Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Show that unless \(\ell=s . L\) and \(S\) cannot be exactly opposite: that is, show that at its minimum possible value, for which \(j=\ell-s\), the magnitude \(J\) of the total angular momenturn is strictly greater than the difference \(L-S \mid\) berween the magnitudes of the orbital and intrinsic angular momentum vectors.

Short Answer

Expert verified
Upon applying the triangle inequality, it is verified that the total angular momentum is always greater than the difference between the orbital and the intrinsic angular momenta, hence confirming the inquiry from the exercise. The only exception occurs when the two momenta are diametrically opposed.

Step by step solution

01

Understand the Terms

Firstly, it is key to understand that angular momentum is a measure of the amount of rotation an object has, taking into account its mass, shape, and speed. There are two forms here: intrinsic angular momentum \(S\), inherent to an object, and orbital angular momentum \(L\), which relates to the object's motion about a point outside itself.
02

Showing that \(J\) is always greater than \(|L - S|\)

Recall that \(J = |\ell - s|\) is the magnitude of the total angular momentum, and that \(\ell = s . L\) is given in the problem. Substituting the latter equation into the former, we get \(J = |\ell - s| = |s.L - s| = |s (L - 1)|\). Now, there is a rule that for any vectors \(A\) and \(B\), \(|A - B| \geq |A| - |B|\), known as the triangle inequality. Applying this to \(s (L - 1)\), we get \(|s (L - 1)| \geq |s| - |L - 1|\). Since \(s\) and \((L - 1)\) are both positive, we can remove the absolute value signs, giving \(s (L - 1) \geq s - (L - 1) = J\), re-arranging for \(J\), we obtain \(s (L - 1) + (L - 1) \geq J\), thus proving that at a minimum, \(J\) is always greater than \(|L - S|\).
03

Interpretation

In essence, this result confirms the original premise: regardless of their orientations, the magnitude of the total angular momentum will always be greater than the difference in magnitude between the orbital and intrinsic angular momenta, barring the case where they are exactly opposite. This represents a key governing principle in the dynamics of angularly rotating objects.

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.

Orbital Angular Momentum
Orbital angular momentum is a fundamental concept in physics, encapsulating the idea of rotational motion for objects in orbit. In classical mechanics, it represents an object's momentum attributed to its rotation around a fixed center of mass, much akin to how a planet revolves around a star.

For an object revolving in a two-dimensional plane, the magnitude of its orbital angular momentum, denoted by the symbol \(L\), can be expressed as the product of the object's mass \(m\), its velocity \(v\), and the perpendicular distance from the center of rotation to the line of motion, known as the moment arm \(r\). The formula is simply \( L = mvr \).

This physical quantity becomes even more intriguing when we travel into the realm of quantum mechanics. Here, \(L\) is quantized, meaning it can only take specific discrete values, and it plays a pivotal role in determining the energy levels of electrons in an atom. An understanding of orbital angular momentum is crucial as it helps explain the structure of the atom, the nature of chemical bonds, and the spectral lines of elements.
Intrinsic Angular Momentum
Intrinsic angular momentum, more famously known as spin, represents the internal rotation of a particle on its own axis. Unlike orbital angular momentum, intrinsic angular momentum does not involve the motion of particles around an external point. Instead, it is a fundamental property of particles just like mass or charge, and every elementary particle has a specific 'spin' value.

In quantum mechanics, the spin \(S\) is characterized by its own set of quantum numbers, and it greatly influences a particle’s quantum behavior, particularly in magnetic environments. An electron, for instance, can have a spin of \(+1/2\) or \(–1/2\). The concept of spin is necessary to understand phenomena such as the Pauli exclusion principle, which explains the arrangement of electrons in an atom and is the foundation for the entire field of chemistry and solid-state physics.
Vector Magnitudes
In physics, a vector represents a quantity that possesses both magnitude and direction. Common examples include force, velocity, and acceleration. The magnitude of a vector is its 'length' in a spatial sense, and it is always a non-negative number.

To calculate the magnitude of a vector \( \vec{A} \), which has components \(A_x, A_y\), and \(A_z\) in three-dimensional space, one would use the formula \( |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \). Understanding vector magnitudes is crucial because they tell us about the 'size' of the vector quantity, irrespective of the direction in which it's pointing. This measure is essential when combining multiple vectors to determine the net effect, such as the total displacement or force acting on an object.
Triangle Inequality
The triangle inequality theorem is a principle that states that the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the remaining side. This concept is also applied in the context of vector algebra in physics.

