Derive (in Cartesian coordinates) the quantum mechanical operators for the three components of angular momentum starting from the classical definition of angular momentum, \(l=r \times p\). Show that any two of the components do not mutually commute, and find their commutator.

Short Answer

Expert verified
The angular momentum components in quantum mechanics are given by \( \hat{L}_x = -i\hbar(y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}) \), \( \hat{L}_y = -i\hbar(z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z}) \), and \( \hat{L}_z = -i\hbar(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}) \). The commutator between any two angular momentum components, such as \( [\hat{L}_x, \hat{L}_y] \), is non-zero and is given by \( i\hbar \hat{L}_z \), indicating that they do not commute.

Step by step solution

01

Expressing angular momentum in terms of position and momentum operators

Start by expressing the classical definition of angular momentum in vector form, where \( \textbf{l} = \textbf{r} \times \textbf{p} \). In quantum mechanics, these are represented by operators. The position operator \( \textbf{r} \) has components \( \hat{x}, \hat{y}, \hat{z} \) and the momentum operator \( \textbf{p} \) has components \( -i\hbar\frac{\partial}{\partial x}, -i\hbar\frac{\partial}{\partial y}, -i\hbar\frac{\partial}{\partial z} \) where \( i \) is the imaginary unit and \( \hbar \) is the reduced Planck's constant. Substituting these into the cross product gives us the components of the angular momentum operator in Cartesian coordinates.
02

Calculating the Lx component

Calculate the first component of angular momentum, \( L_x = y p_z - z p_y \) using the operators for \( y, z, p_y, p_z \) which are \( \hat{y}, \hat{z}, -i\hbar\frac{\partial}{\partial y}, -i\hbar\frac{\partial}{\partial z} \), respectively. This results in \( \hat{L}_x = -i\hbar(y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}) \) as the x-component of the angular momentum operator.
03

Calculating the Ly and Lz components

Similarly, calculate the remaining components \( L_y \) and \( L_z \) using the same substitution method. This gives us \( \hat{L}_y = -i\hbar(z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z}) \) and \( \hat{L}_z = -i\hbar(x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}) \) as the y- and z-components of the angular momentum operator.
04

Calculating the commutator of Lx and Ly

To show that two components do not commute, calculate the commutator \( [\hat{L}_x, \hat{L}_y] \). This is done by evaluating \( \hat{L}_x\hat{L}_y - \hat{L}_y\hat{L}_x \) using the operator expressions derived for \( \hat{L}_x \) and \( \hat{L}_y \).
05

Simplification of the commutator

Simplify the expression obtained in the previous step by performing the differentiation and simplifying the terms, making use of the properties of partial derivatives and the commutativity of position coordinates. This will lead to the final expression of the commutator.
06

Result for the commutator

The final result of the commutator \( [\hat{L}_x, \hat{L}_y] = i\hbar \hat{L}_z \) shows that the angular momentum components do not commute and gives the explicit form for the commutator of these two components.

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.

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

Key Concepts

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

Quantum Mechanical Operators
In quantum mechanics, quantum mechanical operators play a pivotal role. They are mathematical objects associated with observable physical quantities such as position, momentum, and in our case, angular momentum. Unlike classical physics, where these quantities are simply numerical values, in quantum mechanics they are represented by operators that act on wave functions (the state of a quantum system). For example, the position operator in Cartesian coordinates is denoted by components \( \hat{x}, \hat{y}, \hat{z} \) and the momentum operator by \( -i\hbar\frac{\partial}{\partial x}, -i\hbar\frac{\partial}{\partial y}, -i\hbar\frac{\partial}{\partial z} \), with \(-i\) being the imaginary unit and \(\hbar\) the reduced Planck's constant.

The operators have to adhere to specific rules and principles. One of these principles is the commutation relation, which will tell us if two physical observables can be measured precisely at the same time. The use of these operators allows us to calculate and predict outcomes of quantum systems, making them fundamental to understanding quantum theory.
Commutation of Angular Momentum Components
The commutation of angular momentum components is a crucial concept in quantum mechanics that refers to the inability of certain pairs of angular momentum components to be measured simultaneously with arbitrary precision. This is related to the Heisenberg uncertainty principle. For example, to check if the \(x\)- and \(y\)-components of angular momentum commute, we calculate the commutator \( [\hat{L}_x, \hat{L}_y] \).

