Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 4

Show that (a) \([A, B]=-[B, A],\) (b) \(\left[A^{m}, A^{n}\right]=0\) for all \(m, n,\) (c) \(\left[A^{2}, B\right]=A[A, B]+[A, B] A\) (d) \([A,[B, C]]+[B,[C, A]]+[C,[A, B]]=0\)

Short Answer

Expert verified
In this problem, we've confirmed that the properties (a) \([A, B] = -[B, A]\), (b) \([A^{m}, A^{n}] = 0\) for all \(m, n,\), (c) \([A^{2}, B] = A[A, B] + [A, B]A\), (d) \([A,[B, C]]+[B,[C, A]]+[C,[A, B]]=0\) hold for the operators \(A\), \(B\), and \(C\).

Step by step solution

01

Verify Property (a)

To confirm property (a), start with the right-hand side of the equation \(-[B, A]\). This equals to \(-(BA - AB)\) which then simplifies to \(AB - BA\), also known as \([A, B]\). Thus, we've proven that \([A, B] = -[B, A]\).
02

Verify Property (b)

For property (b), it's desirable to know that a commutator of an operator with itself is always zero, because \(A^nA^m - A^mA^n\) ends up being \(0\). Therefore, we can tell that \([A^{m}, A^{n}] = 0\).
03

Verify Property (c)

For property (c), we'll work on the right-hand side of the equation \(A[A, B] + [A, B]A\). This simplifies to \(A(AB - BA) + (AB - BA)A\). Then, \(ABA - A^2B + ABA - BA^2\). After rearranging terms, you'll get \(A^2B - BA^2\), which is \([A^{2}, B]\). Therefore, we've proven that \([A^{2}, B] = A[A, B] + [A, B]A\).
04

Verify Property (d)

For property (d), we need to consider that \([A,[B, C]] + [B,[C, A]] + [C,[A, B]]\) equals to \(ABC-ACB+BCA-BAC+CAB-CBA\), after applying the definitions of the commutators. Adding these terms up, we may notice that each term appears once positively and once negatively, so their sum equals to zero. Hence, \([A,[B, C]]+[B,[C, A]]+[C,[A, B]]=0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Evaluate the commutators (a) \(\left[(1 / x), p_{x}\right]\) (b) \(\left[(1 / x), p_{x}^{2}\right]\) (c) \(\left[x p_{y}-y p_{x}, y p_{z}-z p_{y}\right]\) (d) \(\left[x^{2}\left(\partial^{2} / \partial y^{2}\right), y(\partial / \partial x)\right]\)

A particle in an infinite one-dimensional system was described by the wavefunction \(\psi(x)=\mathrm{Ne}^{-x^{2} / 2 r^{2}}\). Normalize this function. Calculate the probability of finding the particle in the range \(-\Gamma \leq x \leq \Gamma\). Hint. The integral encountered in the second part is the error function. It is available in mathematical software.

Confirm that the operators (a) \(T=-\left(\hbar^{2} / 2 m\right)\left(d^{2} / d x^{2}\right)\) and (b) \(l_{z}=(\hbar / \mathrm{i})(\mathrm{d} / \mathrm{d} \varphi)\) are Hermitian. Hint. Consider the integrals \(\int_{0}^{L} \psi_{a}^{*} T \psi_{b} \mathrm{d} x\) and \(\int_{0}^{2 \pi} \psi_{a}^{*} l_{z} \psi_{b} \mathrm{d} \varphi\) and integrate by parts.

Evaluate the commutator \(\left[l_{y}\left[l_{y}, l_{z}\right]\right]\) given that \(\left[l_{x}, l_{y}\right]=i \hbar l,\left[l_{y}, l_{z}\right]=i \hbar l_{x},\) and \(\left[l_{z}, l_{x}\right]=i \hbar l_{y}\)

Use the momentum representation and a general function \(f\left(p_{x}\right)\) of the linear momentum to confirm that the position and momentum operators in this representation do not commute, and find the value of their commutator.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free