When dealing with vector magnitudes, the triangle inequality implies that the magnitude of the sum of two vectors will never be less than the magnitude of the difference between them. In mathematical terms, for vectors \( \vec{A} \) and \( \vec{B} \), the following always holds true: \( |\vec{A} + \vec{B}| \geq ||\vec{A}| - |\vec{B}|| \).

Application in Angular Momentum

In the realm of angular momentum, the triangle inequality is used to understand the relationship between various types of angular momentum. For instance, when we discuss the total angular momentum \(J\) of a system, this value can never be less than the difference in magnitudes of the orbital \(L\) and intrinsic \(S\) angular momenta, reaffirming the notion that individual contributions to a system's total momentum are intricately linked.

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

All other things being equal, should the spin-orbit interaction be a larger or smaller effect in hydrogen as \(n\) increases? Justify your answer.

As is done for helium in Table 8.3 , determine for a carton atom the various states allowed according to \(\angle S\) coupling. The coupling is between carbon's two \(2 p\) electrons (its filled 2 s subshell not participating). one of which always remains in the \(2 p\) state. Consider cases in which the other is as high as the \(3 d\) level. (Note: When both electrons are in the \(2 p\), the exclusion principle Iestricts the number of states. The only allowed states are those in which \(s_{r}\) and \(\ell_{T}\) are both even or both odd.)

Figure 8.3 shows the Stern-Gerlach apparans. It reveals that spin- \(\frac{1}{2}\) particles have just two possible spin states. Assume that when these rwo beams are separated inside the channel (though still near its centerline). we can choose to block one or the other for study. Now a second such apparatus is added after the first. Their channels are aligned. but the second one is rotated about the \(x\) -axis by an angle \(\phi\) from the first. Suppose we block the spin-down beam in the first apparatus, allowing only the spin-up beam into the second. There is no wave function for spin. but we can still talk of a probability amplitude, which we square to give a probability. After the first apparatus' spin-up beam passes chrough the second apparatus, the probability amplitude is \(\cos (\phi / 2) T_{2 n d}+\sin (\phi / 2) b_{2 n d}\). where the arrows indicate the two possible findings for spin in the second apparatus. (a) What is the probability of finding the particle spin up in the second apparatus? Of finding it spin down? Argue thatthese probabilities make sense individually for representative values of \(\phi\) and that their sum is also sensible. (b) By contrasting this spin prohability anplitude with a sporial probability amplitude, such as \(\psi(x)=A e^{-b x^{2}}\), argue that although the surhitrariness of \(\phi\) gives the spin case an infinite number of values. it is still justified to refer to it as 8 "two-state system." while the spatial case is an infinite.state system.

The 21 cm Line: One of the most important windows to the mysteries of the cosmos is the 21 cm line. With it, astronomers map hydrogen throughout the universe. An important trait is that it involves a highly forbidden transition that is, accordingly, quite long-lived. But it is also an excellent example of the coupling of angular momenta. Hydrogen's ground state has no spin-orbit interaction - for \(\boldsymbol{\theta}=0,\) there is no orbit. However, the proton and electron magnetic moments do interact. Consider the following simple model. (a) The proton sees itself surrounded by a spherically symmetric cloud of Is electron, which has an intrinsic magnetic dipole moment/spin that, of course, has a direction. For the purpose of investigating its effect on the proton, treat this dispersed magnetic moment as behaving effectively like a single loop of current whose radius is \(a_{0}\), then find the magnetic tield at the middle of the loop in tenns of \(e, h, m_{e}, \mu_{0},\) and \(a_{0} .\) (b) The proton sits right in the middle of the electron's magnetic moment. Like the electron. the proton is a spin- \(\frac{1}{2}\) particle, with only two possible orientations in a magnetic field. Noting, however, that its spin and magnetic moment are parallel rather than opposite, would the interaction energy be lower with the proton's spin aligned or antialigned with that of the electron'? (c) For the proton, \(g_{p}\) is \(5.6 .\) Obtain a rough value for the energy difference between the two orientations. (d) What would be the wavelength of a photon that carries away this energy difference?

The spin-orbit interaction splits the hydrogen 4 f state into many. (a) Identify these states and rank them in order of increasing energy. (b) If a weak external magnetic field were now introduced (weak enough that it does not disturb the spin-orbit coupling), into how many difierent energies would each of these states be split?

See all solutions

Recommended explanations on Physics 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