A resulting commutator that equals zero indicates that the operators commute and thus can be measured exactly at the same time. However, for angular momentum components, as shown in the exercise's final result \( [\hat{L}_x, \hat{L}_y] = i\hbar \hat{L}_z \), we find that they do not commute. The non-zero commutator implies an intrinsic quantum mechanical property called 'quantum uncertainty' and forces a limit on the precision of measuring two angular momentum components simultaneously. This commutation property is one of the distinctive aspects that separates quantum mechanics from classical physics.
Cartesian Coordinates in Quantum Mechanics
The use of Cartesian coordinates in quantum mechanics is very similar to their use in classical mechanics – it is a way to describe the position of points in space using three axes (\(x, y, z\)). In quantum mechanics, the position of a particle is described by the position operator with respect to these axes. When dealing with the angular momentum, expressed as \( \textbf{l} = \textbf{r} \times \textbf{p} \) in classical mechanics, we adopt these Cartesian components to define the quantum angular momentum operators.

However, it's important to note that position and momentum are no longer mere values but operators acting on wave functions. This change from classical to quantum description requires a new framework of mathematics, including operator algebra and eigenvalue equations, which are vital to performing physical and measurable predictions in quantum systems.
Derivatives in Quantum Operator Expressions
In the expression for momentum operators, the appearance of derivatives in quantum operator expressions is particularly noteworthy. These derivatives act on the wave function (a function describing the quantum state of a system) and help determine how this state changes with respect to space. In our angular momentum operator example, partial derivatives indicate how the wave function's spatial variation contributes to the overall angular momentum.

Specifically, for the angular momentum operator components \( \hat{L}_x, \hat{L}_y, \hat{L}_z \), derivatives with respect to the coordinates are taken, highlighting the orientation of angular momentum in space. Understanding these derivative operations is fundamental for analyzing quantum systems, especially when applying operators to wave functions in order to extract physical information like probability densities and expectation values.

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

Is the Schrodinger equation for a particle on an elliptical ring of semimajor axes \(a\) and \(b\) separable? Hint. Although \(\mathrm{r}\) varies with angle \((\phi\), the two are related by \(r^{2}=a^{1} \sin ^{2} \Phi+b^{2} \cos ^{2} \Phi\).

Here we explore further the idea introduced in Impact \(19.2\) that quantum mechanical effects need to be invoked in the description of the electronic properties of metallic nanocrystals, here modelled as three-dimensional boxes. (a) Set up the Schrodinger equation for a particle of mass \(m\) in a threedimensional rectangular box with sides \(L_{1}, L_{2}\), and \(L_{3}\). Show that the Schrodinger equation is separable. (b) Show that the wavefunction and the energy are defined by three quantum numbers. (c) Specialize the result from part (b) to an electron moving in a cubic box of side \(L=5 \mathrm{~nm}\) and draw an energy diagram resembling Fig. \(9.2\) and showing the first 15 energy levels. Note that each energy level may consist of degenerate energy states. (d) Compare the energy level diagram from part (c) with the energy level diagram for an electron in a one-dimensional box of length \(L=5 \mathrm{~nm}\). Are the energy levels become more or less sparsely distributed in the cubic box than in the one-dimensional box?

We normally think of the one-dimensional well as being horizontal. Suppose it is vertical; then the potential energy of the particle depends on \(\mathrm{x}\) because of the presence of the gravitational field. Calculate the first-order correction to the zero-point energy, and evaluate it for an electron in a box on the surface of the Earth. Account for the result. Hint. The energy of the particle depends on its height as \(m g h\), where \(g=9.81 \mathrm{~m} \mathrm{~s}-{ }^{2}\). Because \(g\) is so small, the energy correction is small; but it would be significant if the box were near a very massive star.

Calculate the energies of the first four rotational levels of \({ }^{1} \mathrm{H}^{127} \mathrm{I}\) free to rotate in three dimensions, using for its moment of inertia \(I=\mu R^{2}\), with \(\mu=\) \(m_{\mathrm{H}} m_{1} /\left(m_{\mathrm{H}}+m_{1}\right)\) and \(R=160 \mathrm{pm}\).

The potential energy of the rotation of one \(\mathrm{CH}\), group relative to its neighbour in ethane can be expressed as \(V(\Psi)=V_{0} \cos 3 \phi\). Show that for small displacements the motion of the group is harmonic and calculate the energy of excitation from \(v=0\) to \(v=1\). What do you expect to happen to the energy levels and wavefunctions as the excitation increases?

See all solutions

Recommended explanations on Chemistry Textbooks

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