need to calculate \( [\hat{A}, \hat{B}]f(x) = (\hat{A}\hat{B}f(x) - \hat{B}\hat{A}f(x)) \) where \( f(x) \) is any arbitrary function. Calculate \( \hat{B}f(x) = xf(x) \). Then \( \hat{A}\hat{B}f(x) = \frac{d^2}{dx^2}(xf(x)) = 2\frac{df}{dx} + x\frac{d^2f}{dx^2} \).Calculate \( \hat{A}f(x) = \frac{d^2f}{dx^2} \) and then \( \hat{B}\hat{A}f(x) = x\frac{d^2f}{dx^2} \).Thus, the commutator is \( [\hat{A}, \hat{B}]f(x) = (2\frac{df}{dx} + x\frac{d^2f}{dx^2}) - x\frac{d^2f}{dx^2} = 2\frac{df}{dx} \).Result: \( [\hat{A}, \hat{B}] = 2\frac{d}{dx} \).
03

Evaluating Case (b)

For case (b), the operators are \( \hat{A} = \frac{d}{dx} - x \) and \( \hat{B} = \frac{d}{dx} + x \). Calculate the commutator as follows: \( \hat{A}f(x) = \frac{df}{dx} - xf(x) \) and \( \hat{B}f(x) = \frac{df}{dx} + xf(x) \).Now, \( \hat{A}\hat{B}f(x) = (\frac{d}{dx} - x)(\frac{d}{dx} + x)f(x) = (\frac{d^2}{dx^2}f(x) + x\frac{df}{dx} + \frac{df}{dx}x - x^2f(x)) = \frac{d^2 f}{dx^2} + 2x \frac{df}{dx} + (1-x^2)f(x) \).\( \hat{B}\hat{A}f(x) = (\frac{d}{dx} + x)(\frac{d}{dx} - x)f(x) = \frac{d^2}{dx^2}f(x) - x \frac{df}{dx} - \frac{df}{dx}x - x^2f(x) = \frac{d^2 f}{dx^2} - 2x \frac{df}{dx} - (1+x^2)f(x) \).The commutator is then \( [\hat{A}, \hat{B}]f(x) = (\frac{d^2 f}{dx^2} + 2x \frac{df}{dx} + (1-x^2)f(x)) - (\frac{d^2 f}{dx^2} - 2x \frac{df}{dx} - (1+x^2)f(x)) = 4x \frac{df}{dx} + 2f(x)\).Result: \( [\hat{A}, \hat{B}] = 4x \frac{d}{dx} + 2 \).
04

Evaluating Case (c)

For case (c), the operators are \( \hat{A} = \int_0^x dx' \) and \( \hat{B} = \frac{d}{dx} \). We find \( \hat{B}f(x) = \frac{df}{dx} \) and \( \hat{A}\hat{B}f(x) = \int_0^x \frac{df}{dx} dx = f(x) - f(0) \).\( \hat{A}f(x) = \int_0^x f(x) dx \) and then \( \hat{B}\hat{A}f(x) = \frac{d}{dx}\int_0^x f(x) dx = f(x) \).The commutator is \( [\hat{A}, \hat{B}]f(x) = (f(x) - f(0)) - f(x) = -f(0) \).Result: \( [\hat{A}, \hat{B}] = -f(0) \).
05

Evaluating Case (d)

For case (d), the operators are \( \hat{A} = \frac{d^2}{dx^2} - x \) and \( \hat{B} = \frac{d}{dx} + x^2 \). Calculate each part. \( \hat{B}f(x) = \frac{df}{dx} + x^2f(x) \) and \( \hat{A}\hat{B}f(x) = (\frac{d^2}{dx^2} - x)(\frac{df}{dx} + x^2f(x)) = \frac{d^3f}{dx^3} + 2x \frac{d^2f}{dx^2} + x \frac{df}{dx} - 2x^2 \frac{df}{dx} - x^3f(x) \).\( \hat{A}f(x) = \frac{d^2 f}{dx^2} - xf(x)\) and then \( \hat{B}\hat{A}f(x) = (\frac{d}{dx} + x^2)(\frac{d^2 f}{dx^2} - xf(x)) = \frac{d^3f}{dx^3} + 2x \frac{df^2}{dx^2} - 3x^2 \frac{df}{dx} - xf(x) \).The commutator is \( [\hat{A}, \hat{B}]f(x) = \left( \frac{d^3f}{dx^3} + 2x \frac{d^2f}{dx^2} + x \frac{df}{dx} - 2x^2 \frac{df}{dx} - x^3f(x) \right) - \left( \frac{d^3f}{dx^3} + 2x \frac{d^2f}{dx^2} - 3x^2 \frac{df}{dx} - xf(x) \right)= 4x^2 \frac{df}{dx} - (x^3 - 1)f(x) \).Result: \( [\hat{A}, \hat{B}] = 4x^2 \frac{d}{dx} - (x^3 -1) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Commutator
In quantum mechanics and other areas of physics, the commutator provides a measure of how much two operators differ when applied in sequence. It is defined mathematically as \([\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\). This is crucial because it shows whether the operations described by the two operators can be considered independent or if they affect each other when applied sequentially.
  • For example, if the commutator is zero, i.e., \([\hat{A}, \hat{B}] = 0\), it means the operators commute, indicating that they do not interfere with each other's operations.
  • If the commutator is non-zero, there is an indication of dependency, often pointing towards some underlying symmetry or conservation law in the physical system being described.
Understanding the commutator is fundamental in quantum mechanics as it assists in describing the behavior of quantum systems, and also plays a critical role in the famous Heisenberg Uncertainty Principle. Knowing how operators relate to each other through their commutator can provide insight into the possible measurement outcomes and limitations in quantum systems.
Operators
Operators in quantum mechanics are mathematical objects that act on the wave functions of a system to extract physical information, such as energy or position. They play an essential role in defining the states and evolution of quantum systems.
  • Operators can be linear, meaning that they satisfy conditions like \(\hat{O}(af + bg) = a\hat{O}f + b\hat{O}g\), where \(a\) and \(b\) are constants, and \(f\) and \(g\) are functions.
  • Many operators, such as the momentum and Hamiltonian operator, are derived from classical physics and modified for the quantum context.
In the exercise above, operators such as \(\frac{d}{dx}\), which represents differentiation, are used to manipulate functions, and the integral operator, \(\int_0^x dx\), accumulates the area under a curve. These operators showcase the different ways to interact with and change functions, reflecting the dynamic nature of quantum particles.
Differential Equations
Differential equations are equations that involve functions and their derivatives. They are crucial in quantum mechanics as they often describe how quantum states change over time or space. The Schr枚dinger equation, a fundamental equation in quantum mechanics, is a type of differential equation.
  • The second derivative operator \(\frac{d^2}{dx^2}\) is used to create or solve differential equations, influencing wave functions and their behavior.
  • Differential equations can be solved to uncover functions that describe particle movement or probabilities in quantum systems.
In the context of the commutator exercise, differential operators like \(\frac{d}{dx}\) and \(\frac{d^2}{dx^2}\) are used to probe how changes in one variable affect another, reflecting the continuous and often complex nature of quantum mechanics.
Linear Algebra
Linear algebra forms the mathematical backbone of quantum mechanics, providing tools and frameworks, such as vector spaces, matrices, and linear mappings, to describe quantum states and transformations. Quantum observables and operations can be depicted using vectors and matrices.
  • Linear operators, a key concept in linear algebra, align well with the principles of quantum mechanics, allowing for operations that obey linearity.
  • Concepts like eigenvalues and eigenvectors help define quantum states and possible measurement outcomes.
In the solved exercise, linear algebra principles are helpful in simplifying operations and understanding the symmetry properties of quantum systems. Working with operators using the language of linear algebra makes it possible to transform and manipulate complex quantum states precisely and efficiently, contributing significantly to our ability to predict and understand quantum phenomena.

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Most popular questions from this chapter

Which of the following wave functions are normalized over the indicated two- dimensional intervals? a. \(e^{-\left(x^{2}+y^{2}\right) / 2}\) \(0 \leq x<\infty\) \(0 \leq y<\infty\) b. \(e^{-(x+y) / 2}\) \(0 \leq x<\infty\) \(0 \leq y<\infty\) c. \(\left(\frac{4}{a b}\right)^{1 / 2} \sin \frac{\pi x}{a} \sin \frac{\pi y}{b}\) \(0 \leq x \leq a\) \(0 \leq y \leq b\) Normalize those that aren't.

Which of the following candidates for wave functions are normalizable over the indicated intervals? a. \(e^{-x^{2} / 2}(-\infty, \infty)\) b. \(e^{x}(0, \infty)\) c. \(e^{i \theta}(0,2 \pi)\) d. \(\sinh x(0, \infty)\) e. \(x e^{-x}(0, \infty)\) Normalize those that can be normalized. Are the others suitable wave functions?

In ordinary algebra, \((P+Q)(P-Q)=P^{2}-Q^{2} .\) Expand \((\hat{P}+\hat{Q})(\hat{P}-\hat{Q}) .\) Under what conditions do we find the same result as in the case of ordinary algebra?

Problem \(4-35\) gives that the probability of a particle of relative energy \(E / V_{0}\) will penetrate a rectangular potential barrier of height \(V_{0}\) and thickness \(a\) is $$T=\frac{1}{1+\frac{\sinh ^{2}\left[v_{0}^{1 / 2}(1-r)^{1 / 2}\right]}{4 r(1-r)}}$$ where \(v_{0}=2 m V_{0} a^{2} / \hbar^{2}\) and \(r=E / V_{0} .\) What is the limit of \(T\) as \(r \rightarrow 1 ?\) Plot \(T\) against \(r\) for \(v_{0}=1 / 2,1,\) and \(2 .\) Interpret your results.

Can the angular momentum and kinetic energy of a particle be measured simultaneously to arbitrary precision?
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